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  • Let $a(n)$ be A234289 i.e. integer sequence with exponential generating function $$ A(x)=1+A(x)^2\int \frac{1}{A(x)}\,dx $$ The sequence begins with $$ 1, 1, 3, 17, 147, 1729, 25827, 468593, 10012083, 246287521, 6856204803, 213102768977 $$
  • Let $$ R(n,q)=(q+1)R(n-1,q+1)+\sum\limits_{j=0}^{q-1}R(n,j), \\ R(0,q)=1 $$

I conjecture that $$R(n,0)=a(n).$$

Here is the PARI/GP prog to check it numerically:

a(n)=local(A=1+x); for(i=1, n, A=1+A^2*intformal(1/(A+x*O(x^n)))); n!*polcoeff(A, n)
R_upto(n)=my(v1, v2, v3); v1=vector(n+1, i, 1); v2=v1; v3=vector(n+1, i, 0); v3[1]=1; for(i=1, n, for(q=0, n-i, v2[q+1]=(q+1)*v1[q+2] + sum(j=0, q-1, v2[j+1])); v1=v2; v3[i+1]=v1[1]); v3
test(n)=vector(n+1, i, a(i-1))==R_upto(n)

Is there a way to prove it?

Improve this question
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    $\begingroup$ $\int$ stands for $\int_0^x$, I presume? $\endgroup$
    – Pietro Majer
    Commented Jul 31, 2023 at 15:48
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    $\begingroup$ Equivalently, $A'=\dfrac{A^2}{2-A}$. $\endgroup$
    – Alexander Burstein
    Commented Aug 1, 2023 at 22:03

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