Hypergeometric closed form for the excursions of the second Kreweras-like model

Consider the model

S={2ˉ0,1ˉ1ˉ,02ˉ,11}.\mathcal S=\{\bar 2 0,\bar 1\bar 1,0\bar 2,11\}.

Let Q(0,0;t1/2)Q(0,0;t^{1/2}) be its excursion generating function after extracting the even subsequence. The hypergeometric closed-form conjecture.

Q(0,0;t1/2)=13t112t6t(2F1(16,13;1;108t(1+4t)2(112t)3)+2F1(16,23;1;108t(1+4t)2(112t)3)).Q(0,0;t^{1/2})=\frac{1}{3t}-\frac{\sqrt{1-12t}}{6t}\left({}_2F_1\left(\frac16,\frac13;1;\frac{-108t(1+4t)^2}{(1-12t)^3}\right)+{}_2F_1\left(-\frac16,\frac23;1;\frac{-108t(1+4t)^2}{(1-12t)^3}\right)\right).

The formula is inferred from the guessed differential operator and its hypergeometric solutions, so it remains conjectural rather than proved in the source.

Sources & referencesView supporting material

Primary source

Alin Bostan, Mireille Bousquet-Mélou and Stephen Melczer, “Counting walks with large steps in an orthant”, arXiv:1806.00968 (2018).

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