Open Access

Closed expressions for averages of set partition statistics

  • Bobbie Chern1,
  • Persi Diaconis2,
  • Daniel M Kane3 and
  • Robert C Rhoades4Email author
Research in the Mathematical Sciences20141:2

DOI: 10.1186/2197-9847-1-2

Received: 17 January 2014

Accepted: 7 February 2014

Published: 17 June 2014

Abstract

In studying the enumerative theory of super characters of the group of upper triangular matrices over a finite field, we found that the moments (mean, variance, and higher moments) of novel statistics on set partitions of [n]={1,2,,n} have simple closed expressions as linear combinations of shifted bell numbers. It is shown here that families of other statistics have similar moments. The coefficients in the linear combinations are polynomials in n. This allows exact enumeration of the moments for small n to determine exact formulae for all n.

Background

The set partitions of [ n]={1,2,,n} (denoted Π(n)) are a classical object of combinatorics. In studying the character theory of upper triangular matrices (see section ‘Set partitions, enumerative group theory, and super characters’ for background) we were led to some unusual statistics on set partitions. For a set partition λ of n, consider the dimension exponent (Table 1).
d ( λ ) : = i = 1 ( M i m i + 1 ) n https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equa_HTML.gif
where λ has blocks, M i and m i are the largest and smallest elements of the i th block. How does d(λ) vary with λ? As shown below, its mean and second moment are determined in terms of the Bell numbers B n
λ Π ( n ) d ( λ ) = 2 B n + 2 + ( n + 4 ) B n + 1 λ Π ( n ) d 2 ( λ ) = 4 B n + 4 ( 4 n + 15 ) B n + 3 + ( n 2 + 8 n + 9 ) B n + 2 ( 4 n + 3 ) B n + 1 + n B n . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equb_HTML.gif
Table 1

A table of the dimension exponent f ( n ,0, d )

nd

0

1

2

3

4

5

6

7

8

9

10

11

12

0

1

            

1

1

            

2

2

            

3

4

1

           

4

8

4

3

          

5

16

12

13

9

2

        

6

32

32

42

42

35

12

8

      

7

64

80

120

145

159

133

86

52

32

6

   

8

128

192

320

440

559

600

591

440

380

248

164

48

30

The right hand sides of these formulae are linear combinations of Bell numbers with polynomial coefficients. Dividing by B n and using asymptotics for Bell numbers (see section ‘Asymptotic analysis’) in terms of α n , the positive real solution of u e u =n+1 (so α n = log(n)− log log(n)+) gives
E ( d ( λ ) ) = α n 2 α n 2 n 2 + O n α n VAR ( d ( λ ) ) = α n 2 7 α n + 17 α n 3 ( α n + 1 ) 2 n 3 + O n 2 α n . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equc_HTML.gif

This paper gives a large family of statistics that admit similar formulae for all moments. These include classical statistics such as the number of blocks and number of blocks of size i. It also includes many novel statistics such as d(λ) and c k (λ), the number of k crossings. The number of two crossings appears as the intertwining exponent of super characters.

Careful definitions and statements of our main results are in section ‘Statement of the main results’. Section ‘Set partitions, enumerative group theory, and super characters’ reviews the enumerative and probabilistic theory of set partitions, finite groups, and super characters. Section ‘Computational results’ gives computational results; determining the coefficients in shifted Bell expressions involves summing over all set partitions for small n. For some statistics, a fast new algorithm speeds things up. Proofs of the main theorems are in sections ‘Proofs of recursions, asymptotics, and Theorem 3’ and ‘Proofs of Theorems 1 and 2’. Section ‘More data’ gives a collection of examples - moments of order up to six for d(λ) and further numerical data. In a companion paper [1], the asymptotic limiting normality of d(λ), c2(λ), and some other statistics is shown.

Statement of the main results

Let Π(n) be the set partitions of [n]={1,2,,n} (so |Π(n)|=B n , the n th Bell number). A variety of codings are described in section ‘Set partitions, enumerative group theory, and super characters’. In this section, λΠ(n) is described as λ=B1|B2||B with B i B j =, i = 1 B i = [ n ] https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq1_HTML.gif. Write i λ j if i and j are in the same block of λ. It is notationally convenient to think of each block as being ordered. Let First (λ) be the set of elements of [n] which appear first in their block and L a s t(λ) be the set of elements of [n] which occur last in their block. Finally, let A r c(λ) be the set of distinct pairs of integers (i,j) which occur in the same block of λ such that j is the smallest element of the block greater than i. As usual, λ may be pictured as a graph with vertex set [n] and edge set A r c(λ).

For example, the partition λ=1356|27|4, represented in Figure 1, has F i r s t(λ)={1,2,4}, L a s t(λ)={6,7,4}, and A r c(λ)={(1,3),(3,5),(5,6),(2,7)}.
https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Fig1_HTML.jpg
Figure 1

An example partition λ =1356|27|4.

A statistic on λ is defined by counting the number of occurrences of patterns. This requires some notation.

Definition 1

  1. (i)

    A pattern P _ https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq2_HTML.gif of length k is defined by a set partition P of [k] and subsets F ( P _ ) , L ( P _ ) [ k ] https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq3_HTML.gif, and A ( P _ ) , C ( P _ ) { [ k ] × [ k ] : i < j } https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq4_HTML.gif. Let P _ = ( P , F , L , A , C ) https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq5_HTML.gif.

     
  2. (ii)

    An occurrence of a pattern P _ https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq6_HTML.gif of length k in λΠ(n) is s=(x 1,,x k ) with x i [ n] such that

     
  3. 1.

    x 1<x 2<<x k .

     
  4. 2.

    x i λ x j if and only if i P j.

     
  5. 3.

    x i F i r s t(λ) if i F ( P _ ) https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq7_HTML.gif.

     
  6. 4.

    x i L a s t(λ) if i L ( P _ ) https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq8_HTML.gif.

     
  7. 5.

    (x i ,x j )A r c(λ) if ( i , j ) A ( P _ ) https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq9_HTML.gif.

     
  8. 6.

    x j x i =1 if ( i , j ) C ( P _ ) https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq10_HTML.gif.

     
Write s P _ λ https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq11_HTML.gif if s is an occurrence of P _ https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq12_HTML.gif in λ.
  1. (iii)
    A simple statistic is defined by a pattern P _ https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq13_HTML.gif of length k and Q Z [ y 1 , , y k , m ] https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq14_HTML.gif. If λΠ(n) and s = ( x 1 , , x k ) P _ λ https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq15_HTML.gif, write Q ( s ) = Q y i = x i , m = n https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq16_HTML.gif. Let
    f ( λ ) = f P _ , Q ( λ ) : = s P _ λ Q ( s ) . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equd_HTML.gif
     
Let the degree of a simple statistic f P _ , Q https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq17_HTML.gif be the sum of the length of P _ https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq18_HTML.gif and the degree of Q.
  1. (iv)

    A statistic is a finite Q https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq19_HTML.gif-linear combination of simple statistics. The degree of a statistic is defined to be the minimum over such representations of the maximum degree of any appearing simple statistic.

     

Remark 1

In the notation above, F ( P _ ) https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq20_HTML.gif is the set of firsts elements, L ( P _ ) https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq21_HTML.gif is the set of lasts, A is the arc set of the pattern, and C ( P _ ) https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq22_HTML.gif is the set of consecutive elements.

Examples
  1. 1.
    Number of blocks in λ:
    ( λ ) = 1 x n x is smallest element in its block 1 . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Eque_HTML.gif
     
Here, P _ https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq23_HTML.gif is a pattern of length 1, F ( P _ ) = { 1 } https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq24_HTML.gif, L ( P _ ) = A ( P _ ) = C ( P _ ) = https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq25_HTML.gif, and Q(y,m)=1. Similarly, the n th moment of (λ) can be computed using
( λ ) k = f P _ k , 1 ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equf_HTML.gif
where P _ k https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq26_HTML.gif is the pattern of length k corresponding to P, the partition of [k] into blocks of size 1, with F ( P _ k ) = { 1 , 2 , , k } https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq27_HTML.gif, and L ( P _ k ) = A ( P _ k ) = C ( P _ k ) = https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq28_HTML.gif.
  1. 2.
    Number of blocks of size i: define a pattern P _ i https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq29_HTML.gif of length i by: (1) all elements of [i] are equivalent, (2) F ( P _ i ) = { 1 } https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq30_HTML.gif, (3) L ( P _ i ) = { i } https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq31_HTML.gif, (4) A ( P _ i ) = { ( 1 , 2 ) , , ( i 1 , i ) } https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq32_HTML.gif, and (5) C ( P _ i ) = https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq33_HTML.gif. Then,
    X i ( λ ) : = f P _ i , 1 ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equ1_HTML.gif
    (1)
     
is the number of i blocks in λ (if i=1, A ( P _ 1 ) = https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq34_HTML.gif). Similarly, the moments of the number of blocks of size i is a statistic. See Theorem 1.
  1. 3.
    k crossings: a k crossing [2] of a λΠ(n) is a sequence of arcs (i t ,j t )1≤tkA r c(λ) with
    i 1 < i 2 < < i k < j 1 < j 2 < < j k . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equg_HTML.gif
     

The statistic c r k (λ) which counts the number of k crossings of λ can be represented by a pattern P _ = ( P , F , L , A , C ) https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq35_HTML.gif of length 2k with (1) i P k+i for i=1,,k, (2) F ( P _ ) = L ( P _ ) = https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq36_HTML.gif, (3) A ( P _ ) = { ( 1 , k + 1 ) , ( 2 , k + 2 ) , , ( k , 2 k ) } https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq37_HTML.gif, and (4) C ( P _ ) = https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq38_HTML.gif.

