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  • Let $a(n)$ be A235129 i.e. an integer sequence such that its exponential generating function $A(x)$ satisfies $$ A'(x) = 1 + A(x)\exp(A(x)) $$ The sequence begins with $$ 1, 1, 3, 12, 64, 424, 3358, 30952, 325488, 3845724, 50437624, 727094704, 11427436072, 194468970904 $$

  • Also, $a(n)$ is the number of increasing trees on vertex set $[n]$ in which vertices of out-degree $r$ come in $r$ varieties for $r\geqslant 1$ or, more picturesquely, each non-leaf vertex has a favorite child.

  • Let $$ R(n, q) = R(n-1, q+1) + (q+1)\sum\limits_{j=0}^{q}\binom{q}{j}R(n-1, j), \\ R(0, q) = [q = 0] $$ Here square bracket denotes Iverson bracket.

I conjecture that $$R(n-1,0)=a(n).$$

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

a(n)=local(A=x); for(i=1, n, A=intformal(1+A*exp(A+x*O(x^n)))); n!*polcoeff(A, n)
R_upto(n) = n--; my(v1, v2, v3, v4); v1 = vector(n + 1, i, i--; i == 0); 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?

Improve this question
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    $\begingroup$ Don't know if this is useful but it seems that the ordinary generating function $\sum_qR(n,q)t^q$ is $(1-t)^{-2}(1-2t)^{-3}\cdots(1-nt)^{-(n+1)}$ times a polynomial in $t$ $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Dec 21, 2023 at 15:14
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    $\begingroup$ Again, don't know if this can be used but denoting $f_n(t)=\sum_{q\geqslant0}R(n,q)t^q$,$$f_n(t)=\frac{f_{n-1}(t)-f_{n-1}(0)}t+\frac d{dt}\left(\frac t{1-t}f_{n-1}\left(\frac t{1-t}\right)\right)$$ $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Dec 23, 2023 at 20:53
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    $\begingroup$ Or else,$$f_n(t)=\frac{f_{n-1}(t)-f_{n-1}(0)}t+\frac1{(1-t)^2}f_{n-1}\left(\frac t{1-t}\right)+\frac t{(1-t)^3}f_{n-1}'\left(\frac t{1-t}\right)$$ $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Dec 23, 2023 at 20:56
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    $\begingroup$ Or also, $F(z,t)=\sum_{n,q\geqslant0}R(n,q)z^nt^q$ satisfies$$(t-z)F(z,t)=tF(0,t)-zF(z,0)+\frac{zt}{(1-t)^2}F\left(z,\frac t{1-t}\right)+\frac{zt^2}{(1-t)^5}\frac{\partial F}{\partial t}\left(z,\frac t{1-t}\right)$$ $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Dec 23, 2023 at 21:28
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    $\begingroup$ (And $F(0,t)=1$ by definition) $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Dec 24, 2023 at 3:42

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