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  "body": "\n## Statement\n\nLet $s, t \\in \\C$ with $\\abs{s} \u003c 1$ and $t \\ne 0$.\nThen\n$$\\prod_{n=1}^{\\infty} \\left(1 - s^n\\right) \\left(1 - s^n t\\right) \\left(1 - s^{n-1} t^{-1}\\right) \\left(1 - s^{2n-1} t^2\\right) \\left(1 - s^{2n-1} t^{-2}\\right) = \\sum_{n=-\\infty}^{\\infty} s^{(3n^2 + n)/2} \\left( t^{3n} - t^{-3n-1} \\right).$$\n\n## Remarks\n\nBoth sides are holomorphic in $t$ on $\\C \\setminus \\set{0}$, and the [[infinite product|RZ0Z2BTMT4]] converges because $\\abs{s} \u003c 1$.\nProof idea: as for the [[Jacobi triple product|JFP2DGMNXT]], one checks that both sides $F$ satisfy $F(st) = s^{-2} t^{-3} F(t)$.\nThis relation links the Laurent coefficients of index $k$ and $k + 3$, so it fixes them from three of them, and these are compared directly.\nExample: at $t = 1$ the factor $1 - s^0 t^{-1}$ of index $n = 1$ vanishes, so the left side is $0$.\nOn the right the terms $s^{(3n^2+n)/2}(1 - 1)$ also vanish, so both sides agree.\nThe exponents $(3n^2 + n)/2$ are the generalized pentagonal numbers, as in [[Euler's pentagonal number theorem|R587SR9GNV]].\n\n## Explanations\n\n### L1 human\n\nLike the triple product identity, this identity turns an infinite product into a sum, but now each step of the product has five factors.\nThe sum on the right is very sparse: only exponents of the form $(3n^2 + n)/2$ appear.\nIt is used to prove identities for partitions and for theta functions.\n\n### L2 human\n\nThe identity has been rediscovered many times, and there are many equivalent forms obtained by changes of variables.\nIt can be derived from the [[Jacobi triple product|JFP2DGMNXT]], and it is a source of identities for theta functions and for the [[Dedekind eta function|MX3CVYMNDV]].\n",
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