Let
- $B_n$ be the $n$-th Bernoulli number.
- $T(n,k)$ be an integer coefficients such that $T(n,k) = \nu_k$ where we start with vector $\nu$ of a fixed length $n$ with elements $\nu_1=1$, $\nu_i=0$ for $i>1$ and for
i=1..n-1, forj=i+1..napply $\nu_j := \nu_j + \nu_{j-1}$ and fork=n-1..i(with decreasing step equals $1$) also apply $\nu_k := \nu_k + \nu_{k+1}$.
I conjecture that for $1 \leqslant k \leqslant n$ we have $$\frac{(-1)^{k-1}(2n+1)T(n,k)}{2(4^k-1)\binom{2n+1}{2k}} = B_{2k}.$$
Here is the PARI/GP program to check it numerically:
upto(n) = {my(A = 1, B = 1, v1);
v1 = vector(n, i, 0); v1[1] = 1;
for(i=1, n-1, for(j=i+1, n,
v1[j] = v1[j] + v1[j-1]);
forstep(k=n-1, i, -1,
v1[k] = v1[k] + v1[k+1]);
v1[i] = (-1)^(i-1)*(2*n+1)*
v1[i]/(2*((A*=4)-1)*
(B *= (2*n+2-2*i+1)*(2*n+2-2*i)
/(2*i*(2*i-1)))));
v1[n] *= (-1)^(n-1)/(2*((A*=4)-1));
v1}
test(n) = upto(n) == vector(n, i, bernfrac(2*i))
Note that for n=5000 vector of bernfrac computes in $7$ min $20$ sec, while upto computes in $3$ min $35$ sec.
Is there a way to prove it?