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Consider the triple-indexed sequence of integers defined by \begin{align} \label{coefficientsV} \nonumber f(\alpha,\beta,\gamma) &:=(2\alpha+8\beta+12\gamma-1)\cdot f(\alpha-1,\beta,\gamma) - 2(\alpha+1)\cdot f(\alpha+1,\beta-1,\gamma) \\ &\qquad \ \ \ \ \ - {\color{red}8}(\beta+1)\cdot f(\alpha,\beta+1,\gamma-1)- 12(\gamma+1)\cdot f(\alpha,\beta-2,\gamma+1), \end{align} where $\alpha, \beta, \gamma\geq0$. To list them as triangular arrays, choose $(\alpha,\beta,\gamma)$ such that $\alpha+2\beta+3\gamma=t$ for a given $t\geq0$.

To seed the recursion, we let $f(0, 0, 0):=1,$ and we let $f(\alpha,\beta,\gamma):=0$ if any of the arguments are negative. Here we list the ``first few'' values: \begin{align*} f(1,0,0)=1, \ f(0,1,0)=-2,\ f(0,0,1)=16, \ f(1,1,0)=-30, \ f(1,0,1)=448,\dots. \end{align*}

QUESTION. Is it possible to give some closed formula or a generating function representation of the above sequence?

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  • $\begingroup$ What do you mean by "set the condition $\alpha+2\beta+3\gamma=t$ for a given $t\geq0$"? $\endgroup$ Commented Nov 17, 2024 at 19:23
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    $\begingroup$ But $\alpha+2\beta+3\gamma$ always is equal to something. Another thing is that LHS of your relation has only triples for which this guy equals $t-1$, thus it makes reasonable to consider the sequence of generating polynomials over such triangles. $\endgroup$ Commented Nov 17, 2024 at 20:24
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    $\begingroup$ I get $f(0,0,1)=6$ and $f(1,0,1)=168$, please check. $\endgroup$ Commented Nov 18, 2024 at 19:56
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    $\begingroup$ Seems to be $\ldots -8(\beta+1) \ldots$ instead of $\ldots -3(\beta+1)\ldots$. $\endgroup$ Commented Nov 18, 2024 at 20:04
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    $\begingroup$ You are absolutely right, we need $-8(\beta+1)$. $\endgroup$ Commented Nov 19, 2024 at 0:49

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I assume a typo as suggested in my comments, such that \begin{align} f(\alpha,\beta,\gamma) &=(2\alpha+8\beta+12\gamma-1) \, f(\alpha-1,\beta,\gamma) \\ {}&- 2(\alpha+1) \, f(\alpha+1,\beta-1,\gamma) \\ {}&- 8(\beta+1) \, f(\alpha,\beta+1,\gamma-1) \\ {}&- 12(\gamma+1) \, f(\alpha,\beta-2,\gamma+1). \end{align} In direction $\alpha$ the recursion can be summed. Noting that \begin{align} \frac{(\alpha+1)f(\alpha+1,\beta,\gamma)}{f(\alpha,\beta,\gamma)} \end{align} is a slow growing integer sequence for all $\alpha,\beta,\gamma\geq0$, Mathematica's FindSequenceFunction[] can be guided to find the closed form \begin{align}\label{eq:1}\tag{1} \frac{(\alpha+1)f(\alpha+1,\beta,\gamma)}{f(\alpha,\beta,\gamma)} =(1 + \alpha + 2\beta + 3\gamma) \, (1 + 2\alpha + 4\beta + 6\gamma) \, , \end{align} such that \begin{align}\label{eq:2a}\tag{2a} \frac{f(\alpha,\beta,\gamma)}{f(0,\beta,\gamma)} &= \prod_{\alpha'=0}^{\alpha-1} \frac{(1 + \alpha' + 2\beta + 3\gamma) \, (1 + 2\alpha' + 4\beta + 6\gamma)}{1+\alpha'} \\ \label{eq:2b}\tag{2b} &= \frac{\Gamma(1 + 2\alpha + 4\beta + 6\gamma)} {2^\alpha \, \Gamma(1+\alpha)\,\Gamma(1 + 4\beta + 6\gamma)}\\ \label{eq:2c}\tag{2c} &= \frac{(1 + 4\beta + 6\gamma)_{2\alpha}} {2^\alpha \, \alpha!}\,, \end{align} with Pochhammer symbol $(a)_n$.

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  • $\begingroup$ If one substitutes this in the original recurrence and denotes $g(b,c)=f(0,b,c)$, we get, I believe, $g(b,c)=-(4b+6c-2)(4b+6c-3)g(b-1,c)-8(b+1)g(b+1,c-1)-12(c+1)g(b-2,c+1)$ - perhaps Mathematica can do miracles for this recurrence once again? $\endgroup$ Commented Nov 19, 2024 at 5:36
  • $\begingroup$ Yes, that's true, but MMA can not solve this recursion. $\endgroup$ Commented Nov 19, 2024 at 5:39
  • $\begingroup$ Can you solve the original recursion in the direction of $\beta$ and $\gamma$ also? Perhaps knowing all of them would help somehow. Or is $\alpha$ somehow special? $\endgroup$ Commented Nov 20, 2024 at 15:33
  • $\begingroup$ It does seem like $\alpha$ is more special among the others, I think. $\endgroup$ Commented Nov 20, 2024 at 17:42

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