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Working on suitable closed form for A329369, I discovered very useful coefficients, which have the following recurrence relation:

$$T(0,1)=T(0,2)=1$$ $$T(n,1)=1, n>0$$ $$T(0,k)=0, k>2$$ $$T(2n+1,k)=kT(n,k)+T(n,k-1)$$ $$T(2n,k)=kT(n,k)+T(n,k-1)-\frac{1}{k-1}(T(2n,k-1)+T(n,k-1))$$

It looks like $T(n,k)$ is always integer.

Also I recognize Stirling numbers of the second kind in $T(2^n-1,k)$.

Is there any way to get a closed form from this recurrence relation?

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  • $\begingroup$ @PeterTaylor, thank you for comment! Done (also sign was fixed). Please try to compute coefficients again. $\endgroup$
    – Notamathematician
    Commented Nov 22, 2022 at 11:17
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    $\begingroup$ I am not sure whether I got your recurrence right, but if I did, the generating function $T_3(x)$ of $T(n, 3)$ satisfies an equation of the form $p(x)T_3(x^2) + q(x) T_3(x) = r(x)$ for polynomials $p, q, r$ of degree $5$, $4$ and $3$ respectively. $\endgroup$
    – Martin Rubey
    Commented Nov 22, 2022 at 12:42
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    $\begingroup$ The number of non-zero $T(n, k)$ for given $n$ is oeis.org/A063787, right? $\endgroup$
    – Martin Rubey
    Commented Nov 22, 2022 at 12:46
  • $\begingroup$ @MartinRubey, thank you for comment! Number of non-zero $T(n,k)$ for a given $n$ is $\operatorname{wt}(n)+2$ where $\operatorname{wt}(n)$ is the binary weight of $n$ (OEIS A000120). Also I will try to check your first comment later. $\endgroup$
    – Notamathematician
    Commented Nov 22, 2022 at 12:58
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    $\begingroup$ You have certainly looked at the scatterplot oeis.org/A329369/graph, which shows nice, quite straightforward self-similar patterns. :) $\endgroup$
    – Wolfgang
    Commented Dec 1, 2022 at 14:41

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