12
$\begingroup$

First of all my apologies if this question is well known or obvious: this is not in my area of research.

Let $T(x)=\sum_{n=0}^\infty t_nx^n$, where $t_n$ is the number of partitions $\lambda$ of $n$ into $m$ parts where, if $(m-i)\times (m-i)$ is the size of the Durfee square of $\lambda$, then the partition on the right of this Durfee square has at most $m-i-1$ parts.

As usual denote by $(x)_j:=(1-x)(1-x^2)\cdots (1-x^j)$.

If I am not mistaken $$ T(x)= \sum_{i=0}^{m-1} \frac{(x)_{m-1} x^{(m-i)^2+i}} {(x)_{m-i-1}^2(x)_i}.$$ In fact, given a positive integer $N$, the coefficient of degree $N$ in $$\frac{(x)_{m-1}x^i}{(x)_{m-i-1}(x)_{i}}$$ is the number of partitions of $N$ into exactly $i$ parts each of size at most $m-i$ (and this accounts for the partition at the bottom of the Durfee square because we are counting partitions with exactly $m$ parts). Similarly, the coefficient of degree $N$ of $$\frac{1}{(x)_{m-i-1}}$$ is the number of partitions of $N$ into at most $m-i-1$ parts (and this accounts for the partition on the right of the Durfee square because we are requiring that this piece has at most $m-i-1$ parts).

Computational evidence shows that $$T=x^m\sum_{i=1}^m\frac{(-1)^{i-1}x^{i(i-1)/2}}{(x)_{m-i}}$$ but I have no idea why this is true.

$\endgroup$
  • $\begingroup$ yes, I know that. I am not counting all $(m-i)\times (m-i)$ partitions with given Durfee square, but only the partitions having on the right of their Durfee square at most $m-i-1$ parts. $\endgroup$ – Pablo Spiga Dec 28 '15 at 5:20
  • $\begingroup$ how do you get number of partitions of N into exactly i parts each of size at most m−i'? i get $(x(1-x^{m-i})/(1-x))^i$. $\endgroup$ – JonMark Perry Jan 6 '16 at 2:04
  • $\begingroup$ I do not know how you get your formula, but it is incorrect: as you can check by direct inspection using small values of $m$ and $i$. I give you a reference (this will be clearer) for mine. $\endgroup$ – Pablo Spiga Jan 6 '16 at 16:31
  • $\begingroup$ I do not know how you get your formula: it is incorrect, check small values of $m$ and $i$. Hardy and Wright, An introduction to the Theory of Numbers, formula 19.3.2: you see that $1/(x)_m$ is the generating function for the number of partitions in at most $m$ parts (and also for the number of partitions having parts of size at most $m$). Now looking at: Andrews, The theory of Partition, Theorem 3.1, the generating function for the number of partitions in at most a parts of size at most b is $(x)_{a+b}/((x)_a(x)_b)$. From this you can deduce the one you are asking for. $\endgroup$ – Pablo Spiga Jan 6 '16 at 16:54
  • 2
    $\begingroup$ It seems to me that the coefficient of $x^N$ in the latter generating formula also counts the partitions of $N$ in exactly $m$ parts where the smaller lacking part is even, that is of the form $N=1+1+3+4+7\dots$ or $N=1+2+2+2+3+4+5+7\dots$. $\endgroup$ – Pietro Majer Jan 10 '16 at 8:03
6
+50
$\begingroup$

Lemma. Fix $n$ and $m$. Consider pairs of partitions $(\lambda,\mu)$ such that $\lambda$ has $m$ parts and $|\lambda|+|\mu|=n$. Let $A$ be the number of pairs for which $\max(\lambda)>\max(\mu)$ (where $\max(\emptyset)=-\infty$). Let $B$ be number of pairs for which $\lambda$ satisfies this condition with Durfee square, which we may rephrase as '$\lambda_i=i$ for some $i$, where $\lambda=(\lambda_1\geqslant \lambda_2\geqslant \dots)$'. Then $A=B$.

We construct a bijection. Take partitions $\lambda=(\lambda_1\geqslant \dots \geqslant \lambda_m>0)$ and $\mu=(\mu_1\geqslant\dots)$ such that $\lambda_1>\mu_1$ (or $\mu$ is empty). If $\lambda_i=i$ for some index $i$, take the same pair of partitions. If not, then there exists $i$ such that $\lambda_i\geqslant i+1$, $\lambda_{i+1}\leqslant i$. Take partitions $(\lambda_2-1,\dots,\lambda_{i}-1,i,\lambda_{i+1},\dots,\lambda_m)$ and $(\lambda_1-1,\mu_1,\dots)$.

Now reduction.

At first, let's count all the other partitions onto $m$ parts. These are partitions for which $k$'s largest part never equals $k$, $k=1,2,\dots$. We want to prove that their generating function equals $$ x^m\sum_{i=0}^m\frac{(-1)^{i}x^{i(i-1)/2}}{(x)_{m-i}} $$ (I have used that total number of partitions onto $m$ parts have generating function $x^m/(x)_m$, that is seen by duality.) Call such partitions interesting.

