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What is the Generating Functiongenerating function for Skewskew Young Diagramsdiagrams?

The Problemproblem

This strikes me as a very natural problem which should have been asked (and solved?) already.

For each positive integer k, find a nice expression for the following generating function in the variable x: $$ \sum_{\lambda/\mu} x^{|\lambda|}. $$

Here $\lambda$ ranges over all partitions and $\mu$ over those partitions contained in $\lambda$ for which the skew Young diagram $\lambda/\mu$ has k nodes, i.e., each partition of n is weighted by the number of partitions of $n-k$ it contains.

Examples: k=1, the function is $\frac{x}{(1-x)}P(x) $, where $P(x)=\prod_{i=1}^\infty(1-x^i)^{-1}$ is the partition generating function. So in this case I'm just enumerating partitions by the number of removable nodes. The formula is equivalent to the well-known fact that every partition has one more addable than removable node.

I've computed the cases k=2,3,4 also (k=4 was painful - I broke it into 14 possible types of skew-diagrams). For k=2 the generating function is $\frac{ x^2(2-x)}{(1-x)(1-x^2)}P(x).$

It seems plausible that there is a polynomial $F_k(x)$ of degree at most $k(k-1)/2$ (with leading coefficient $\pm 1$) so that the power series is $$ \frac{ x^k F_k(x)}{(1-x)(1-x^2)..(1-x^k)}P(x). $$

If $F_k(x)$ exists, it's easy to see that it must have lowest terms $p_k+p_{k+1}x+2(p_{k+2}-1)x^2+...$, where $p_n$=number of partitions of n, after which the terms depend on congruences for k. This suggests that its complicated. Perhaps there is no nice expression for $F_k(x)$. Even knowing whether $F_k(x)$ exists is of interest to me. Maybe there is a neater way of expressing the entire generating function?

Motivation

The coefficient of $x^n$ in the generating function is the dimension of the centre of a certain subalgebra of the complex group algebra of the symmetric group of degree n. This is ${\mathbb C}S_n^{S_{n-k}}$, the centralizer of the subgroup $S_{n-k}$ in ${\mathbb C}S_n$. It is easy to see that this has as ${\mathbb C}$-basis the $S_{n-k}$-orbit sums in $S_n$. The centre is indexed by pairs $(\chi,\phi)$, where $\chi$ is an irreducible character of $S_n$, and $\phi$ is an irreducible character of $S_{n-k}$ occuring in the restriction of $\chi$. The formulation above is then an easy consequence of the parametrization of irreducible characters of $S_n$, and the classic branching rule.

Literature

Yoshiaki Ueno, On the Generating Functions of the Young Lattice, J. Algebra 116 (1988) 261--270.

This gives a generating polynomial for the partitions contained in a given partition $\lambda$ in terms of a determinant involving Gaussian coefficients. It's a beautiful result, but it did not give me any insight into my problem.

What is the Generating Function for Skew Young Diagrams?

The Problem

This strikes me as a very natural problem which should have been asked (and solved?) already.

For each positive integer k, find a nice expression for the following generating function in the variable x: $$ \sum_{\lambda/\mu} x^{|\lambda|}. $$

Here $\lambda$ ranges over all partitions and $\mu$ over those partitions contained in $\lambda$ for which the skew Young diagram $\lambda/\mu$ has k nodes, i.e., each partition of n is weighted by the number of partitions of $n-k$ it contains.

Examples: k=1, the function is $\frac{x}{(1-x)}P(x) $, where $P(x)=\prod_{i=1}^\infty(1-x^i)^{-1}$ is the partition generating function. So in this case I'm just enumerating partitions by the number of removable nodes. The formula is equivalent to the well-known fact that every partition has one more addable than removable node.

