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.



Generating function for certain partitions (with a restriction on the Durfee square)
**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 $|\lambda|+|\mu|$.


