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A twist on just unfolded recursive summation formula. Let polynomials in nonnegative integer variables $t_1,t_2,\dots$ be defined by the recurrence: \begin{split} g_0 &= 1, \\ g_k(t_1,t_2,\dots,t_k) &= \sum_{j=0}^{t_1} g_{k-1}(t_2+j,t_3+j,\dots,t_k+j)\qquad (k\geq 1). \end{split} Can $g_k(t_1,t_2,\dots,t_k)$ be explicitly expressed in terms of $t_1,t_2,\dots,t_k$ or via known combinatorial entities?


First few values are: \begin{split} g_0 &= 1,\\ g_1(t_1) &= 1+t_1,\\ g_2(t_1,t_2) &= (1+t_1)(1+t_2) + \frac{t_1(1+t_1)}2,\\ g_3(t_1,t_2,t_3) &= \frac12 \left(t_1^{2}+\left(2 t_2 + t_3 +3\right) t_1 +\left(1+t_2 \right) \left(t_2 +2 t_3 +2\right)\right) \left(1+t_1 \right) \end{split}

In the case of equal arguments, the values $\big(g_k(t,t,\dots,t)\big)_{k\geq 0}$ apparently represent the row sums of the $t$-th power of the matrix $T$ defined in OEIS A097712 and also the $t$-th column of table in OEIS A125860.

In particular, $\big(g_k(1,1,\dots,1)\big)_{k\geq 0}$ form OEIS A016121 defined as the number of tuples $(a_1=1, a_2, ..., a_k)$ satisfying $a_i \leq a_{i+1} \leq 2a_i$ for each $i=1,2,\dots,k-1$.

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    $\begingroup$ It should be $a(n+1)=g_n(1,2,3,4,\ldots)$ in A008934. $\endgroup$
    – BillyJoe
    May 11 at 21:58
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    $\begingroup$ More generally a conjecture is $g_n(k-1, k, k+1, \ldots )=$ column $k$ of A093729, for $k \ge 0$. $\endgroup$
    – BillyJoe
    May 12 at 19:41
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    $\begingroup$ A generalization of this function and the one in the linked question could be: $h_k(t_1,t_2,\dots,t_k):=\sum_{j_1=0}^{t_1} \sum_{j_2=0}^{t_2+s_1(j_1)} \sum_{j_3=0}^{t_2+s_2(j_1,j_2)} \dots \sum_{j_k=0}^{t_k+s_{k-1}(j_1,\ldots,j_{k-1})} 1$. For $g_k$ we have $s_n(j_1,\ldots, j_n) = \sum_{m=1}^n j_m$. Another interesting case is $s_n(j_1,\ldots, j_n) = rj_{n-1}$, for example with $r=2$. $\endgroup$
    – BillyJoe
    May 13 at 8:11
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    $\begingroup$ It seems that B.W.J. Irwin tried some of these nested sums: see A254439, A002449 and this that includes a generalization and several example sequences, A101481, A132616. Maybe one day I will write a program to check if there is some other sequence left at the OEIS that can be expressed in this way. $\endgroup$
    – BillyJoe
    May 13 at 8:20

1 Answer 1

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I originally saw the idea that inspired me to this answer here.

From the statement it follows that

$$ g_k(t_1, t_2, \dots, t_k) - g_k(t_1 - 1, t_2, \dots, t_k) = g_{k-1}(t_2+t_1, t_3+t_1, \dots, t_k+t_1). $$

Generally, one can express shift in the argument of analytic function with the operator exponent

$$ f(x+a) = e^{a\frac{\partial}{\partial x}}f(x), $$

Let's denote $D_k = \frac{\partial}{\partial t_k}$, then the recurrence above rewrites as

$$ (1-e^{-D_1}) g_k = e^{t_1 (D_2+D_3+\dots + D_k)}g_{k-1}, $$

We can't invert $(1-e^{-D_1})$ operator directly because it doesn't distinguish between $0$ and $1$.

However we know that $g_k(0, t_2, \dots, t_k) = g_{k-1}(t_2, \dots, t_k)$, hence we can just solve the equation above for higher degrees of $t_1$, meaning that it's sufficient to find $D_1 g_k$:

$$ D_1 g_k = \frac{D_1}{1-e^{-D_1}}e^{t_1(D_2+\dots+D_k)} g_{k-1} $$

Now let $I_k$ be the integrating operator over $t_k$ such that $[t_k^0]I_k f(t_k, \dots)=0$, we can express $g_k$ as

$$ g_k(t_1, t_2, \dots, t_k) = \left[1 + I_1 \cdot \frac{D_1}{1-e^{-D_1}} \cdot e^{t_1(D_2+\dots+D_k)} \right]g_{k-1}(t_2, \dots, t_k). $$

Substituting it for $g_{k-1}$ and up to the very end, we get

$$ g_k(t_1, t_2, \dots, t_k) = \prod\limits_{j=1}^k \left(1+I_j \cdot \frac{D_j}{1-e^{-D_j}} \cdot e^{t_j(D_{j+1}+\dots+D_k)} \right) \cdot 1, $$

assuming that the product unravels left-to-right as $j$ grows and then is applied to the constant $1$.

Example for $k=1$

The expression in this case is simply

$$ g_1(t_1) = \left(1+I_1 \frac{D_1}{1-e^{-D_1}}\right) \cdot 1 $$

The exponent operator rewrites as

$$ \frac{D_1}{1-e^{-D_1}} = \frac{1}{\sum\limits_{k=1}^\infty \frac{(-D_1)^{k-1}}{k!}} = \frac{1}{1-\frac{D_1}{2}+O(D_1^2)} = 1+\frac{D_1}{2}+O(D_1^2). $$

When it is applied to $1$, it yields $1$, then $I_1 \cdot 1 = t_1$, as the integration makes sure that it is $0$ for $t_1=0$.

Then, adding $1$ the result is, indeed, $g_1 = 1+t_1$.

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  • $\begingroup$ Thank you! This is quite a interesting but somewhat unusual representation, and I do not see right away what kind of benefits it provides. Say, can it be used to speed up evaluation of $g_k$? $\endgroup$ May 24 at 1:59
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    $\begingroup$ I think, computationally you could use it to get the explicit representation of $g_k$ as a multivariate polynomial of $t_1, \dots, t_k$. Generally you can compute the result of applying the polynomial $g(D)$ of $D=\frac{\partial}{\partial x}$ to the polynomial $f(x)$ as $$ g(D) f(x) = [g(x^{-1})\{f(x)\}], $$ where $[\cdot]$ and $\{\cdot\}$ are linear transforms such that $[x^k]=\frac{x^k}{k!}$ and $\{x^k\}=k!x^k$. I wrote a bit more about it some time ago in this article for the univariate case. $\endgroup$ May 24 at 2:11

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