I've computed the cases k=2,3,4 also (k=4 was painful - I broke it into 14 possible types of skew-diagrams). For k=2 the generating function is $\frac{ x^2(2-x)}{(1-x)(1-x^2)}P(x).$

It seems plausible that there is a polynomial $F_k(x)$ of degree at most $k(k-1)/2$ (with leading coefficient $\pm 1$) so that the power series is $$ \frac{ x^k F_k(x)}{(1-x)(1-x^2)..(1-x^k)}P(x). $$

If $F_k(x)$ exists, it's easy to see that it must have lowest terms $p_k+p_{k+1}x+2(p_{k+2}-1)x^2+...$, where $p_n$=number of partitions of n, after which the terms depend on congruences for k. This suggests that its complicated. Perhaps there is no nice expression for $F_k(x)$. Even knowing whether $F_k(x)$ exists is of interest to me. Maybe there is a neater way of expressing the entire generating function?

Motivation

The coefficient of $x^n$ in the generating function is the dimension of the centre of a certain subalgebra of the complex group algebra of the symmetric group of degree n. This is ${\mathbb C}S_n^{S_{n-k}}$, the centralizer of the subgroup $S_{n-k}$ in ${\mathbb C}S_n$. It is easy to see that this has as ${\mathbb C}$-basis the $S_{n-k}$-orbit sums in $S_n$. The centre is indexed by pairs $(\chi,\phi)$, where $\chi$ is an irreducible character of $S_n$, and $\phi$ is an irreducible character of $S_{n-k}$ occuring in the restriction of $\chi$. The formulation above is then an easy consequence of the parametrization of irreducible characters of $S_n$, and the classic branching rule.

Literature

Yoshiaki Ueno, On the Generating Functions of the Young Lattice, J. Algebra 116 (1988) 261--270.

This gives a generating polynomial for the partitions contained in a given partition $\lambda$ in terms of a determinant involving Gaussian coefficients. It's a beautiful result, but it did not give me any insight into my problem.

What is the generating function for skew Young diagrams?

The problem

This strikes me as a very natural problem which should have been asked (and solved?) already.

For each positive integer k, find a nice expression for the following generating function in the variable x: $$ \sum_{\lambda/\mu} x^{|\lambda|}. $$

Here $\lambda$ ranges over all partitions and $\mu$ over those partitions contained in $\lambda$ for which the skew Young diagram $\lambda/\mu$ has k nodes, i.e., each partition of n is weighted by the number of partitions of $n-k$ it contains.

Examples: k=1, the function is $\frac{x}{(1-x)}P(x) $, where $P(x)=\prod_{i=1}^\infty(1-x^i)^{-1}$ is the partition generating function. So in this case I'm just enumerating partitions by the number of removable nodes. The formula is equivalent to the well-known fact that every partition has one more addable than removable node.

I've computed the cases k=2,3,4 also (k=4 was painful - I broke it into 14 possible types of skew-diagrams). For k=2 the generating function is $\frac{ x^2(2-x)}{(1-x)(1-x^2)}P(x).$

It seems plausible that there is a polynomial $F_k(x)$ of degree at most $k(k-1)/2$ (with leading coefficient $\pm 1$) so that the power series is $$ \frac{ x^k F_k(x)}{(1-x)(1-x^2)..(1-x^k)}P(x). $$

If $F_k(x)$ exists, it's easy to see that it must have lowest terms $p_k+p_{k+1}x+2(p_{k+2}-1)x^2+...$, where $p_n$=number of partitions of n, after which the terms depend on congruences for k. This suggests that its complicated. Perhaps there is no nice expression for $F_k(x)$. Even knowing whether $F_k(x)$ exists is of interest to me. Maybe there is a neater way of expressing the entire generating function?

Motivation

The coefficient of $x^n$ in the generating function is the dimension of the centre of a certain subalgebra of the complex group algebra of the symmetric group of degree n. This is ${\mathbb C}S_n^{S_{n-k}}$, the centralizer of the subgroup $S_{n-k}$ in ${\mathbb C}S_n$. It is easy to see that this has as ${\mathbb C}$-basis the $S_{n-k}$-orbit sums in $S_n$. The centre is indexed by pairs $(\chi,\phi)$, where $\chi$ is an irreducible character of $S_n$, and $\phi$ is an irreducible character of $S_{n-k}$ occuring in the restriction of $\chi$. The formulation above is then an easy consequence of the parametrization of irreducible characters of $S_n$, and the classic branching rule.

