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Please note that this question differs from one of the previous questions of mine.

  • Let $f(n)$ be an arbitrary function with integer values.
  • Let $c_n$ be an arbitrary integer sequence.
  • Let $a(n)$ be an integer sequence such that $$ \sum\limits_{n=0}^{\infty}c_n x^n=\sum\limits_{n=0}^{\infty}a(n)x^n\prod\limits_{k=1}^{n+1}(1-f(k)x^k) $$
  • Start with vector $\nu$ of length $n$ with elements $\nu_i = c_{i-1}$ (that is, $\nu = \{c_0, c_1, \dotsc, c_{n-1}\}$) and for $i$ from $1$ to $\left\lfloor\frac{n}{2}\right\rfloor$ and for $j$ from $2i$ to $n$ apply $\nu_{j} = f(i)\nu_{j-i} + \nu_{j}$.

I conjecture that after the whole transform we have vector $\nu$ with elements $\nu_i = a(i-1)$.

In particular, we have A209405 for $f(n)=c_n=1$.

Here is the PARI/GP program to check it numerically:

f(n) = n
c(n) = n+1
upto1(n) = my(v1); v1 = vector(n, i, c(i-1)); for(i=1, n\2, for(j=2*i, n, v1[j] += f(i)*v1[j-i])); v1
h(n, x) = my(v1); v1 = upto1(n+1); sum(i=0, n, v1[i+1]*x^i*prod(k=1, i+1, 1-f(k)*x^k)) + x*O(x^n)
test1(n) = my(x = 'x); Vec(h(n, x)) == vector(n+1, i, c(i-1))

Is there a way to prove it?

Improve this question
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    $\begingroup$ You may even replace $1-f(k)x^k$ to a more general $1-f(k)x^{t_k}$ for positive integers $t_1,t_2,...$, and on $i$-th step of your algorithm for $\nu$ put $\nu_j=\nu_{j-t_i}f(i)+\nu_j$ for $j\geqslant i+t_i$ $\endgroup$
    – Fedor Petrov
    Commented Aug 21 at 14:14
  • $\begingroup$ @FedorPetrov, nice observation! Do you have a proof? $\endgroup$
    – Notamathematician
    Commented Aug 21 at 14:33
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    $\begingroup$ Lirerally the same as in your previous question $\endgroup$
    – Fedor Petrov
    Commented Aug 21 at 14:59
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    $\begingroup$ (on a side note, couldn’t $f(n), c_n, a(n)$ be all functions denoted as functions, or all sequences denoted as sequences?) $\endgroup$
    – Pietro Majer
    Commented Aug 21 at 19:50
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    $\begingroup$ Factors $1-f(i)x-g(i)x^2$? $\endgroup$
    – Fedor Petrov
    Commented Aug 22 at 9:16

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