A plane partition is a subset of $\mathbb Z\_{\geqslant0}^3$ s.t. if it contains $(i+1,j,k)$ or $(i,j+1,k)$ or $(i,j,k+1)$ it also contains $(i,j,k)$.



> 
> What is the generating function $R(q)$ of (volumes of) plane partitions *not containing the cell $(1,1,1)$*?
> 
> 
> 


Since a plane partition containing $(1,1,1)$ contains a $2\times2\times2$ cube, mod $q^8$ this g.f. coincides with MacMahon’s
$$
M(q)=\frac1{\prod(1-q^i)^i}=1+q+3q^2+6q^3+13q^4+24q^5+48q^6+86q^7+160q^8+282q^9+500q^{10}+\ldots
$$
and computing a couple more terms
$$
R(q)=1+q+3q^2+6q^3+13q^4+24q^5+48q^6+86q^7+159q^8+279q^9+488q^{10}+\ldots
$$
[OEIS doesn’t know](http://oeis.org/search?q=1%2C+3%2C+6%2C+13%2C+24%2C+48%2C+86%2C+159%2C+279%2C488) this sequence, but in the form
$$
R(q)=1+\frac{q-3q^3+6q^6-10q^{10}+\ldots}{[(1-q)(1-q^2)(1-q^3)\ldots]^3}.
$$
a pattern is evident.


So actually the question is



> 
> How to prove that $R(q)=1+\frac1{(q)\_\infty^3}\sum\limits\_i(-1)^i\frac{i(i-1)}2q^{\tfrac{i(i-1)}2}$?
> 
> 
> 


AFAIK the statement can be found somewhere in the literature on representation theory of quantum toroidal something. But surely there is a well-known combinatorial (in the broad sense, not necessarily bijective) proof?


**Context 1.** G.f. for plane partitions not containing $(1,1,0)$ is $1/(q)\_\infty^2\cdot\sum(-1)^iq^{\frac{i(i+1)}2}$. See the subsection about «V-shaped partitions» in Stanley's Enumerative Combinatorics for a (simple) combinatorial proof. Perhaps there is something like that proof in «(1,1,1)-case».


**Context 2.** G.f. for plane partitions not containing $(1,1,1)$ but intersecting axes at $(i,0,0)$, $(0,j,0)$ and $(0,0,k)$ resp. is $\genfrac[]{0pt}{}{i+j}i\genfrac[]{0pt}{}{j+k}j\genfrac[]{0pt}{}{k+i}kq^{i+j+k+1}$. So maybe it’s possible to compute $R(q)$ as $1+\sum\_n q^{n+1}\sum\_{i+j+k=n}\genfrac[]{0pt}{}{i+j}i\genfrac[]{0pt}{}{j+k}j\genfrac[]{0pt}{}{k+j}k$. And for the internal sum I know the answer at least at $q=1$: <https://math.stackexchange.com/q/177209/>



Plane partitions not containing (1,1,1)
It is indeed possible to get the result from your "Context 2".


Correcting typos, we have
$$R(q)=1+\sum\_{i,j,k=0}^\infty\frac{(q)\_{i+j}(q)\_{i+k}(q)\_{j+k}}{(q)\_i^2(q)\_j^2(q)\_k^2}\,q^{i+j+k+1} $$
(where $(q)\_k=\prod\_{j=1}^k(1-q^j)$). By the $q$-binomial theorem, 
$$\frac{(q)\_{i+j}}{(q)\_i(q)\_j}=\sum\_{x=0}^i\frac{1}{(q)\_x(q)\_{i-x}}(-1)^xq^{\binom x2+(j+1)x}. $$
Using this, and the expansions obtained by cyclically permuting $(i,j,k)$,
we get
\begin{align\*}R(q)-1&=\sum\_{i,j,k=0}^\infty\sum\_{x,y,z=0}^{i,j,k}\frac{(-1)^{x+y+z}q^{\binom x2+\binom y2+\binom z3+(j+1)x+(k+1)y+(i+1)z+i+j+k+1}}{(q)\_x(q)\_{i-x}(q)\_y(q)\_{j-y}(q)\_z(q)\_{k-z}}\\ 
&=\sum\_{x,y,z=0}^\infty \frac{(-1)^{x+y+z}q^{\binom x2+\binom y2+\binom z2+(y+2)x+(z+2)y+(x+2)z+1}}{(q)\_x(q)\_y(q)\_z}\\
&\quad\times\sum\_{i=0}^\infty\frac{q^{(1+z)i}}{(q)\_i}\sum\_{j=0}^\infty\frac{q^{(1+x)j}}{(q)\_j}\sum\_{k=0}^\infty\frac{q^{(1+y)k}}{(q)\_k}\\
&=\frac 1{(q)\_\infty^3}\sum\_{x,y,z=0}^\infty {(-1)^{x+y+z}q^{\binom x2+\binom y2+\binom z2+(y+2)x+(z+2)y+(x+2)z+1}} ,\end{align\*}
where we first replaced
 $(i,j,k)$ by $(i+x,j+y,k+z)$ and then computed the inner sums using again a special case of the $q$-binomial theorem.
We now observe that
$$\binom x2+\binom y2+\binom z2+(y+2)x+(z+2)y+(x+2)z+1=\binom{x+y+z+2}2. $$
Since there are $\binom n2$ solutions $(x,y,z)$ to $x+y+z+2=n$, we get indeed
$$R(q)=1+\frac 1{(q)\_\infty^3}\sum\_{n=0}^\infty(-1)^n\binom n2q^{\binom n2}. $$