Literature

Yoshiaki Ueno, On the Generating Functions of the Young Lattice, J. Algebra 116 (1988) 261--270.

This gives a generating polynomial for the partitions contained in a given partition $\lambda$ in terms of a determinant involving Gaussian coefficients. It's a beautiful result, but it did not give me any insight into my problem.

improved in-line equation formatting; added tag "young-tableaux"
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#The Problem#

The Problem

This strikes me as a very natural problem which should have been asked (and solved?) already.

For each positive integer kk, find a nice expression for the following generating function in the variable xx: $$ \sum_{\lambda/\mu} x^{|\lambda|}. $$

Here \lambda$\lambda$ ranges over all partitions and \mu$\mu$ over those partitions contained in \lambda$\lambda$ for which the skew Young diagram \lambda/\mu$\lambda/\mu$ has kk nodes, i.e., each partition of nn is weighted by the number of partitions of n-k$n-k$ it contains.

Examples: k=1, the function is $\frac{x}{(1-x)}P(x) $, where $P(x)=\prod_{i=1}^\infty(1-x^i)^{-1}$ is the partition generating function. So in this case I'm just enumerating partitions by the number of removable nodes. The formula is equivalent to the well-known fact that every partition has one more addable than removable node.

I've computed the cases k=2,3,4 also (k=4 was painful - I broke it into 14 possible types of skew-diagrams). For k=2 the generating function is $\frac{ x^2(2-x)}{(1-x)(1-x^2)}P(x).$

It seems plausible that there is a polynomial F_k(x)$F_k(x)$ of degree at most k(k-1)/2$k(k-1)/2$ (with leading coefficient \pm 1$\pm 1$) so that the power series is $$ \frac{ x^k F_k(x)}{(1-x)(1-x^2)..(1-x^k)}P(x). $$

If $F_k(x)$ exists, it's easy to see that it must have lowest terms $p_k+p_{k+1}x+2(p_{k+2}-1)x^2+...$, where p_n=number$p_n$=number of partitions of nn, after which the terms depend on congruences for kk. This suggests that its complicated. Perhaps there is no nice expression for $F_k(x)$. Even knowing whether $F_k(x)$ exists is of interest to me. Maybe there is a neater way of expressing the entire generating function?

#Motivation#

Motivation

The coefficient of $x^n$ in the generating function is the dimension of the centre of a certain subalgebra of the complex group algebra of the symmetric group of degree nn. This is ${\mathbb C}S_n^{S_{n-k}}$, the centralizer of the subgroup $S_{n-k}$ in ${\mathbb C}S_n$. It is easy to see that this has as ${\mathbb C}$-basis the $S_{n-k}$-orbit sums in $S_n$. The centre is indexed by pairs $(\chi,\phi)$, where $\chi$ is an irreducible character of $S_n$, and $\phi$ is an irreducible character of $S_{n-k}$ occuring in the restriction of $\chi$. The formulation above is then an easy consequence of the parametrization of irreducible characters of $S_n$, and the classic branching rule.

#Literature#

Literature

Yoshiaki Ueno, On the Generating Functions of the Young Lattice, J. Algebra 116 (1988) 261--270.

This gives a generating polynomial for the partitions contained in a given partition \lambda$\lambda$ in terms of a determinant involving Gaussian coefficients. It's a beautiful result, but it did not give me any insight into my problem.

#The Problem#

This strikes me as a very natural problem which should have been asked (and solved?) already.

For each positive integer k, find a nice expression for the following generating function in the variable x: $$ \sum_{\lambda/\mu} x^{|\lambda|}. $$

Here \lambda ranges over all partitions and \mu over those partitions contained in \lambda for which the skew Young diagram \lambda/\mu has k nodes i.e. each partition of n is weighted by the number of partitions of n-k it contains.

