• Let $f(n)$ be an arbitrary function with integer values.
• Let $a(n)$ be an integer sequence with ordinary generating function $\frac{1}{G(0,x)}$ where $G(0,x)$ is a continued fraction such that $$ G(k,x) = 1 - \cfrac{f(k+1)x}{G(k+1, x)}. $$ Note that $$ G(0, x) = 1 - \cfrac{f(1)x}{1 - \cfrac{f(2)x}{1 - \cfrac{f(3)x}{1 - \cfrac{f(4)x}{\ddots}}}}. $$
• Let $b(n)$ be an integer sequence such that $$ b(n) = \sum\limits_{i=0}^{n-1}a(n-i-1)b(i), \\ b(0) = 1. $$
• Let $c(n)$ be $\nu_n$ (after the whole transform) where we 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 $[\nu_i, \nu_j] = [\nu_i + f(j-i)\nu_j, \nu_i + f(j-i)\nu_j]$.

I conjecture that $$ b(n) = c(n). $$

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

f(n) = n
c(n) = my(v1); v1 = vector(n, i, 1); for(i=1, n-1, for(j=i+1, n, A = v1[i] + f(j-i)*v1[j]; v1[i] = A; v1[j] = A)); v1[n]
upto1(n) = my(v1); v1 = vector(n, i, c(i))
h(n,x) = my(CF = 1); for(i=1, n, CF = 1 - f(n - i + 1)*x/CF + x*O(x^n)); 1/CF
upto2(n) = my(v1); v1 = Vec(h(n,x)); v2 = vector(n+1, i, 0); v2[1] = 1; for(i=1, n, v2[i+1] = sum(j=0, i-1, v2[j+1]*v1[i-j])); v2 = vector(n, i, v2[i+1])
test(n) = upto1(n) == upto2(n)

UPD:

Given conjecture can be reformulated as follows:

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 $1$ to $n-i$ apply $\nu_{i+j} = \nu_{i+j-1} + f(j)\nu_{i+j}$.

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

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

f(n) = n
upto1(n) = my(v1); v1 = vector(n, i, 1); for(i=1, n-1, for(j=1, n-i, v1[i+j] = v1[i+j-1] + f(j)*v1[i+j])); v1
h(n,x) = my(CF = 1); for(i=1, n, CF = 1 - f(n - i + 1)*x/CF + x*O(x^n)); 1/CF
upto2(n) = my(v1); v1 = Vec(h(n,x)); v2 = vector(n+1, i, 0); v2[1] = 1; for(i=1, n, v2[i+1] = sum(j=0, i-1, v2[j+1]*v1[i-j])); v2 = vector(n, i, v2[i+1])
test(n) = upto1(n) == upto2(n)

Is there a way to prove it?