Partitions with c r2(λ)=0 are in bijection with Dyck paths and so are counted by the Catalan numbers C n = 1 n + 1 2 n n https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq39_HTML.gif (see Stanley’s second volume on enumerative combinatorics [3]). Partitions without crossings have proved themselves to be very interesting.

Crossing seems to have been introduced by Krewaras [4]. See Simion’s [5] for an extensive survey and Chen et al. [2] and Marberg [6] for more recent appearances of this statistic. The statistic c r2(λ) appears as the intersection exponent in section ‘Super character theory’.
  1. 4.
    Dimension exponent: the dimension exponent described in the introduction is a linear combination of the number of blocks (a simple statistic of degree 1), the last elements of the blocks (a simple statistic of degree 2), and the first elements of the blocks (a simple statistic of degree 2). Precisely, define f firsts ( λ ) : = f P _ , Q ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq40_HTML.gif where P _ https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq41_HTML.gif is the pattern of length 1, with F ( P _ ) = { 1 } https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq42_HTML.gif, L ( P _ ) = A ( P _ ) = C ( P _ ) = https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq43_HTML.gif, and Q(y,m)=y. Similarly, let f lasts ( λ ) : = f P _ , Q ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq44_HTML.gif where P _ https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq45_HTML.gif is the pattern of length 1, with L ( P _ ) = { 1 } https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq46_HTML.gif, F ( P _ ) = A ( P _ ) = C ( P _ ) = https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq47_HTML.gif, and Q(y,m)=y. Then,
    d ( λ ) = f lasts ( λ ) f firsts ( λ ) + ( λ ) n. https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equh_HTML.gif
     
  2. 5.
    Levels: the number of levels in λ, denoted f levels(λ), (see page 383 of [7] or Shattuck [8]) is the number of i such that i and i+1 appear in the same block of λ. We have
    f levels ( λ ) = f P _ , Q ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equi_HTML.gif
     
where P _ https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq48_HTML.gif is a pattern of length 2 with C ( P _ ) = A ( P _ ) = { ( 1 , 2 ) } https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq49_HTML.gif, L ( P _ ) = F ( P _ ) = https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq50_HTML.gif, and Q=1.
  1. 6.

    The maximum block size of a partition is not a statistic in this notation.

     

The set of all statistics on n = 0 Π ( n ) Q https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq51_HTML.gif is a filtered algebra.

Theorem 1.

Let https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq52_HTML.gif be the set of all set partition statistics thought of as functions f : n Π ( n ) Q https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq53_HTML.gif. Then https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq54_HTML.gif is closed under the operations of pointwise scaling, addition and multiplication. In particular, if f 1 , f 2 S https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq55_HTML.gif and a Q https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq56_HTML.gif, then there exist partition statistics g a ,g+,g so that for all set partitions λ,
a f 1 ( λ ) = g a ( λ ) f 1 ( λ ) + f 2 ( λ ) = g + ( λ ) f 1 ( λ ) · f 2 ( λ ) = g ( λ ) . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equj_HTML.gif

Furthermore, deg(g a )≤ deg(f1), deg(g+)≤ max(deg(f1), deg(f2)), and deg(g)≤ deg(f1)+ deg(f2). In particular, https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq57_HTML.gif is a filtered Q https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq58_HTML.gif-algebra under these operations.

Remark 2.

Properties of this algebra remain to be discovered.

Definition 2.

A shifted Bell polynomial is any function R : N Q https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq59_HTML.gif that is zero or can be expressed in the form
R ( n ) = I j K Q j ( n ) B n + j https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equk_HTML.gif

where I , K Z https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq60_HTML.gif and each Q j ( x ) Q [ x ] https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq61_HTML.gif such that Q I (x)≠0 and Q K (x)≠0. i.e., it is a finite sum of polynomials multiplied by shifted Bell numbers. Call K the upper shift degree of R and I the lower shift degree of R.

Remark 3.

The representation of a shifted Bell polynomial is unique. This can be understood by considering the asymptotics of each individual term as n.

Our first main theorem shows that the aggregate of a statistic is a shifted Bell polynomial.

Theorem 2.

For any statistic, f of degree N, there exists a shifted Bell polynomial R such that for all n≥1
M ( f ; n ) : = λ Π ( n ) f ( λ ) = R ( n ) . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equl_HTML.gif
Moreover,
  1. 1.

    the upper shift index of R is at most N and the lower shift index is bounded below by −k, where k is the length of the pattern associated f.

     
  2. 2.

    the degree of the polynomial coefficient of B n+Nj in R is bounded by j for jN and by j−1 for j>N.

     

The following collects the shifted Bell polynomials for the aggregates of the statistics given previously.

Examples
  1. 1.
    Number of blocks in λ:
    M ( ; n ) = B n + 1 B n . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equm_HTML.gif
     
This is elementary and is established in Proposition 1
  1. 2.
    Number of blocks of size i:
    M ( X i ; n ) = n i B n i . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equn_HTML.gif
     
This is also elementary and is established in Proposition 1.
  1. 3.
    Two crossings: Kasraoui [9] established
    M ( cr 2 ; n ) = 1 4 5 B n + 2 + ( 2 n + 9 ) B n + 1 + ( 2 n + 1 ) B n . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equo_HTML.gif
     
  2. 4.
    Dimension exponent:
    M ( d ; n ) = 2 B n + 2 + ( n + 4 ) B n + 1 . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equp_HTML.gif
     
This is given in Theorem 1.
  1. 5.
    Levels: Shattuck [8] showed that
    M ( f levels ; n ) = 1 2 ( B n + 1 B n B n 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equq_HTML.gif
     

It is amusing that this implies that B3nB3n+1≡1 (mod 2) and B3n+2≡0 (mod 2) for all n≥0.

Remark 4.

Chapter 8 of Mansour’s book [7] and the research papers [911] contain many other examples of statistics which have shifted Bell polynomial aggregates. We believe that each of these statistics is covered by our class of statistics.

Set partitions, enumerative group theory, and super characters

This section presents background and a literature review of set partitions, probabilistic and enumerative group theory, and super character theory for the upper triangular group over a finite field. Some sharpenings of our general theory are given.

Set partitions

Let Π(n,k) denote the set partitions of n labeled objects with k blocks and Π(n)= k Π(n,k); so |Π(n,k)|=S(n,k) is the Stirling number of the second kind and |Π(n)|=B n is the n th Bell number. The enumerative theory and applications of these basic objects is developed in the studies of Graham et al. [12], Knuth [13], Mansour [7], and Stanley [14]. There are many familiar equivalent codings.

 Equivalence relations on n objects
1 | 2 | 3 , 12 | 3 , 13 | 2 , 1 | 23 , 123 https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equr_HTML.gif
 Binary, strictly upper triangular zero-one matrices with no two ones in the same row or column (equivalently, rook placements on a triangular Ferris board (Riordan [15]).
0 0 0 0 0 0 0 0 0 , 0 1 0 0 0 0 0 0 0 , 0 0 1 0 0 0 0 0 0 , 0 0 0 0 0 1 0 0 0 , 0 1 0 0 0 1 0 0 0 https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equs_HTML.gif

 Arcs on n points

 Restricted growth sequences a1,a2,…,a n ; a1=0,aj+1≤1+ max(a1,…,a j ) for 1≤j<n (Knuth [13], p. 416)
012 , 001 , 010 , 011 , 000 https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equt_HTML.gif
https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equu_HTML.gif
https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equv_HTML.gif

 Semi-labeled trees on n+1 vertices

 Vacillating tableau: a sequence of partitions λ0,λ1,,λ2n with λ0=λ2n= and λ2i+1 is obtained from λ2i by doing nothing or deleting a square and λ2i is obtained from λ2i−1 by doing nothing or adding a square (see [2]).

The enumerative theory of set partitions begins with Bell polynomials. Let B n , k ( w 1 , , w n ) = λ Π ( n , k ) w i X i ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq62_HTML.gif with X i (λ) the number of blocks in λ of size i; so set B n ( w 1 , , w n ) = k B n , k ( w 1 , , w n ) https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq63_HTML.gif and B ( t ) = n = 0 B n ( w _ ) t n n ! . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq64_HTML.gif A classical version of the exponential formula gives
B ( t ) = e n = 1 w n t n n ! . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equ2_HTML.gif
(2)
These elegant formulae have been used by physicists and chemists to understand fragmentation processes ([16] for extensive references). They also underlie the theory of polynomials of binomial type [17, 18], that is, families P n (x) of polynomials satisfying
P n ( x + y ) = P k ( x ) P n k ( y ) . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equw_HTML.gif

These unify many combinatorial identities, going back to Faa de Bruno’s formula for the Taylor series of the composition of two power series.

