- Let $a(n,p,q)$ be the family of integer sequences such that exponential generating functions for it satisfy
$$ A_1(x)=\exp\left(x + p\int\int (A_1(x))^q \, dx \, dx\right). $$
- Let $b(n,p,q)$ be the family of integer sequences such that exponential generating functions for it satisfy
$$ A_2(x)=(1 + \int A_2(x) \, dx)(1 + p\int (A_2(x))^q \, dx). $$
- Let
$$ T_1(n,k,p,q) = (q(k-1)+1)T_1(n-1,k,p,q) + p(n-2k+3)T_1(n-1,k-1,p,q), \\ T_1(n,1,p,q) = [n > 0], \\ T_1(n,0,p,q) = T_1(0,k,p,q) = 0. $$
Here square bracket denotes Iverson bracket.
- Let
$$ T_2(n,k,p,q) = (q(k-1)+1)T_2(n-1,k,p,q) + p(n+(q-2)(k-2)-1)T_2(n-1,k-1,p,q), \\ T_2(n,1,p,q) = [n > 0], \\ T_2(n,0,p,q) = T_2(0,k,p,q) = 0. $$
I conjecture that for $n>0$
$$ \sum\limits_{k=1}^{n}T_1(n,k,p,q) = a(n,p,q), \\ \sum\limits_{k=1}^{n}T_2(n,k,p,q) = b(n-1,p,q). $$
Here is the PARI/GP program to check it numerically:
a(n,p,q) = local(A=1+x); for(i=1, n, A=exp(x+p*intformal(intformal(A^q+x*O(x^n))))); n!*polcoeff(A, n)
b(n,p,q) = my(A=1); for(i=1, n, A = (1 + intformal( A^1 )) * (1 + p*intformal( A^q +x*O(x^n))) ); n!*polcoeff(A, n)
upto1(n,p,q) = my(v1); v1 = vector(n, i, a(i,p,q))
upto2(n,p,q) = my(v1); v1 = vector(n, i, b(i-1,p,q))
upto3(n,p,q) = my(v1); v1 = vector(n, i, 0); v1[1] = 1; v2 = v1; for(i=2, n, v1 = vector(n, j, if(j==1, 1, (q*(j-1)+1)*v1[j] + p*(i-2*j+3)*v1[j-1])); v2[i] = vecsum(v1)); v2
upto4(n,p,q) = my(v1); v1 = vector(n, i, 0); v1[1] = 1; v2 = v1; for(i=2, n, v1 = vector(n, j, if(j==1, 1, (q*(j-1)+1)*v1[j] + p*(i+(q-2)*(j-2)-1)*v1[j-1])); v2[i] = vecsum(v1)); v2
test1(n,p,q) = upto1(n,p,q) == upto3(n,p,q)
test2(n,p,q) = upto2(n,p,q) == upto4(n,p,q)
Is there a way to prove it?
