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I'm trying to come up with the formula describing the number of paths on hexagonal lattice of length $2n$ that start at the origin $O$ and go back to $O$ but doing so for the first time at step $2n$ (i.e. first return to the origin).
Suppose I already have a formula for the number of such paths but without the condition of returning for the first time at step $2n$. Is there a way to go through this formula to the one I'm looking for? For example, through generating functions, etc?

A. G
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  • Assume that two sequences satisfy $f(n)=g(n)f(0)+g(n-1)f(1)+\dots+g(1)f(n-1)$ for all $n \ge 1$, with the initial condition $f(0)=g(0)=1$. Then it is not hard to see that their generated functions satisfy $F(x)-1=F(x)G(x)-F(x)$, i.e., $G(x) = 2-F^{-1}(x)$. – Arsenii Sagdeev Jul 12 '22 at 19:55
  • @ArseniiSagdeev, but is there actually a way to sum the series $F(x)$ in order to find $G(x)$ (the formula for $g(n)$ is quite complex)? And if so, would it help to find an analytical formula for $f(n)$? – A. G Jul 12 '22 at 20:22
  • A similar question was recently asked at https://math.stackexchange.com/questions/4489077/probability-that-random-walk-returns-to-starting-vertex-in-at-most-20-moves – Henry Jul 13 '22 at 00:17
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    If there are $f(n)$ paths of length $n$ from O to O, then I think the number of paths first coming back to O at $n$ is $f(n) - \sum_{i+j=n}f(i)f(j) + \sum_{i+j+k=n}f(i)f(j)f(k) - \dots$ –  Jul 13 '22 at 01:22
  • @MattF. Oh, I wasn't aware about such a nice expression of the inverse series! Tried to google it and found several other expressions (https://mathoverflow.net/questions/53384 https://math.stackexchange.com/questions/1264615 https://math.stackexchange.com/questions/710252), but from my point of view all of them looks kind of sophisticated, though of course they are all equivalent... – Arsenii Sagdeev Jul 13 '22 at 07:23
  • @A.G The desired sequence $g(n)$ is https://oeis.org/A094060 Looks like no nice expression for $G(x)$ is known (since there is no nice expression for $F(x)$ from https://oeis.org/A002898 as well) – Arsenii Sagdeev Jul 13 '22 at 07:27
  • @A.G However, I'd expect that the connection between $F(x)$ and $G(x)$ allows one to deduce the asymptotic of $g(n)$ from the fact that $f(n) \sim c3^n/n$. For instance, $F(1/3-0)=+\infty$ and thus $G(1/3)=2 < \infty$. Hence, $g(n) < (3-\varepsilon+o(1))^n$ for some $\varepsilon>0$ – Arsenii Sagdeev Jul 13 '22 at 07:34
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    @ArseniiSagdeev, I guess the links for the sequences of numbers you provided describe slightly different walks, i.e. the walks that move on a tile space, not along the edges... Probably, I should have written the question more clearly, but I'm looking for paths that go along the edges (between vertices) on a hexagonal lattice. – A. G Jul 13 '22 at 09:06
  • @A.G Have you calculated some initial values of your sequence? What is the first moment when it is not coincide with https://oeis.org/A094060? – Arsenii Sagdeev Jul 13 '22 at 11:15
  • Let $f(n)$ give the sequence A002893 referenced in the question; then by considering the first return of each path in $f(n)$, the desired sequence satisfies $g(n) = f(n) - \sum_{m=1}^{n-1} g(m) f(n-m)$ for $n > 0$. If $F(x)$ is the ogf for A002893 then the ogf of $g$ is $G(x) = 1 - \frac{1}{F(x)}$ – Peter Taylor Jul 13 '22 at 11:59
  • @ArseniiSagdeev, first of all, it is not visible there that all odd terms are 0, and, secondly, after 6 the next term should be 30. So I guess, the sequence just describes a slightly different walk (which is not through the vertices). – A. G Jul 13 '22 at 21:47
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    No, it's over a different graph. There are two interpretations of the term "hexagonal lattice": one is a lattice in the vector algebra sense and gives a graph where each vertex has degree six; the other gives the honeycomb graph where each vertex has degree 3. The desired walks in the latter aren't in OEIS, but this Sage code calculates them. – Peter Taylor Jul 14 '22 at 13:58
  • @PeterTaylor, thanks for mentioning, that's exactly what I meant moving via edges or tiles, probably quite ambiguously. And thank you for the link. – A. G Jul 18 '22 at 09:28

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