There is a healthy algebraic theory of set partitions. The partition algebra of [19] is based on a natural product on Π(n) which first arose in diagonalizing the transfer matrix for the Potts model of statistical physics. The set of all set partitions n Π ( n ) https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq65_HTML.gif has a Hopf algebra structure which is a general object of study in [20].

Crossings and nestings of set partitions is an emerging topic, see [2, 21, 22] and their references. Given λΠ(n) two arcs (i1,j1) and (i2,j2) are said to cross if i1<i2<j1<j2 and nest if i1<i2<j2<j1. Let c r(λ) and n e(λ) be the number of crossings and nestings. One striking result: the crossings and nestings are equi-distributed ([21] Corollary 1.5), they show
λ Π ( n ) x cr ( λ ) y ne ( λ ) = λ Π ( n ) x ne ( λ ) y cr ( λ ) . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equx_HTML.gif

As explained in section ‘Super character theory’, crossings arise in a group theoretic context and are covered by our main theorem. Nestings are also a statistic.

This crossing and nesting literature develops a parallel theory for crossings and nestings of perfect matchings (set partitions with all blocks of size 2). Preliminary works suggest that our main theorem carry over to matchings with B n reduced to (2n)!/2 n n!.

Turn next to the probabilistic side: what does a ‘typical’ set partition ‘look like’? For example, under the uniform distribution on Π(n)

 What is the expected number of blocks?

 How many singletons (or blocks of size i) are there?

 What is the size of the largest block?

The Bell polynomials can be used to get moments. For example:

Proposition 1.

  1. (i)
    Let (λ) be the number of blocks. Then
    m ( ; n ) : = λ Π ( n ) ( λ ) = B n + 1 B n m ( 2 ; n ) = B n + 2 3 B n + 1 + B n m ( 3 ; n ) = B n + 3 6 B n + 2 + 8 B n + 1 B n + 1 B n https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equy_HTML.gif
     
  2. (ii)
    Let X 1(λ) be the number of singleton blocks, then
    m ( X 1 ; n ) = nB n 1 m ( X 1 2 ; n ) = nB n 1 + n ( n 1 ) B n 2 https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equz_HTML.gif
     

In accordance with our general theorem, the right hand sides of (i) and (ii) are shifted Bell polynomials. To make contact with the results shown previously, there is a direct proof of these classical formulae.

Proof.

Specializing the variables in the generating function (2) gives a two variable generating functions for :
n = 0 λ Π ( n ) y ( λ ) x n n ! = n 0 0 S ( n , ) y x n n ! = e y ( e x 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equaa_HTML.gif
Differentiating with respect to y and setting y=1 shows that m(;n) is the coefficient of x n n ! https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq66_HTML.gif in ( e x 1 ) e e x 1 https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq67_HTML.gif. Noting that
∂x e e x 1 = e x e e x 1 = n = 0 B n + 1 x n n ! https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equab_HTML.gif

yields m()=Bn+1B n . Repeated differentiation gives the higher moments.

For X1, specializing variables gives
n = 0 λ Π ( n ) y X 1 ( λ ) x n n ! = e e x 1 x + yx . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equac_HTML.gif

Differentiation with respect to y and settings y=1 readily yields the claimed results.

The moment method may be used to derive limit theorems. An easier, more systematic method is due to Fristedt [23]. He interprets the factorization of the generating function B(t) in (2) as a conditional independence result and uses ‘dePoissonization’ to get results for finite n. Let X i (λ) be the number of blocks of size i. Roughly, his results say that { X i } i = 1 n https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq68_HTML.gif are asymptotically independent and of size (log(n)) i /i!. More precisely, let α n satisfy α n e α n = n + 1 https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq69_HTML.gif (so α n = log(n)− log log(n)+o(1)). Let β i = α n i / i ! https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq70_HTML.gif then
P X i β i β i x = Φ ( x ) + o ( 1 ) https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equad_HTML.gif

where Φ ( x ) = 1 2 π x e u 2 / 2 du. https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq71_HTML.gif Fristedt also has a description of the joint distribution of the largest blocks.

Remark 5.

It is typical to expand the asymptotics in terms of u n where u n e u n = n https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq72_HTML.gif. In this notation, u n and α n differ by O(1/n).

The number of blocks (λ) is asymptotically normal when standardized by its mean μ n n log ( n ) https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq73_HTML.gif and variance σ n 2 n log 2 ( n ) https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq74_HTML.gif. These are precisely given by Proposition 1. Refining this, Hwang [24] shows
P μ n σ n x = Φ ( x ) + O log ( n ) n . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equae_HTML.gif

Stam [25] has introduced a clever algorithm for random uniform sampling of set partitions in Π(n). He uses this to show that if W(i) is the size of the block containing i, 1≤ik, then for k finite and n large W(i) are asymptotically independent and normal with mean and variance asymptotic to α n . In [1], we use Stam’s algorithm to prove the asymptotic normality of d(λ) and c r2(λ).

Any of the codings previously mentioned lead to distribution questions. The upper triangular representation leads to the study of the dimension and crossing statistics, the arc representation suggests crossings, nestings, and even the number of arcs, i.e. n(λ). Restricted growth sequences suggest the number of zeros, the number of leading zeros, largest entry. See Mansour [7] for this and much more. Semi-labeled trees suggest the number of leaves, length of the longest path from root to leaf, and various measures of tree shape (e.g., max degree). Further probabilistic aspects of uniform set partitions can be found in [16, 26].

Probabilistic group theory

One way to study a finite group G is to ask what ‘typical’ elements ‘look like’. This program was actively begun by Erdös and Turan [2733] who focused on the symmetric group S n . Pick a permutation σ of n at random and ask the following:

 How many cycles in σ? (about logn)

 What is the length of the longest cycle? (about 0.61n)

 How many fixed points in σ? (about 1)

 What is the order of σ? (roughly e ( log n ) 2 / 2 https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq75_HTML.gif)

In these and many other cases, the questions are answered with much more precise limit theorems. A variety of other classes of groups have been studied. For finite groups of Lie type, see [34] for a survey and [35] for wide-ranging applications. For p groups, see [36].

One can also ask questions about ‘typical’ representations. For example, fix a conjugacy class C (e.g., transpositions in the symmetric group), what is the distribution of χ ρ (C) as ρ ranges over irreducible representations [34, 37, 38]. Here, two probability distributions are natural, the uniform distribution on ρ and the Plancherel measure ( Pr ( ρ ) = d ρ 2 / | G | https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq76_HTML.gif with d ρ the dimension of ρ). Indeed, the behavior of the ‘shape’ of a random partition of n under the Plancherel measure for S n is one of the most celebrated results in modern combinatorics. See Stanley’s study [39] for a survey with references to the work of Kerov and Vershik [40], Logan and Shepp [41], Baik et al. [42], and many others.

The previous discussion focuses on finite groups. The questions make sense for compact groups. For example, pick a random matrix from Haar measure on the unitary group U n and ask: what is the distribution of its eigenvalues? This leads to the very active subject of random matrix theory. We point to the wonderful monographs of Anderson et al. [43], and Forrester [44] which have extensive surveys.

Super character theory

Let G n (q) be the group of n×n matrices which are upper triangular with ones on the diagonal over the field ?? q https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq77_HTML.gif. The group G n (q) is the Sylow p subgroup of GL n ( ?? q ) https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq78_HTML.gif for q=p a . Describing the irreducible characters of G n (q) is a well-known wild problem. However, certain unions of conjugacy classes, called superclasses, and certain characters, called supercharacters, have an elegant theory. In fact, the theory is rich enough to provide enough understanding of the Fourier analysis on the group to solve certain problems, see the work of Arias-Castro et al. [45]. These superclasses and supercharacters were developed by André [4648] and Yan [49]. Supercharacter theory is a growing subject. See [6, 5054] and their references.

For the groups G n (q), the supercharacters are determined by a set partition of [ n] and a map from the set partition to the group ?? q https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq79_HTML.gif. In the analysis of these characters, there are two important statistics, each of which only depends on the set partition. The dimension exponent is denoted d(λ), and the intertwining exponent is denoted i(λ).

Indeed, if χ λ and χ μ are two supercharacters, then
dim ( χ λ ) = q d ( λ ) and χ λ , χ μ = δ λ , μ q i ( λ ) . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equaf_HTML.gif
While d(λ) and i(λ) were originally defined in terms of the upper triangular representation (for example, d(λ) is the sum of the horizontal distance from the ‘ones’ to the super diagonal), their definitions can be given in terms of blocks or arcs:
d ( λ ) : = e f Arc ( λ ) f e 1 https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equ3_HTML.gif
(3)
and
i ( λ ) : = e 1 < e 2 < f 1 < f 2 e 1 f 1 Arc ( λ ) e 2 f 2 Arc ( λ ) 1 https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equ4_HTML.gif
(4)

Remark 6.