Examples: k=1, the function is $\frac{x}{(1-x)}P(x) $, where $P(x)=\prod_{i=1}^\infty(1-x^i)^{-1}$ is the partition generating function. So in this case I'm just enumerating partitions by the number of removable nodes. The formula is equivalent to the well-known fact that every partition has one more addable than removable node.

I've computed the cases k=2,3,4 also (k=4 was painful - I broke it into 14 possible types of skew-diagrams). For k=2 the generating function is $\frac{ x^2(2-x)}{(1-x)(1-x^2)}P(x).$

It seems plausible that there is a polynomial F_k(x) of degree at most k(k-1)/2 (with leading coefficient \pm 1) so that the power series is $$ \frac{ x^k F_k(x)}{(1-x)(1-x^2)..(1-x^k)}P(x). $$

If $F_k(x)$ exists, it's easy to see that it must have lowest terms $p_k+p_{k+1}x+2(p_{k+2}-1)x^2+...$, where p_n=number of partitions of n, after which the terms depend on congruences for k. This suggests that its complicated. Perhaps there is no nice expression for $F_k(x)$. Even knowing whether $F_k(x)$ exists is of interest to me. Maybe there is a neater way of expressing the entire generating function?

#Motivation#

The coefficient of $x^n$ in the generating function is the dimension of the centre of a certain subalgebra of the complex group algebra of the symmetric group of degree n. This is ${\mathbb C}S_n^{S_{n-k}}$, the centralizer of the subgroup $S_{n-k}$ in ${\mathbb C}S_n$. It is easy to see that this has as ${\mathbb C}$-basis the $S_{n-k}$-orbit sums in $S_n$. The centre is indexed by pairs $(\chi,\phi)$, where $\chi$ is an irreducible character of $S_n$, and $\phi$ is an irreducible character of $S_{n-k}$ occuring in the restriction of $\chi$. The formulation above is then an easy consequence of the parametrization of irreducible characters of $S_n$, and the classic branching rule.

#Literature#

Yoshiaki Ueno, On the Generating Functions of the Young Lattice, J. Algebra 116 (1988) 261--270.

This gives a generating polynomial for the partitions contained in a given partition \lambda in terms of a determinant involving Gaussian coefficients. It's a beautiful result, but it did not give me any insight into my problem.

The Problem

This strikes me as a very natural problem which should have been asked (and solved?) already.

For each positive integer k, find a nice expression for the following generating function in the variable x: $$ \sum_{\lambda/\mu} x^{|\lambda|}. $$

Here $\lambda$ ranges over all partitions and $\mu$ over those partitions contained in $\lambda$ for which the skew Young diagram $\lambda/\mu$ has k nodes, i.e., each partition of n is weighted by the number of partitions of $n-k$ it contains.

Examples: k=1, the function is $\frac{x}{(1-x)}P(x) $, where $P(x)=\prod_{i=1}^\infty(1-x^i)^{-1}$ is the partition generating function. So in this case I'm just enumerating partitions by the number of removable nodes. The formula is equivalent to the well-known fact that every partition has one more addable than removable node.

I've computed the cases k=2,3,4 also (k=4 was painful - I broke it into 14 possible types of skew-diagrams). For k=2 the generating function is $\frac{ x^2(2-x)}{(1-x)(1-x^2)}P(x).$

It seems plausible that there is a polynomial $F_k(x)$ of degree at most $k(k-1)/2$ (with leading coefficient $\pm 1$) so that the power series is $$ \frac{ x^k F_k(x)}{(1-x)(1-x^2)..(1-x^k)}P(x). $$

If $F_k(x)$ exists, it's easy to see that it must have lowest terms $p_k+p_{k+1}x+2(p_{k+2}-1)x^2+...$, where $p_n$=number of partitions of n, after which the terms depend on congruences for k. This suggests that its complicated. Perhaps there is no nice expression for $F_k(x)$. Even knowing whether $F_k(x)$ exists is of interest to me. Maybe there is a neater way of expressing the entire generating function?

