Let $T(n,k)$ be a triangle of coefficients such that $T(n,k)\geqslant0$ for $n>0$, $0<k\leqslant n$, $0$ otherwise. Also $$T(2n+1,1)=\frac{1}{2n+1}, T(2n,1)=0$$ $$T(n,k)=\frac{1}{n}(T(n-1,k-1)+(n-2)(T(n-2,k)+\frac{T(n-2,k-1)}{n-1}))$$ Let $$P(n,m)=m\sum\limits_{k=1}^{m}n^{k-1}T(m,k)(-1)^{m+k}$$ I conjecture that $$P(n,m)=2^{n-1}, 0<n\leqslant m$$ To verify it one may use this pari prog:

T(n)=my(v, v1, v2); v=vector(n, i, 0); v[1]=1; v1=v; if(n>1, v[1]=0; v[2]=1/2); v2=v; for(i=3, n, v[1]=if(i%2,1/i); for(j=2, i, v[j]=(v2[j-1]+(i-2)*(v1[j]+v1[j-1]/(i-1)))/i); v1=v2; v2=v); v
P(n, m)=my(A=T(m)); m*sum(k=1, m, n^(k-1)*A[k]*(-1)^(m+k))

Is there a way to prove it?