Notice that i(λ)=c r2(λ) is the number of two crossings which were introduced in the previous sections.

Our main theorem shows that there are explicit formulae for every moment of these statistics. The following represents a sharpening using special properties of the dimension exponent.

Theorem 3.

For each k{0,1,2, }, there exists a closed form expression
M ( d k ; n ) : = λ Π ( n ) d ( λ ) k = P k , 2 k ( n ) B n + 2 k + P k , 2 k 1 ( n ) B n + 2 k 1 + + P k , 0 ( n ) B n https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equag_HTML.gif
where each Pk,2kj is a polynomial with rational coefficients. Moreover, the degree of Pk,2kj is
j j k k j k 2 j > k . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equah_HTML.gif
For example,
λ Π ( n ) d ( λ ) = 2 B n + 2 + ( n + 4 ) B n + 1 λ Π ( n ) d ( λ ) 2 = 4 B n + 4 ( 4 n + 15 ) B n + 3 + ( n 2 + 8 n + 9 ) B n + 2 ( 4 n + 3 ) B n + 1 + nB n https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equai_HTML.gif

Remark 7.

See section ‘More data’ for the moments with k≤6, and see [55] for the moments with k≤22. The first moment may be deduced easily from results of Bergeron and Thiem [56]. Note that they seem to have an index which differs by one from ours.

Remark 8.

Theorem 3 is stronger than what is obtained directly from Theorem 2. For example, the lower shift index is 0, while the best that can be obtained from Theorem 2 is a lower shift index of −k. This theorem is proved by working directly with the generating function for a generalized statistic on ‘marked set partitions’. These set partitions are introduced in section ‘Computational results’.

Asymptotics for the Bell numbers yield the following asymptotics for the moments. The following result gives some asymptotic information about these moments.

Theorem 4.

Let α n = log(n)− log log(n)+o(1)be the positive real solution of u e u =n+1. Then
E ( d ( λ ) ) = α n 2 α n 2 n 2 + O n α n 1 . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equaj_HTML.gif
Let S k ( d ; n ) : = 1 B n λ Π ( n ) ( d ( λ ) M ( d ; n ) / B n ) k https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq80_HTML.gif be the symmetrized moments of the dimension exponent. Then
S 2 ( d ; n ) = α n 2 7 α n + 17 α n 3 ( α n + 1 ) n 3 + O n 2 α n 1 S 3 ( d ; n ) = 881 3 244 α n + 145 α n 2 83 3 α n 3 + 2 α n 4 n 4 α n 4 ( α n + 1 ) 3 + O n 3 α n 1 https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equak_HTML.gif

Remark 9.

Asymptotics for S k (d;n) with k=1,2,3,4,5,6 and with further accuracy are in section ‘More data’.

Analogous to these results for the dimension exponent are the following results for the intertwining exponent.

Theorem 5.

For each k{0,1,2, } there exists a closed form expression
M ( i k ; n ) : = λ Π ( n ) i ( λ ) k = Q k , 2 k ( n ) B n + 2 k + + Q k , 0 ( n ) B n + + Q k , k ( n ) B n k https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equal_HTML.gif

where each Qk,2kj is a polynomial with rational coefficients. Moreover, the degree of Qk,2kj is bounded by j.

For example,
M ( i ; n ) = 1 4 ( 2 n + 1 ) B n + ( 2 n + 9 ) B n + 1 5 B n + 2 M ( i 2 ; n ) = 1 144 ( 36 n 2 + 24 n 23 ) B n + ( 72 n 2 + 72 n 260 ) B n + 1 + ( 36 n 2 + 156 n + 489 ) B n + 2 ( 180 n + 814 ) B n + 3 + 225 B n + 4 . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equam_HTML.gif

Remark 10.

The expression for M(i;n)=M(c r2;n) was established first by Kasraoui (Theorem 2.3 of [9]).

Remark 11.

Theorem 5 is deduced directly from Theorem 2. The shifted Bell polynomials for M(i k ;n) for k≤5 are given in section ‘More data’, and see [55] for the aggregates with k≤12.

Remark 12.

Amusingly, the formula for M(i;n) implies that the sequence { B n } n = 0 https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq81_HTML.gif taken modulo 4 is periodic of length 12 beginning with {1,1,2,1,3,0,3,1,0,3,3,2}. Similarly, the formula for M(i2;n) shows that the sequence is periodic modulo 9 (respectively 16) with period 39 (respectively 48). For more about such periodicity, see the papers of Lunnon et al. [57] and Montgomery et al. [58].

In analogy with Theorem 4, there is the following asymptotic result.

Theorem 6.

With α n as above,
E ( i ( λ ) ) = 2 α n 5 4 α n 2 n 2 + O n α n 1 . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equan_HTML.gif
Let S k ( i ; n ) = 1 B n λ Π ( n ) i ( λ ) M ( i , n ) / B n k https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq82_HTML.gif. Then,
S 2 ( i ; n ) = 3 α n 2 22 α n + 56 9 α n 3 ( α n + 1 ) n 3 + O n 2 α n 1 S 3 ( i ; n ) = ( α n 5 ) ( 4 α n 3 31 α n 2 + 100 α n + 99 ) 8 α n 4 ( α n + 1 ) 3 n 4 + O n 3 α n 3 https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equao_HTML.gif

Theorems 3 and 5 show that there will be closed formulae for all of the moments of these statistics. Moreover, these theorems give bounds for the number of terms in the summand and the degree of each of the polynomials. Therefore, to compute the formulae, it is enough to compute enough values for M(d k ;n) or M(i k ;n) and then to do linear algebra to solve for the coefficients of the polynomials. For example, M(d;n) needs P1,2(n) which has degree at most 0, P1,1(n) which has degree at most 1, and P1,0(n) which has degree at most 0. Hence, there are four unknowns, and so only M(d;n) for n=1,2,3,4 are needed to derive the formula for the expected value of the dimension exponent.

Computational results

Enumerating set partitions and calculating these statistics would take time O(B n ) (see Knuth’s volume [13] for discussion of how to generate all set partitions of fixed size, the book of Wilf and Nijenhuis [59], or the website [60] of Ruskey). This section introduces a recursion for computing the number of set partitions of n with a given dimension or intertwining exponent in time O(n4). The recursion follows by introducing a notion of ‘marked’ set partitions. This generalization seems useful in general when computing statistics which depend on the internal structure of a set partition.

The results may then be used with Theorems 3 and 5 to find exact formulae for the moments. Proofs are given in section ‘Proofs of recursions, asymptotics, and Theorem 3’.

For a set partition λ, mark each block either open or closed. Call such a partition a marked set partition. For each marked set partition λ of [ n], let o(λ) be the number of open blocks of λ and (λ) be the total number of blocks of λ. (Marked set partitions may be thought of as what is obtained when considering a set partition of a potentially larger set and restricting it to [n]. The open blocks are those that will become larger upon adding more elements of this larger set, while the closed blocks are those that will not.) With this notation, define the dimension of λ with blocks B1,B2, by
d ~ ( λ ) = B j B j is closed max ( B j ) B j min ( B j ) + ( λ ) + n ( o ( λ ) 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equ5_HTML.gif
(5)
It is clear that if o(λ)=0, then λ may be thought of as a usual ‘unmarked’ set partition and d ~ ( λ ) = d ( λ ) https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq83_HTML.gif is the dimension exponent of λ. Define
f ( n ; A , B ) : = λ Π ( n ) : o ( λ ) = A and d ~ ( λ ) = B https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equ6_HTML.gif
(6)

Theorem 7.

For n>0
f ( n ; A , B ) = f ( n 1 ; A 1 , B A + 1 ) + f ( n 1 ; A , B A ) + Af ( n 1 ; A , B A + 1 ) + ( A + 1 ) f ( n 1 ; A + 1 , B A ) . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equap_HTML.gif

with initial condition f(0;A,B)=0 for all (A,B)≠(0,0) and f(0;0,0)=1.

Therefore, to find the number of partitions of [ n] with dimension exponent equal to k, it suffices to compute f(n,0,k) for k and n. Figure 2 gives the histograms of the dimension exponent when n=20 and n=100. With increasing n, these distributions tend to normal with mean and variance given in Theorem 4. This approximation is already apparent for n=20.
https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Fig2_HTML.jpg
Figure 2

Histograms of the dimension exponent counts for n =20 and n =100.

It is not necessary to compute the entire distribution of the dimension index to compute the moment formulae for the dimension exponent. Namely, it is better to implement the following recursion for the moments.

Corollary 1.

Define M k ( d ; n , A ) : = λ Π ( n ) o ( λ ) = A d ( λ ) k https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq84_HTML.gif. Then
M k ( d ; n , A ) = j = 0 k k j ( A 1 ) k j M j ( d ; n 1 , A 1 ) + j = 0 k k j A k j M j ( d ; n 1 , A ) + A j = 0 k k j ( A 1 ) k j M j ( d ; n 1 , A ) + ( A + 1 ) j = 0 k k j A k j M j ( d ; n 1 , A + 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equaq_HTML.gif

To compute M(d k ;n), then for each m<n, this recursion allows us to keep only k values rather than computing all O(m·m2) values of f(m,A,B). To find the linear relation of Theorem 3, only O(k·k2) values of M k (d;n,A) are needed.

