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  • Let $a(n)$ be A112487 i.e. an integer sequence with exponential generating function $$ A(x)=\exp\left(\int (A(x)+A(x)^2)\,dx\right), \\ A(0)=1 $$ However, the definition in the name of the sequence is $$ a(n)=\sum\limits_{k=0}^{n}E_2(n,k)2^k $$ where $E_2(n,k)$ is A340556, the second-order Eulerian numbers with the following recurrence: $$ E_2(n, k) = kE_2(n-1, k) + (2n-k)E_2(n-1, k-1), \\ E_2(0,0) = 1 $$
  • Let $b(n)$ be an integer sequence with generating function $\frac{1}{G(0)}$ where $$ G(j)=\cfrac{1-\cfrac{(j+1)x}{G(j+1)}}{1+\cfrac{(j+1)x}{G(j+1)}} $$ Here $G(0)$ is something which might be called multi-continued fraction.

I conjecture that $$a(n)=b(n).$$

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

a(n)=local(A=1+x); for(i=1, n, A=exp(intformal(A+A^2)+x*O(x^n))); n!*polcoeff(A, n)
b(n)=local(CF=1+x*O(x)); for(j=0, n, CF=(1 - (n-j+1)*x/CF)/(1 + (n-j+1)*x/CF)); polcoeff(1/CF, n, x);
test(n)=a(n)==b(n)

Is there a way to prove it?

Improve this question
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    $\begingroup$ Definition of $b(n)$ is unclear, no initial condition for the recurrence for $G$ is given. $\endgroup$
    – Max Alekseyev
    Commented Jul 17, 2023 at 2:46
  • $\begingroup$ @MaxAlekseyev, thank you for comment! $G(0)$ is completely defined by a given recurrence. Just put $j=0$ and then expand $G(1)$ using $G(2)$, then expand $G(2)$ using $G(3)$ and so on. $\endgroup$
    – Notamathematician
    Commented Jul 17, 2023 at 5:36
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    $\begingroup$ That expansion gives an expression of the form $G(j) = \frac{G(j+k)p_k(j,x) + q_k(j,x)}{G(j+k)r_k(j,x) + s_k(j,x)}$ where $p_k, r_k$ have constant coefficient $1$ and $q_k, s_k$ have constant coefficient $0$. How do you ever extract the linear coefficient of $G(0)$? $\endgroup$
    – Peter Taylor
    Commented Jul 17, 2023 at 14:53
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    $\begingroup$ Let $F(j)=-\frac{1}{G(j)}$, then $F(j)=1-\frac{2}{1+(j+1)x F(j+1)}$ and $(b_n)_{n\ge 0}$ has the generating function $-F(0)$. $\endgroup$
    – Alexander Burstein
    Commented Jul 17, 2023 at 18:20
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    $\begingroup$ @Notamathematician Let's check: $$F(j)=-\frac{1}{G(j)}=\frac{-1-\frac{(j+1)x}{G(j+1)}}{1-\frac{(j+1)x}{G(j+1)}}=\frac{-1+(j+1)xF(j+1)}{1+(j+1)xF(j+1)}=1-\frac{2}{1+(j+1)xF(j+1)}$$ and $(b_n)$ has generating function $\frac{1}{G(0)}=-F(0)$. $\endgroup$
    – Alexander Burstein
    Commented Jul 17, 2023 at 21:46

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