cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A385907 E.g.f. A(x) satisfies: 1 = Sum_{n>=0} (1/(1-x)^n - A(x))^n / n!.

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%I A385907 #8 Jul 24 2025 14:05:26
%S A385907 1,1,3,23,352,8514,283631,12112806,635825408,39879146040,
%T A385907 2927184461577,247446468028660,23783152246963000,2571886518254945436,
%U A385907 310182327935629919849,41412336266329078729672,6081427517910705699609120,976832191726100607511719504,170781177018035313244639721409
%N A385907 E.g.f. A(x) satisfies: 1 = Sum_{n>=0} (1/(1-x)^n - A(x))^n / n!.
%C A385907 In general, the following sums are equal:
%C A385907 (C.1) Sum_{n>=0} (q^n + p)^n * r^n/n!,
%C A385907 (C.2) Sum_{n>=0} q^(n^2) * exp(p*q^n*r) * r^n/n!;
%C A385907 here, q = 1/(1-x) with p = -A(x), r = 1.
%H A385907 Paul D. Hanna, <a href="/A385907/b385907.txt">Table of n, a(n) for n = 0..260</a>
%F A385907 E.g.f. A(x) satisfies:
%F A385907 (1) 1 = Sum_{n>=0} (1/(1-x)^n - A(x))^n / n!.
%F A385907 (2) 1 = Sum_{n>=0} 1/(1-x)^(n^2) * exp(-A(x)/(1-x)^n) / n!.
%e A385907 E.g.f.: A(x) = 1 + x + 3*x^2/2! + 23*x^3/3! + 352*x^4/4! + 8514*x^5/5! + 283631*x^6/6! + 12112806*x^7/7! + 635825408*x^8/8! + 39879146040*x^9/9! + 2927184461577*x^10/10! + ...
%e A385907 where
%e A385907 1 = 1 + (1/(1-x) - A(x)) + (1/(1-x)^2 - A(x))^2/2! + (1/(1-x)^3 - A(x))^3/3! + (1/(1-x)^4 - A(x))^4/4! + (1/(1-x)^5 - A(x))^5/5! + (1/(1-x)^6 - A(x))^6/6! + ...
%e A385907 also
%e A385907 1 = exp(-A(x)) + 1/(1-x)*exp(-A(x)/(1-x)) + 1/(1-x)^4*exp(-A(x)/(1-x)^2)/2! + 1/(1-x)^9*exp(-A(x)/(1-x)^3)/3! + 1/(1-x)^16*exp(-A(x)/(1-x)^4)/4! + 1/(1-x)^25*exp(-A(x)/(1-x)^5)/5! + ...
%o A385907 (PARI) {a(n) = my(A=[1]); for(i=1, n, A=concat(A, 0); A[#A] = polcoeff( sum(m=0, #A, (1/(1-x)^m - Ser(A))^m/m! ), #A-1) ); n!*A[n+1]}
%o A385907 for(n=0, 20, print1(a(n), ", "))
%Y A385907 Cf. A326097.
%K A385907 nonn
%O A385907 0,3
%A A385907 _Paul D. Hanna_, Jul 24 2025