In analogy, there is a recursion for the intertwining exponent.

Let f(i)(n,A,B) be the number of marked partitions of [ n] with intertwining weight equal to B and with A open sets where the intertwining weight is equal to the number of interlaced pairs ij and k where k is in a closed set plus the number of triples i,k,j such that ij and k is in an open set.

Theorem 8.

With the notation above, the following recursion holds
f ( i ) ( n + 1 , A , B ) = f ( i ) ( n , A , B ) + f ( i ) ( n , A 1 , B ) + j = 0 A f ( i ) ( n , A + 1 , B j ) + j = 0 A 1 f ( i ) ( n , A , B j ) . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equar_HTML.gif
This recursion allows the distribution to be computed rapidly. Figure 3 gives the histograms of the intertwining exponent when n=20 and n=100. Again, for increasing n, the distribution tends to normal with mean and variance from Theorem 6. The skewness is apparent for n=20.
https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Fig3_HTML.jpg
Figure 3

Histograms of the intertwining exponent counts for n =20 and n =100.

Proofs of recursions, asymptotics, and Theorem 3

This section gives the proofs of the recursive formulae discussed in Theorems 7 and 8 Additionally, this section gives a proof of Theorem 3 using the three-variable generating function for f(n,A,B). Finally, it gives an asymptotic expansion for Bn+k/B n with k fixed and n. This asymptotic is used to deduce Theorems 4 and 6.

Recursive formulae

This subsection gives the proof of the recursions for f(n,A,B) and f(i)(n,A,B) given in Theorems 7 and 8 The recursion is used in the next subsection to study the generating function for the dimension exponent.

Proof of Theorem 7 The four terms of the recursion come from considering the following cases: (1) n is added to a marked partition of [ n−1] as a singleton open set, (2) n is added to a marked partition of [ n−1] as a singleton closed set, (3) n is added to an open set of a marked partition of [ n−1] and that set remains open, and (4) n is added to an open set of a marked partition of [ n−1] and that set is closed.

Proof of Theorem 8 The argument is similar to that of Theorem 7. The same four cases arise. However, when adding n to an open set, the statistic may increase by any value j and it does so in exactly one way.

The generating function for f(n,A,B)

This section studies the generating function for f(n,A,B) and deduces Theorem 3. Let
F ( X , Y , Z ) : = n , A , B 0 f ( n ; A , B ) X n n ! Y A Z B https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equ7_HTML.gif
(7)
be the three-variable generating function. Theorem 7 implies that
∂X F ( X , Y , Z ) = ( 1 + Y ) ( F ( X , YZ , Z ) + F Y ( X , YZ , Z ) ) , https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equ8_HTML.gif
(8)

where F Y denotes ∂Y F https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq85_HTML.gif.

Then, F(X,0,Z) is the generating function for the distribution of d(λ), i.e.,
F ( X , 0 , Z ) = n = 0 λ Π ( n ) Z d ( λ ) X n n ! . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equas_HTML.gif
Thus, the generating function for the k th moment is
n 0 M ( d k ; n ) X n n ! = ( Z ∂Z ) k F ( X , Y , Z ) Z = 1 , Y = 0 . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equat_HTML.gif
Consider
F k ( X , Y ) : = ( Z ∂Z ) k F ( X , Y , Z ) Z = 1 . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equ9_HTML.gif
(9)

So F k ( X , 0 ) = M ( d k ; n ) X n n ! https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq86_HTML.gif.

Lemma 1

In the notation above,
∂X ( 1 + Y ) ∂Y F n ( X , Y ) = ( 1 + Y ) k > 0 n k Y ∂Y k 1 + ∂Y F n k ( X , Y ) . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equau_HTML.gif

Proof.

From (8),
∂X F n ( X , Y ) = ( 1 + Y ) k n k Y ∂Y k 1 + ∂Y F n k ( X , Y ) https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equav_HTML.gif
Hence, solving for F n gives
∂X ( 1 + Y ) ∂Y F n ( X , Y ) = ( 1 + Y ) k > 0 n k Y ∂Y k 1 + ∂Y F n k ( X , Y ) . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equ10_HTML.gif
(10)
Throughout the remainder Y=e α −1. Abusing notation, let
G k ( X , α ) : = G k ( X , Y ) : = F k ( X , Y ) exp ( ( 1 + Y ) ( e X 1 ) ) . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equaw_HTML.gif
The following lemma gives an expression for G k (X,α) in terms of a differential operators. Define the operators
R : = ∂X ∂α S : = e α T : = ∂α + e X + α . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equax_HTML.gif

Lemma 2.

Clearly, G0(X,Y)=1. Moreover,
G k ( X , α ) = a , b , c C a , b , c k S a T b X c 1 , https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equay_HTML.gif

Proof.

  1. (10)
    is equivalent to
    ∂X + ( 1 + Y ) e X ( 1 + Y ) ∂Y + e X G n ( X , Y ) = ( 1 + Y ) k > 0 n k Y ∂Y + e X 1 k ∂Y + e X G n k https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equaz_HTML.gif
     
Now
∂X ∂α G k ( X , α ) = > 0 k ( 1 e α ) ∂α + e X + α e α k ∂α + e X + α G k ( X , α ) https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equba_HTML.gif
where a e α has been commuted through. Then,
R G k ( X , α ) = > 0 k ( T T S 1 S ) T G k . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equ11_HTML.gif
(11)
Since G k (0,α)=0 for k>0,
G k ( X , α ) = 0 X > 0 k ( T T S 1 S ) T G k ( t , X + α t ) dt. https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equ12_HTML.gif
(12)
From this
G k ( X , α ) = a , b , c C a , b , c k S a T b X c 1 , https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equbb_HTML.gif
for some constants
C a , b , c k . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equbc_HTML.gif

The next lemma evaluates the terms in the summation of Lemma 2, thus yielding a generating function for G k (X,Y) which resembles that for the Bell numbers.

Lemma 3.

( T 1 ) α = 0 exp ( e X 1 ) = n 0 B n + X n n ! . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equbd_HTML.gif

Proof.

It is easy to see by induction on that T 1 is a polynomial in eX+α. Thus,
T 1 = ∂X + e X + α 1 . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Eqube_HTML.gif
Hence
T 1 α = 0 = ∂X + e X 1 . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equbf_HTML.gif
From this, it is easy to see that
T 1 α = 0 exp ( e X 1 ) = X exp ( e X 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equbg_HTML.gif

And the result follows.

Lemmas 2 and 3 readily yield the following expression for the moments of the dimension exponent as a shifted Bell polynomial.

Lemma 4.

For each k≥0 and n≥0
M ( d k ; n ) = a , b , c C a , b , c k n ( n 1 ) ( n c + 1 ) B n + b c . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equbh_HTML.gif

Theorem 3 needs some further constraints on the degrees of terms in this polynomial. The following lemma yields the claimed bounds for the degrees.

Lemma 5.

In the notation above, C a , b , c k = 0 https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq87_HTML.gif unless all of the following hold:
  1. 1.

    cb.

     
  2. 2.

    c<b unless a=0.

     
  3. 3.

    b≤2k.

     
  4. 4.

    3cbk.

     
  5. 5.

    3cbk−2 if a≠0.

     

Proof.

Let Ha,b,c(X,α)=S a T b X c 1. Using Equation 12, write C a , b , c k https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq88_HTML.gif in terms of the C a , b , c https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq89_HTML.gif for <k. To do this requires understanding
0 X H a , b , c ( t , X + α t ) dt. https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equbi_HTML.gif

As a first claim: if a=0, then the above is simply 1 c + 1 H 0 , b , c + 1 https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq90_HTML.gif. This is seen easily from the fact that R commutes with T. For a≠0, it is easy to see that this is a linear combination of the H a , b , c https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq91_HTML.gif over cc, and of H 0 , b , 0 https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq92_HTML.gif over bb.

The desired properties can now be proved by induction on k. It is clear that they all hold for k=0. For larger k, assume that they hold for all k, and use Equation 12 to prove them for k.

By the inductive hypothesis, the T Gk are linear combinations of Ha,b,c with c<b. Thus, (TT S−1S) T Gk is a linear combination of Ha,b,c’s with b>c. Thus, by Equation 12, G k is a linear combination of Ha,b,c’s with cb and a=0 or with c<b. This proves properties 1 and 2.

By the inductive hypothesis, the Gk are linear combinations of Ha,b,c with b≤2(k). Thus, TT S−1S T Gk is a linear combination of Ha,b,c’s with b≤2k+1−≤2k. Thus, by Equation 12, G k is a linear combination of Ha,b,c’s with b≤2k. This proves property 3.

