- Let $F(n)$ be A000045 (i.e., Fibonacci numbers). Here $$ F(n) = F(n-1) + F(n-2), \\ F(0) = 0, F(1) = 1. $$
- Let $a(n)$ be A268407 (i.e., number of North-East lattice paths that do not bounce off the diagonal $y = x$ to the right). Here $$ a(n) = \frac{1}{n+1} \left[ \binom{2n}{n} + \sum\limits_{k=0}^{n} (k+1) \binom{2n-k}{n-k} F(k) \right]. $$
- Let $b(n)$ be an integer sequence such that $$ b(n) = a(n-1) + \sum\limits_{k=1}^{n-1} b(n-k)a(k-1). $$
- Start with vector $\nu$ of fixed length $m$ with elements $\nu_i=1$ (that is, $\nu = \{1,1,\dotsc,1\}$) and for $i$ from $1$ to $m-1$, for $j$ from $i+1$ to $m$ consecutively apply $$ \nu_j := \nu_j + \nu_{j-1}, $$ for each $j$ and also apply $$ \nu_j := \nu_j + \nu_i. $$ if $(j-i)$ is odd.
I conjecture that after the whole transform we have $$ \nu_n = b(n). $$
Here is the PARI/GP program to check it numerically:
a(n) = (binomial(2*n, n) + sum(k=0, n, (k+1)*binomial(2*n-k, n-k)*fibonacci(k)))/(n+1)
upto1(n) = my(v1); v1 = vector(n, i, a(i-1)); v2 = vector(n, i, 0); for(i=1, n, v2[i] = v1[i] + sum(k=1, i-1, v2[i-k]*v1[k])); v2
upto2(n) = my(v1); v1 = vector(n, i, 1); for(i=1, n-1, for(j=i+1, n, v1[j] += v1[j-1]; if((j-i)%2, v1[j] += v1[i]))); v1
test1(n) = upto2(n) == upto1(n)
UPD1:
I also conjecture that if we modify last condition as $(j-i)>1$ and $(j-i)$ is odd, then after the whole transform we have $$ \nu_n = a(n-1). $$ Here is the PARI/GP program to check it numerically:
a(n) = (binomial(2*n, n) + sum(k=0, n, (k+1)*binomial(2*n-k, n-k)*fibonacci(k)))/(n+1)
upto3(n) = my(v1); v1 = vector(n, i, 1); for(i=1, n-1, for(j=i+1, n, v1[j] += v1[j-1]; if((j-i)>1 && (j-i)%2, v1[j] += v1[i]))); v1
test2(n) = upto3(n) == vector(n, i, a(i-1))
Note that vector of length $10^4$ terms computes in $50,5$ seconds.
UPD2:
I also conjecture that if we modify basic condition as $\nu_{i,k} = \delta_{i,k}$, then using the modification in UPD1 and after the whole transform we have $$ \nu_{n,k} = T(n-1,k-1). $$
where $T(n,k)$ is A174302.
Is there a way to prove it?