Please note that this question has been completely reworked in order not to overload it with unnecessary and useless information.
- Let $f(n)$ be an arbitrary function with integer values.
- Let $a(n)$ be an integer sequence with generating function $\frac{1}{U(x)}$ such that $$ U(x) = 1 - \cfrac{f(0)x}{1 - \cfrac{f(1)x}{1 - \cfrac{f(2)x}{1 - \cfrac{f(3)x}{\ddots}}}}. $$
- Start with vector $\nu$ of length $n$ with elements $\nu_i = 1$ (that is, $\nu = \{1, 1, \dotsc, 1\}$) and for $i$ from $1$ to $n-1$ and for $j$ from $i+1$ to $n$ apply $\nu_{j} := f(j-i-1)\nu_{j} + [j>(i+1)]\nu_{j-1}$.
Here square bracket denotes Iverson bracket.
I conjecture that after the whole transform we have $$ \nu_n = a(n-1). $$
Here is the PARI/GP program to check it numerically:
f(n) = 2*n + 10
upto1(n) = my(v1); v1 = vector(n, i, 1); for(i=1, n-1, for(j=i+1, n, v1[j] = f(j-i-1)*v1[j] + (j>(i+1))*v1[j-1])); v1
U(n,x) = my(CF = 1); for(i=1, n, CF = 1 - f(n - i)*x/CF + x*O(x^n)); CF
upto2(n) = my(v1); v1 = Vec(1/U(n,x))
test1(n) = upto1(n+1) == upto2(n)
In addition, this question can be rephrased as follows:
- Let $P(n,k)$ be a triangle read by rows such that $$ P(n, k) = \begin{cases} 0 & \textrm{if } k > n \\ 1 & \textrm{if } k = 0 \\ P(n-1, k) + f(n-k)P(n, k-1) & \textrm{otherwise} \end{cases} $$
I conjecture that $$ P(n,n)=a(n). $$
Is there a way to prove it?