Finally, consider the contribution to G k coming from each of the Gk terms. For =1, Gk is a linear combination of Ha,b,c’s with 3cbk−3 if a≠0, 3cbk−1 if a=0. Thus, T Gk is a linear combination of Ha,b,c’s with 3cbk−3 if a≠0, and 3cbk−2 otherwise. Thus, (TT S−1S) T Gk is a linear combination of Ha,b,c’s with 3cbk−3 if a=0, and 3cbk−2 otherwise. Thus, the contribution from these terms to G k is a linear combination of Ha,b,c’s with 3cbk and 3cbk−2 if a≠0. For the terms with >1, Gk is a linear combination of Ha,b,c’s with 3cbk−2 and 3cbk−4 when a≠0. Thus, T Gk is a linear combination of Ha,b,c’s with 3cbk−3, as is (TT S−1S) T Gk. Thus, the contribution of these terms to G k is a linear combination of Ha,b,c’s with 3cbk and 3cbk−3 if a≠0. This proves properties 4 and 5.

This completes the induction and proves the Lemma.

From this Lemma, it is easy to see that
M ( d k ; n ) = = 0 2 k B n + P k , ( n ) https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equbj_HTML.gif

for some polynomials Pk,(n) with deg(Pk,)≤ min(2k,k/2+/2).

Asymptotic analysis

This section presents some asymptotic analysis of the Bell numbers and ratios of Bell numbers. These results yield Theorems 4 and 6. Similar analysis can be found in [13].

Proposition 2.

Let α n be the solution to
u e u = n + 1 https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equbk_HTML.gif
and let
ζ n , k : = e α n 1 + 1 α n + k α n 2 = ( n + 1 ) ( α n + 1 ) + k α n 2 . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equbl_HTML.gif
Then
B n + k = ( n + k ) ! 2 π e ζ n , k 1 2 exp ( e α n ( n + k + 1 ) log ( α n ) ) ( 1 + O ( e α n ) ) . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equbm_HTML.gif
More precisely, for T≥0
B n + k = ( n + k ) ! 2 π e ζ n , k 1 2 exp ( e α n ( n + k + 1 ) log ( α n ) ) × 1 + m = 1 T R m , k ( α n ) 1 n m + O α n n T + 1 . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equbn_HTML.gif
where Rm,k are rational functions. In particular
R 1 , k ( u ) = ( 12 k 2 + 24 k 2 ) + ( 24 k 2 + 24 k + 18 ) u + ( 12 k 2 12 k + 20 ) u 2 + ( 12 k + 3 ) u 3 2 u 4 24 ( u + 1 ) 3 R 2 , k ( u ) = ( 1 , 44 k 4 384 k 3 + 624 k 2 1 , 152 k + 100 ) + ( 576 k 4 576 k 3 + 816 k 2 3 , 264 k 648 ) u 1152 ( u + 1 ) 6 + ( 864 k 4 + 1 , 056 k 3 + 432 k 2 6384 k 1 , 292 ) u 2 1 , 152 ( u + 1 ) 6 + ( 5 , 76 k 4 + 2 , 784 k 3 + 2 , 280 k 2 7 , 440 k 2 , 604 ) u 3 1 , 152 ( u + 1 ) 6 + ( 144 k 4 + 2 , 016 k 3 + 3 , 888 k 2 3 , 552 k 2 , 988 ) u 4 + ( 480 k 3 + 2 , 328 k 2 + 72 k 1 , 800 ) u 5 1 , 152 ( u + 1 ) 6 + ( 480 k 2 + 600 k 551 ) u 6 + ( 144 k 60 ) u 7 + 4 u 8 1 , 152 ( u + 1 ) 6 https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equbo_HTML.gif

Proof.

The proof is very similar to the traditional saddle point method for approximating B n . The idea is to evaluate at the saddle point for B n rather than for Bn+k. We follow the proof in Chapter 6 of [61].

By Cauchy’s formula,
2 πie ( n + k ) ! B n + k = C exp ( e z ) z n k 1 dz https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equbp_HTML.gif

where C encircles the origin once in the positive direction. Deform the path to a vertical line ui to u+i by taking a large segment of this line and a large semi-circle going around the origin. As the radius, say R, is taken to infinity the factor znk−1=O(Rnk−1) and exp(e z ) is bounded in the half-plane.

Choose u=α n and then
2 πe ( n + k ) ! B n + k = exp ( e α n ( n + k + 1 ) log ( α n ) ) exp ( ψ n , k ( y ) ) dy https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equbq_HTML.gif
where
ψ n , k ( y ) = e α n ( e iy 1 ) n + 1 + k e α n log 1 + iy α n 1 . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equbr_HTML.gif
The real part has maxima around y=2π m for each integer m, but using log ( 1 + y 2 α n 2 ) > 1 2 y 2 α n 2 https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq93_HTML.gif for π<y<α n and 1 + y 2 α n 2 > 2 y α n 1 https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq94_HTML.gif for y>α n as in [61] gives
exp ( ψ n , k ( y ) ) dy = π π exp ( ψ n , k ( y ) ) dy + O exp e α n α n . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equbs_HTML.gif
Next, note that
ψ n , k ( y ) = iky α n 1 + n + 1 + k ( n + 1 ) α n n + 1 α n y 2 2 + m > 2 1 m ! + ( 1 ) m n + 1 + k m α n m 1 ( n + 1 ) n + 1 α n ( iy ) m https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equbt_HTML.gif
where n + 1 + k e α n = α n + k e α n https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq95_HTML.gif and e α n = n + 1 α n https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq96_HTML.gif were used. Hence,
ψ n , k y ζ n , k = ik α n ζ n , k y 2 2 + m > 2 1 m ! + ( 1 ) m n + 1 + k m α n m 1 ( n + 1 ) n + 1 α n iy ζ n , k m https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equbu_HTML.gif
Making the change of variables and extending the sum of interval of integration gives
exp ( ψ n , k ( y ) ) dy + O exp e α n α n = e y 2 2 exp ik α n ζ n , k + m > 2 frac 1 m ! + ( 1 ) m n + 1 + k α n m 1 ( n + 1 ) n + 1 α n iy ζ n , k m dy. https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equbv_HTML.gif
Hence, Taylor expanding around y=0 and using
R y k e y 2 2 dy = 0 k 1 ( mod 2 ) 2 π k ! 2 k 2 k 2 ! k 0 ( mod 2 ) https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equbw_HTML.gif

gives the desired result.

For more details, see [61].

Proposition 2 yields
B n + k B n = ( n + k ) ! n ! α n k 1 k α n ( n + 1 ) ( α n + 1 ) 1 2 ( 1 + O ( e α n ) ) . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equ13_HTML.gif
(13)

Direct application of this result gives the results in Theorems 4 and 6.

Proofs of Theorems 1 and 2

This section gives the proofs of Theorems 2 and 1. Theorem 2 implies Theorem 5. A pair of lemmas which will be useful in the proof of Theorem 2:

Lemma 6.

For B n , the Bell numbers, define
g r , d , k , s ( n ) : = n d i = 0 n k n k i B i + s r n k i https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equbx_HTML.gif

where r,d,k,s are non-negative integers. Then, gr,d,k,s(n) is a shifted Bell polynomial of lower shift index −k and upper shift index r+sk.

Proof.

It clearly suffices to prove that gr,0,k,s(n) is a shifted Bell polynomial. Since gr,0,0,s(nk)=gr,0,k,s(n), it suffices to prove that gr,s(n):=gr,0,0,s(n) is a shifted Bell polynomial.

For this, consider the exponential generating function
n = 0 g r , s ( n ) x n n ! = n = 0 i = 0 n n i B i + s r n i x n n ! = a = 0 b = 0 B a + s r b x a + b a ! b ! = s x s n = 0 B n x n n ! e rx = e rx s x s e e x 1 . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equby_HTML.gif

This is easily seen to be equal to e e x 1 https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq97_HTML.gif times a polynomial in e x of degree s+r.

Since g0,s(n)=Bn+s, the generating function n = 0 B n + s x n n ! https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq98_HTML.gif equals e e x 1 https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq99_HTML.gif times a polynomial of exact degree s. From this, for all s,r the space of all polynomials in e x of degree at most s+r times e e x 1 https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq100_HTML.gif is spanned by the set of generating functions n = 0 B n + m x n n ! https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq101_HTML.gif as m runs over all integers 0,1,…,s+r. Since the generating function for gr,s(n) lies in this span,
n = 0 g r , s ( n ) x n n ! = m = 0 s + r β s , r , m n = 0 B n + m x n n ! https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equbz_HTML.gif
for some rational numbers βs,r,m. It follows that for all n,
g r , s ( n ) = m = 0 s + r β s , r , m B n + m . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equca_HTML.gif
For a sequence, r={r0,r1,,r k }, of rational numbers and a polynomial Q Q [ y 1 , , y k , m ] https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq102_HTML.gif define
M ( k , Q , r , n , x ) : = 1 x 1 < x 2 < < x k n Q ( x 1 , , x k , n ) i = 0 k ( x + r i ) x i + 1 x i 1 , https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equ14_HTML.gif
(14)

where x0=0,xk+1=n+1.

Lemma 7.