Multiply by $t^m$ and sum up by $m$. We get a double generating function $f(t,x)=\sum t^{{\rm parts}(\lambda)} x^{|\lambda|}$, summation is taken over all interesting partitions $\lambda$. We have to prove that $$ f(t,x)=\sum_{m\geqslant i\geqslant 0} (tx)^m\frac{(-1)^{i}x^{i(i-1)/2}}{(x)_{m-i}}= \left(\sum_{k\geqslant 0} \frac{(tx)^k}{(x)_k}\right)\left(\sum_{i\geqslant 0} (-1)^i(tx)^i x^{i(i-1)/2}\right). $$ As for the first multiple, it is a double generating function for $t^{\max(\lambda)} x^{|\lambda|}$ taken by all partitions $\lambda$, hence by duality it is the same thing as a double generating function for $t^{{\rm parts}(\lambda)} x^{|\lambda|}$, which is $\prod_{n\geqslant 1} (1-tx^n)^{-1}$. As for the second multiple, it is a part of Jacobi triple product $$ \prod_{n\geqslant 1} (1-tx^{n})(1-t^{-1}x^{n-1})(1-x^n)=\sum_{i=-\infty}^{\infty} (-1)^it^i x^{i(i+1)/2},$$ corresponding to non-negative $i$. Denote by $H(t,x)$ the part of double product $\prod_{n\geqslant 1} (1-tx^{n})(1-t^{-1}x^{n-1})$ with only non-negative powers of $t$. We have to prove that $$ \prod_{n\geqslant 1} (1-tx^n)\cdot H(t,x)=f(t,x)\cdot \prod (1-x^n)^{-1}. $$ Coefficient of $t^mx^n$ in RHS equals the number of pairs of partitions $(\lambda,\mu)$ for which $|\lambda|+|\mu|=n$, $\lambda$ has $m$ parts and is interesting and $\mu$ is arbitrary. By the conjecture it is the same as number of pairs of partitions $(\lambda,\mu)$ for which $|\lambda|+|\mu|=n$, $\lambda$ has $m$ parts and $\mu$ has a part which is not less then $\max(\lambda)$.

Let's obtain the same in the LHS. If $H(t,x)$ were not a part, but the whole product, then a lot would cancel when we multiple $\prod(1-tx^n)^{-1}$ and $\prod (1-tx^n)$. But something still cancels. Namely, we look at our product $$ \prod_{n\geqslant 1} (1+tx^n+t^2x^{2n}+\dots) \prod_{n\geqslant 1} (1-tx^n) \prod_{n\geqslant 1} (1-t^{-1}x^{n-1}) $$ and see what we may take from each bracket. We are conditioned to take at least as many $t$'s from the second product than $t^{-1}$'s from the third. Consider partial involution on the set of our choices: denote by $N$ the maximal index $n$ for which we either take $-tx^{n}$ from the second product or take $t^ax^{na}$ for some $a\geqslant 1$ from the first product. We could take $1$ from the corresponding bracket in the second product and $t^{a+1}x^{n(a+1)}$ from the corresponding bracket in the first product instead. This is a sign-changing involution on the set of choices, but the set of admissible choices is not quite invariant. Namely, if total number of $t$'s from the second product equals total number of $t^{-1}$'s in the third product, we are forbidden to replace $-tx^N$ to 1. This is what remains after removing all pairs of choices formed by involution. To be more precise, what we should choose are some $k$ positive integers $0<a_1<a_2<\dots <a_k$ (second product), some $k$ non-negative integers $0\leqslant b_1<b_2<\dots b_k$ (third product) and some partition with maximal part at most $a_k$ (first product).

As for two choices from the second and from the third products, it is equivalent to the choice of a partition with maximal part $a_k$: for such a partition let $k$ be the size of Durfee square, $a_1,\dots,a_k$ correspond to the part of Young diagram on the one side of its diagonal and $b_1,\dots,b_k$ to the other side.

So, we choose a partition with maximal part $a_k$ (this is $\mu$) and another partition onto $m$ parts with maximal part at most $a_k$. This is $\lambda$. Also, exponent $m$ of $t$ is the number of parts in $\lambda$, and exponent of $x$ is $|\lambda|+|\mu|$.

$\endgroup$
  • $\begingroup$ Thanks a lot for this. It looks very promising: just for the record, I don't have an answer yet and I am still interested very much. I do follow your argument, but remember that this is NOT my area and hence I struggle to understand part of the things you say: so please be patient and give complete comments. One more quick comment: you still don't have a proof right? At the very end when you use the Jacobi triple product you have a summation from -\infty to +\infty and in a sense you need a "geometric" interpretation of the non-negative part. Is that right? $\endgroup$ – Pablo Spiga Jan 9 '16 at 12:57
  • $\begingroup$ I really hope that now everything works. $\endgroup$ – Fedor Petrov Jan 9 '16 at 23:49
  • 2
    $\begingroup$ I think that it does work! Good job and thanks. Also nice comment from Pietro Majer...looks like that you guys have another fish to catch ;-) $\endgroup$ – Pablo Spiga Jan 10 '16 at 15:57

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.