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Notamathematician
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On a number of compositions of $n$ into positive triangular numbers
@AlexanderBurstein, thank you for comment! Why did you decide that? Use the program to make sure that it is a polynomial (but we can simply say that even at the first step we multiply one of the terms by a variable to a positive degree, so that in any case it turns out to be a polynomial).
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On a number of compositions of $n$ into positive triangular numbers
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On a number of compositions of $n$ into positive triangular numbers
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On a number of compositions of $n$ into positive triangular numbers
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On a continued fraction and vector $\nu$ of length $n$
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On a continued fraction and vector $\nu$ of length $n$
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Simple algorithm for A107670
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Simple algorithm for A107670
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Simple algorithm for A107670
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Simple algorithm for A107670
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Intersecting algorithm for A065601
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Intersecting algorithm for A065601
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Elegant algorithm for A140717
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Elegant algorithm for A140717
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Elegant algorithm for A140717
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Elegant algorithm for A140717
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On a $\sum\limits_{n=0}^{\infty}c_n x^n=\sum\limits_{n=0}^{\infty}a(n)x^n\prod\limits_{k=1}^{n+1}(1-f(k)x^k)$ (slightly different question)
@FedorPetrov, unfortunately not. Try it in program.
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On a $\sum\limits_{n=0}^{\infty}c_n x^n=\sum\limits_{n=0}^{\infty}a(n)x^n\prod\limits_{k=1}^{n+1}(1-f(k)x^k)$ (slightly different question)
@FedorPetrov, inspired by your generalizations, I want to ask about the case $\nu_j = \nu_j + f(i)\nu_{j-1} + g(i)\nu_{j-2}$ for $j\geqslant i + 2$. Is it possible to change something in RHS to still receive $c_n$ from $a(n)$?
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On a $\sum\limits_{n=0}^{\infty}c_n x^n=\sum\limits_{n=0}^{\infty}a(n)x^n\prod\limits_{k=1}^{n+1}(1-f(k)x^k)$ (slightly different question)
@FedorPetrov, nice observation! Do you have a proof?
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On a $\sum\limits_{n=0}^{\infty}c_n x^n=\sum\limits_{n=0}^{\infty}a(n)x^n\prod\limits_{k=1}^{n+1}(1-f(k)x^k)$ (slightly different question)
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