Fix k, let Q Z [ y 1 , , y k , m ] https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq103_HTML.gif and r={r0,r1,,r k } be a sequence of rational numbers. As defined above, M(k,Q,r,n,x) is a rational linear combination of terms of the form
F ( n ) G ( x ) ( x + r i ) n k , https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equcb_HTML.gif

where F Q [ n ] , G Q [ x ] https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq104_HTML.gif are polynomials.

Proof.

The proof is by induction on k. If k=0 then definitionally, M(k,Q,r,n,x)=Q(n)(x+r0) n , providing a base case for our result. Assume that the lemma holds for k one smaller. For this, fix the values of x1,…,xk−1 in the sum and consider the resulting sum over x k . Then
M ( k , Q , r , n , x ) = 1 x 1 < x 2 < < x k 1 n 1 i = 0 k 2 ( x + r i ) x i + 1 x i 1 × x k 1 < x k n Q ( x 1 , , x k , n ) ( x + r k 1 ) x k x k 1 1 ( x + r k ) n x k . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equcc_HTML.gif

Consider the inner sum over x k :

If rk−1=r k , then the product of the last two terms is always ( x + r k ) n x k 1 2 https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq105_HTML.gif, and thus the sum is some polynomial in x1,…,xk−1,n times ( x + r k ) n x k 1 2 https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq106_HTML.gif. The remaining sum over x1,…,xk−1 is exactly of the form M(k−1,Q,r,n−1,x), for some polynomial Q, and thus, by the inductive hypothesis, of the correct form.

If rk−1r k , the sum is over pairs of non-negative integers a=x k xk−1−1 and b=nx k −1 summing to nxk−1−2 of some polynomial, Q in a and n and the other x i times (x+rk−1) a (x+r k ) b . Letting y=(x+rk−1) and z=(x+r k ), this is a sum of Q(x i ,n,a)y a z b . Let d be the a degree of Q. Multiplying this sum by (yz)d+1, yields, by standard results, a polynomial in y and z of degree nxk−1−2+(d+1) in which all terms have either y exponent or z exponent at least nxk−1−1. Thus, this inner sum over x k when multiplied by the non-zero constant (rk−1r k )d+1 yields the sum of a polynomial in x,n,x1,…,xk−1 times ( x + r k 1 ) n x k 1 2 https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq107_HTML.gif plus another such polynomial times ( x + r k 1 ) n x k 1 2 https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq108_HTML.gif. Thus, M(k,Q,r,n,x) can be written as a linear combination of terms of the form G(x)M(k−1,Q,r,n,x). The inductive hypothesis is now enough to complete the proof.

Turn next to the proof of Theorem 2. Proof of Theorem 2

It suffices to prove this Theorem for simple statistics. Thus, it suffices to prove that for any pattern P and polynomial Q that
M ( f P , Q ; n ) = λ Π n f P , Q ( λ ) = λ Π n s P λ Q ( s ) https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equcd_HTML.gif
is given by a shifted Bell polynomial in n. As a first step, interchange the order of summation over s and λ above. Hence,
M ( f P , Q ; n ) = s [ n ] k Q ( s ) λ Π ( n ) s P λ 1 . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equce_HTML.gif
To deal with the sum over λ above, first consider only the blocks of λ that contain some element of s. Equivalently, let λ be obtained from λ by replacing all of the blocks of λ that are disjoint from s by their union. To clarify this notation, let Π(n) denote the set of all set partitions of [ n] with at most one marked block. For λΠ(n), say that s P λ if s in an occurrence of P in λ as a regular set partition so that additionally the non-marked blocks of λ are exactly the blocks of λ that contain some element of s. For λΠ(n) and λΠ(n), say that λ is a refinement of λ if the unmarked blocks in λ are all parts in λ, or equivalently, if λ can be obtained from λ by further partitioning the marked block. Denote λ being a refinement of λ as λλ. Thus, in the previous computation of M(fP,Q;n), letting λ be the marked partition obtained by replacing the blocks in λ disjoint from s by their union:
M ( f P , Q ; n ) = s [ n ] k Q ( s ) λ Π ( n ) s P λ λ Π ( n ) λ λ 1 . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equcf_HTML.gif
Note that the λ in the final sum above correspond exactly to the set partitions of the marked block of λ. For λΠ(n), let |λ| be the size of the marked block of λ. Thus,
M ( f P , Q ; n ) = s [ n ] k Q ( s ) λ Π ( n ) s P λ B | λ | . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equcg_HTML.gif

Remark 13.

This is valid even when the marked block is empty.

Dealing directly with the Bell numbers above will prove challenging, so instead compute the generating function
M ( P , Q , n , x ) : = s [ n ] k Q ( s ) λ Π ( n ) s P λ x | λ | . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equch_HTML.gif

After computing this, extract the coefficients of M(P,Q,n,x) and multiply them by the appropriate Bell numbers.

To compute M(P,Q,n,x), begin by computing the value of the inner sum in terms of s=(x1<x2<…<x k ) that preserves the consecutivity relations of P (namely those in C(P)). Denote the equivalence classes in P by 1,2,…,. Let z i be a representative of this i th equivalence class. Then, an element λΠ(n) so that s P λ can be thought of as a set partition of [ n] into labeled equivalence classes 0,1,…,, where the 0th class is the marked block, and the i th class is the block containing x z i https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq109_HTML.gif. Thus, think of the set of such λ as the set of maps g:[ n]→{0,1,…,} so that:
  1. 1.

    g(x j )=i if j is in the i th equivalence class

     
  2. 2.

    g(x)≠i if x<x j , jF(P) and j is in the i th equivalence class

     
  3. 3.

    g(x)≠i if x>x j , jL(P) and j is in the i th equivalence class

     
  4. 4.

    g(x)≠i if x j < x < x j https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq110_HTML.gif, (j,j )A(P) and j,j are in the i th equivalence class

     
It is possible that no such g will exist if one of the latter three properties must be violated by some x=x h . If this is the case, this is a property of the pattern P, and not the occurrence s, and thus, M(fP,Q;n)=0 for all n. Otherwise, in order to specify g, assign the given values to g(x i ) and each other g(x) may be independently assigned values from the set of possibilities that does not violate any of the other properties. It should be noted that 0 is always in this set, and that furthermore, this set depends only which of the x i our given x is between. Thus, there are some sets S0,S1,…,S k {0,1,…,}, depending only on s, so that g is determined by picking functions
{ 1 , , x 1 1 } S 0 , { x 1 + 1 , , x 2 1 } S 1 , , { x k + 1 , , n } S k . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equci_HTML.gif
Thus, the sum over such λ of x | λ | https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq111_HTML.gif is easily seen to be
( x + r 0 ) x 1 1 ( x + r i ) x 2 x 1 1 ( x + r k 1 ) x k x k 1 1 ( x + r k ) n x k , https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equcj_HTML.gif
where r i =|S i |−1 (recall |S i |>0, because 0S i ). For such a sequence, r of rational numbers define
M ( k , Q , r , n , x , C ( P ) ) : = 1 x 1 < x 2 < < x k n x i x j = 1 for ( i , j ) C ( P ) Q ( x 1 , , x k , n ) i = 0 k ( x + r i ) x i + 1 x i 1 , https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equ15_HTML.gif
(15)

where, as in Lemma 7, using the notation x0=0,xk+1=n+1.

Note that the sum is empty if C(P) contains nonconsecutive elements. We will henceforth assume that this is not the case. We call j a follower if either (j−1,j) or (j,j−1) are in C(P). Clearly, the values of all x i are determined only by those x i where i is not a follower. Furthermore, Q is a polynomial in these values and n. If j is the index of the i th non-follower then let y i =x j j+i. Now, sequences of x i satisfying the necessary conditions correspond exactly to those sequences with 1≤y1<y2<<ykfnf where f is the total number of followers. Thus,
M ( k , Q , r , n , x , C ( P ) ) = 1 y 1 < y 2 < < y k f n f Q ~ ( y 1 , , y k , n ) i = 0 k ( x + r i ~ ) y i + 1 y i 1 = M ( k f , Q ~ , r ~ , n f , x ) . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equck_HTML.gif

where the r i ~ https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq112_HTML.gif are modified versions of the r i to account for the change from {x j } to {y i }. In particular, if x j is the (i+1) s t non-follower, then r i ~ = r j 1 https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq113_HTML.gif.

By Lemma 7, M ( k f , Q ~ , r ~ , n f , x ) https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq114_HTML.gif is a linear combination of terms of the form F(n)G(x)(x+r i )nk for polynomials F Q [ n ] https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq115_HTML.gif and G Q [ x ] https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq116_HTML.gif.

Thus, M(fP,Q;n) can be written as a linear combination of terms of the form gr,d,,s(n) where is the number of equivalence classes in P and r,d,s are non-negative integers. Therefore, by Lemma 6 M(fP,Q;n) is a shifted Bell polynomial.