Motivation

The coefficient of $x^n$ in the generating function is the dimension of the centre of a certain subalgebra of the complex group algebra of the symmetric group of degree n. This is ${\mathbb C}S_n^{S_{n-k}}$, the centralizer of the subgroup $S_{n-k}$ in ${\mathbb C}S_n$. It is easy to see that this has as ${\mathbb C}$-basis the $S_{n-k}$-orbit sums in $S_n$. The centre is indexed by pairs $(\chi,\phi)$, where $\chi$ is an irreducible character of $S_n$, and $\phi$ is an irreducible character of $S_{n-k}$ occuring in the restriction of $\chi$. The formulation above is then an easy consequence of the parametrization of irreducible characters of $S_n$, and the classic branching rule.

Literature

Yoshiaki Ueno, On the Generating Functions of the Young Lattice, J. Algebra 116 (1988) 261--270.

This gives a generating polynomial for the partitions contained in a given partition $\lambda$ in terms of a determinant involving Gaussian coefficients. It's a beautiful result, but it did not give me any insight into my problem.

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John Murray
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What is the Generating Function for Skew Young Diagrams?

#The Problem#

This strikes me as a very natural problem which should have been asked (and solved?) already.

For each positive integer k, find a nice expression for the following generating function in the variable x: $$ \sum_{\lambda/\mu} x^{|\lambda|}. $$

Here \lambda ranges over all partitions and \mu over those partitions contained in \lambda for which the skew Young diagram \lambda/\mu has k nodes i.e. each partition of n is weighted by the number of partitions of n-k it contains.

Examples: k=1, the function is $\frac{x}{(1-x)}P(x) $, where $P(x)=\prod_{i=1}^\infty(1-x^i)^{-1}$ is the partition generating function. So in this case I'm just enumerating partitions by the number of removable nodes. The formula is equivalent to the well-known fact that every partition has one more addable than removable node.

I've computed the cases k=2,3,4 also (k=4 was painful - I broke it into 14 possible types of skew-diagrams). For k=2 the generating function is $\frac{ x^2(2-x)}{(1-x)(1-x^2)}P(x).$

It seems plausible that there is a polynomial F_k(x) of degree at most k(k-1)/2 (with leading coefficient \pm 1) so that the power series is $$ \frac{ x^k F_k(x)}{(1-x)(1-x^2)..(1-x^k)}P(x). $$

If $F_k(x)$ exists, it's easy to see that it must have lowest terms $p_k+p_{k+1}x+2(p_{k+2}-1)x^2+...$, where p_n=number of partitions of n, after which the terms depend on congruences for k. This suggests that its complicated. Perhaps there is no nice expression for $F_k(x)$. Even knowing whether $F_k(x)$ exists is of interest to me. Maybe there is a neater way of expressing the entire generating function?

#Motivation#

The coefficient of $x^n$ in the generating function is the dimension of the centre of a certain subalgebra of the complex group algebra of the symmetric group of degree n. This is ${\mathbb C}S_n^{S_{n-k}}$, the centralizer of the subgroup $S_{n-k}$ in ${\mathbb C}S_n$. It is easy to see that this has as ${\mathbb C}$-basis the $S_{n-k}$-orbit sums in $S_n$. The centre is indexed by pairs $(\chi,\phi)$, where $\chi$ is an irreducible character of $S_n$, and $\phi$ is an irreducible character of $S_{n-k}$ occuring in the restriction of $\chi$. The formulation above is then an easy consequence of the parametrization of irreducible characters of $S_n$, and the classic branching rule.

#Literature#

Yoshiaki Ueno, On the Generating Functions of the Young Lattice, J. Algebra 116 (1988) 261--270.

This gives a generating polynomial for the partitions contained in a given partition \lambda in terms of a determinant involving Gaussian coefficients. It's a beautiful result, but it did not give me any insight into my problem.