Let $a(n)$ be A249833 i.e. an integer sequence such that its exponential generating function $A(x)$ satisfies $$ A(x) = 1 + \int A(x) + (A(x))^2\log A(x)\,dx $$ The sequence begins with $$ 1, 1, 2, 7, 34, 210, 1574, 13866, 140340, 1604284, 20439484, 287152488, 4409695952, 73482586464 $$
Let $$ R(n, q) = R(n-1, q+1) + (q+1)\sum\limits_{j=0}^{q}\binom{q}{j}R(n-1, j), \\ R(0, q) = 1 $$
I conjecture that for $n>0$ $$R(n-1,0)=a(n).$$
Here is the PARI/GP program to check it numerically:
a(n)=local(A=1+x); for(i=1, n, A=1+intformal(A+A^2*log(A +x*O(x^n)))); n!*polcoeff(A, n)
R_upto(n) = n--; my(v1, v2, v3, v4); v1 = vector(n + 1, i, 1); v2 = v1; v3 = vector(n + 1, i, 0); v3[1] = 1; v4 = vector(n, i, vector(i, j, (j == 1) || (j == i))); for(i = 3, n, for(j = 2, i - 1, v4[i][j] = v4[i - 1][j] + v4[i - 1][j - 1])); for(i = 1, n, for(q = 0, n - i, v2[q + 1] = v1[q + 2] + (q + 1) * sum(j = 0, q, v4[q + 1][j + 1] * v1[j + 1])); v1 = v2; v3[i + 1] = v1[1]); v3
test(n) = R_upto(n) == vector(n, i, a(i))
Is there a way to prove it?
