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Sep 14, 2020 at 21:47 comment added Grigory M @RichardStanley We also have some formulas for $\pi_{nm}=0$ in arXiv:1512.08779. In that case an answer is given by Brion's theorem. What really puzzles me is that the formula for $R(q)-1$ also looks like some version of Brion's theorem (each triple (x,y,z) in the answer below should correspond to a vertex etc)…
Sep 14, 2020 at 18:13 comment added Richard Stanley Regard a plane partition as a two-dimensional array $(\pi_{ij})_{i,j\geq 1}$ in the usual way. The present question asks to count plane partition $n$ with $\pi_{22}\leq 2$. If $v(n)$ denotes the number of plane partitions of $n$ with $\pi_{22}=0$, then by EC1, Prop. 2.5.1, $\sum_{n\geq 0} v(n)q^n = \left( \sum_{m\geq 0} (-1)^mq^{{m+1\choose 2}}\right)\prod_{i\geq 1}(1-q^i)^{-2}$. This suggests counting plane partitions of $n$ with $\pi_{22}\leq k$ for $k\geq 2$ (and similar problems).
Sep 13, 2020 at 20:11 history edited Sam Hopkins
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Apr 13, 2017 at 12:19 history edited CommunityBot
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Jan 15, 2015 at 18:47 vote accept Grigory M
Jan 13, 2015 at 13:12 answer added Hjalmar Rosengren timeline score: 8
Jan 9, 2015 at 10:43 history edited Grigory M CC BY-SA 3.0
typo in "context 2" fixed
Jan 9, 2015 at 10:39 comment added Grigory M @Hjalmar Yes, of course, thank you (I'll fix the typo)
Jan 9, 2015 at 8:11 comment added Hjalmar Rosengren Am I correct that you mean $R(q)=1+\sum_{n=0}^\infty q^{n+1}\sum_{i+j+k=n}\genfrac[]{0pt}{}{i+j}i\genfrac[]{0pt}{}{j+k}k\genfrac[]{0pt}{}{k+i}k$
Jan 8, 2015 at 15:10 history asked Grigory M CC BY-SA 3.0