The bound for the upper shift index follows from the fact that M(fP,Q;n)=O(n N B n ) and by (13) each term n α Bn+β is of an asymptotically distinct size. To complete the proof of the result it is sufficient to bound the lower shift index of the Bell polynomial. By (15) it is clear the largest power of x in each term is (nk). Thus, from Lemma 6, the resulting shift Bell polynomials can be written with minimum lower shift index −k. This completes the proof.

Next turn to the proof of Theorem 1. To this end, introduce some notation.

Definition 3.

Given three patterns P1,P2,P3, of lengths k1,k2,k3, say that a merge of P1 and P2 onto P3 is a pair of strictly increasing functions m1:[ k1]→[k3], m2:[ k2]→[ k3] so that
  1. 1.

    m 1([k 1])m 2([k 2])=[ k 3]

     
  2. 2.
    m 1 ( i ) P 3 m 1 ( j ) https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equcl_HTML.gif
     
if and only if i P 1 j https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq117_HTML.gif, and m 2 ( i ) P 3 m 2 ( j ) https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq118_HTML.gif if and only if
i P 2 j https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equcm_HTML.gif
  1. 3.

    iF(P 3) if and only if there exists either a jF(P 1) so that i=m 1(j) or a jF(P 2) so that i=m 2(j)

     
  2. 4.

    iL(P 3) if and only if there exists either a jL(P 1) so that i=m 1(j) or a jL(P 2) so that i=m 2(j)

     
  3. 5.

    (i,i )A(P 3) if and only if there exists either a (j,j )A(P 1) so that i=m 1(j) and i =m 1(j ) or a (j,j )A(P 2) so that i=m 2(j) and i =m 2(j )

     
  4. 6.

    (i,i )C(P 3) if and only if there exists either a (j,j )C(P 1) so that i=m 1(j) and i =m 1(j ) or a (j,j )C(P 2) so that i=m 2(j) and i =m 2(j )

     

Such a merge is denoted as m1,m2:P1,P2P3.

Note that the last four properties above imply that given P1 and P2, a merge (including a pattern P3) is uniquely defined by maps m1,m2 and an equivalence relation P 3 https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq119_HTML.gif satisfying (1) and (2) above.

Lemma 8.

Let P1 and P2 be patterns. For any λ there is a one-to-one correspondence:
( s 1 , s 2 ) : s 1 P 1 λ , s 2 P 2 λ P 3 , s 3 P 3 λ , and m 1 , m 2 : P 1 , P 2 P 3 . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equ16_HTML.gif
(16)
Moreover, under this correspondence
Q m 1 , m 2 , Q 1 , Q 2 ( s 3 ) : = Q 1 ( z m 1 ( 1 ) , z m 1 ( 2 ) , , z m 1 ( k 1 ) , n ) Q 2 ( z m 2 ( 1 ) , z m 2 ( 2 ) , , z m 2 ( k 2 ) , n ) = Q 1 ( s 1 ) Q 2 ( s 2 ) . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equ17_HTML.gif
(17)

Proof.

Begin by demonstrating the bijection defined by Eq. 16. On the one hand, given s 3 P 3 λ https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq120_HTML.gif given by z 1 < z 2 < < z k 3 https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq121_HTML.gif and m1,m2:P1,P2P3, define s1 and s2 by the sequences z m 1 ( 1 ) < z m 1 ( 2 ) < < z m 1 ( k 1 ) https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq122_HTML.gif and z m 2 ( 1 ) < z m 2 ( 2 ) < < z m 2 ( k 2 ) https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq123_HTML.gif. It is easy to verify that these are occurrences of the patterns P1 and P2 and furthermore that Eq. 17 holds for this mapping.

This mapping has a unique inverse: Given s1 and s2, note that s3 must equal the union s1s2. Furthermore, the maps m a , for a=1,2, must be given by the unique function so that m a (i)=j if and only if the i t h smallest element of s a equals the j t h smallest element of s3. Note that the union of these images must be all of [k3]. In order for s3 to be an occurrence of P3 the equivalence relation P 3 https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq124_HTML.gif must be that i P 3 j https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq125_HTML.gif if and only if the i t h and j t h elements of s3 are equivalent under λ. Note that since S1 and S2 were occurrences of P1 and P2, that this must satisfy condition (2) for a merge. The rest of the data associated to P3 (namely F(P3),L(P3), A(P3), and C(P3)) is now uniquely determined by m1,m2,P1,P2 and the fact that P3 is a merge of P1 and P2 under these maps. To show that s3 is an occurrence of P3 first note that by construction the equivalence relations induced by λ and P3 agree. If iF(P3), then there is a jF(P a ) with i=m a (j) for some a,j. Since s a is an occurrence of P a , this means that the j t h smallest element of s a in in F i r s t(λ). On the other hand, by the construction of m a , this element is exactly z m a ( j ) = z i https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq126_HTML.gif. This if iF(P3), z i F i r s t(λ). The remaining properties necessary to verify that S3 is an occurrence of P3 follow similarly. Thus, having shown that the above map has a unique inverse, the proof of the lemma is complete.

Recall, the number of singleton blocks is denoted X1 and it is a simple statistic. To illustrate this lemma return to the example of X 1 2 https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq127_HTML.gif discussed prior to the lemma. Let P1=P2 be the pattern of length 1 with A(P1)=ϕ, F(P1)=L(P1)=1. Then there are five possible merges of P1 and P2 into some pattern P3. The first choice of P3 is P1 itself. In which case m1(1)=m2(1)=1. The latter choices of P3 is the pattern of length 2 with F(P3)=L(P3)={1,2},A(P3)=. The equivalence relation on P3 could be either the trivial one or the one that relates 1 and 2 (though in the latter case the pattern P3 will never have any occurrences in any set partition). In either of these cases, there is a merge with m1(1)=1 and m2(1)=2 and a second merge with m1(1)=2 and m2(1)=1. As a result,
M ( X 1 2 ; n ) = λ Π ( n ) X 1 ( λ ) 2 = λ Π ( n ) x 1 x 1 First ( λ ) x 1 Last ( λ ) 1 2 = λ Π ( n ) x 1 x 1 First ( λ ) x 1 Last ( λ ) y 1 y 1 First ( λ ) y 1 Last ( λ ) 1 = 2 λ Π ( n ) x 1 < x 2 x 1 , x 2 First ( λ ) x 1 , x 2 Last ( λ ) 1 + λ Π ( n ) x 1 x 1 First ( λ ) x 1 Last ( λ ) 1 https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equcn_HTML.gif

Proof of Theorem 1 The fact that statistics are closed under pointwise addition and scaling follows immediately from the definition. Similarly, the desired degree bounds for these operations also follow easily. Thus only closure and degree bounds for multiplication must be proved. Since every statistic may be written as a linear combination of simple statistics of no greater degree, and since statistics are closed under linear combination, it suffices to prove this theorem for a product of two simple statistics. Thus let f i be the simple statistic defined by a pattern P i of size k i and a polynomial Q i . It must be shown that f1(λ)f2(λ) is given by a statistic of degree at most k1+k2+ deg(Q1)+ deg(Q2).

For any λ
f 1 ( λ ) f 2 ( λ ) = s 1 P 1 λ , s 2 P 2 λ Q 1 ( s 1 ) Q 2 ( s 2 ) . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equco_HTML.gif

Simplify this equation using Lemma 8, writing this as a sum over occurrences of only a single pattern in λ.

Applying Lemma 8,
f 1 ( λ ) f 2 ( λ ) = s 1 P 1 λ , s 2 P 2 λ Q 1 ( s 1 ) Q 2 ( s 2 ) = P 3 m 1 , m 2 : P 1 , P 2 P 3 s 3 P 3 λ Q m 1 , m 2 , Q 1 , Q 2 ( s 3 ) , = m 1 , m 2 : P 1 , P 2 P 3 f P 3 , Q m 1 , m 2 , Q 1 , Q 2 ( λ ) . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_Equcp_HTML.gif

Thus, the product of f1 and f2 is a sum of simple characters. Note that the quantity is a polynomial of s3 which is denoted Q m 1 , m 2 , Q 1 , Q 2 ( s 3 ) . https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq128_HTML.gif Finally, each pattern P3 has size at most k1+k2 and each polynomial Q m 1 , m 2 , Q 1 , Q 2 https://static-content.springer.com/image/art%3A10.1186%2F2197-9847-1-2/MediaObjects/40687_2014_Article_2_IEq129_HTML.gif has degree at most deg(Q1)+ deg(Q2). Thus the degree of the product is at most the sum of the degrees.

More data

This section contains some data for the dimension and intertwining exponent statistics. The moment formulae of Theorem 3 for k≤22 and the moment formulae for the intertwining exponent for k≤12 have been computed and are available at [55]. Moreover, the values f(n,0,B) for n≤238 and f(i)(n,0,B) for n≤146 are available. These sequences can also be found on Sloane’s Online Encyclopedia of integer sequences [62].

The remainder of this section contains a small amount of data and observations regarding the distributions f(n,0,B) and f(i)(n,0,B) and regarding the shifted Bell polynomials of Theorems 3 and 5.

Dimension index

A couple of easy observations: it is clear that
f ( n , 0 , 0 ) = 2 n 1 .