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.
%I A352448 #28 Mar 24 2025 22:33:59
%S A352448 1,3,22,278,5128,125592,3850000,142013328,6129705088,303238991744,
%T A352448 16920975718144,1051612647426816,72045481821580288,
%U A352448 5394849460316820480,438392509692455286784,38424395486908104071168,3613476161122656804438016
%N A352448 Expansion of e.g.f. LambertW( -2*x/(1-x) ) / (-2*x).
%C A352448 An interesting property of this e.g.f. A(x) is that the sum of coefficients of x^k, k=0..n, in 1/A(x)^n equals zero, for n > 1.
%H A352448 Seiichi Manyama, <a href="/A352448/b352448.txt">Table of n, a(n) for n = 0..344</a>
%F A352448 E.g.f. A(x) = Sum_{n>=0} a(n)*x^n/n! satisfies:
%F A352448 (1) A(x) = LambertW( -2*x/(1-x) ) / (-2*x).
%F A352448 (2) A(x) = exp( 2*x*A(x) ) / (1-x).
%F A352448 (3) A(x) = log( (1-x) * A(x) ) / (2*x).
%F A352448 (4) A( x/(exp(2*x) + x) ) = exp(2*x) + x.
%F A352448 (5) A(x) = (1/x) * Series_Reversion( x/(exp(2*x) + x) ).
%F A352448 (6) Sum_{k=0..n} [x^k] 1/A(x)^n = 0, for n > 1.
%F A352448 (7) [x^(n+1)/(n+1)!] 1/A(x)^n = -2^(n+1) * n for n >= (-1).
%F A352448 a(n) ~ (1 + 2*exp(1))^(n + 3/2) * n^(n-1) / (2^(3/2) * exp(n + 1/2)). - _Vaclav Kotesovec_, Mar 18 2022
%F A352448 a(n) = n! * Sum_{k=0..n} 2^k * (k+1)^(k-1) * binomial(n,k)/k!. - _Seiichi Manyama_, Mar 03 2023
%e A352448 E.g.f.: A(x) = 1 + 3*x + 22*x^2/2! + 278*x^3/3! + 5128*x^4/4! + 125592*x^5/5! + 3850000*x^6/6! + 142013328*x^7/7! + ...
%e A352448 such that A(x) = exp( 2*x*A(x) ) / (1-x), where
%e A352448 exp( 2*x*A(x) ) = 1 + 2*x + 16*x^2/2! + 212*x^3/3! + 4016*x^4/4! + 99952*x^5/5! + 3096448*x^6/6! + 115063328*x^7/7! + ...
%e A352448 Related table.
%e A352448 Another interesting property of the e.g.f. A(x) is illustrated here.
%e A352448 The table of coefficients of x^k/k! in 1/A(x)^n begins:
%e A352448 n=1: [1, -3, -4, -44, -736, -16832, -491168, ...];
%e A352448 n=2: [1, -6, 10, -16, -320, -8064, -249344, ...];
%e A352448 n=3: [1, -9, 42, -78, -48, -1776, -66528, ...];
%e A352448 n=4: [1, -12, 92, -392, 728, -128, -8960, ...];
%e A352448 n=5: [1, -15, 160, -1120, 4600, -8520, -320, ...];
%e A352448 n=6: [1, -18, 246, -2424, 16104, -64752, 119952, ...];
%e A352448 ...
%e A352448 from which we can illustrate that the partial sum of coefficients of x^k, k=0..n, in 1/A(x)^n equals zero, for n > 1, as follows:
%e A352448 n=1:-2 = 1 + -3;
%e A352448 n=2: 0 = 1 + -6 + 10/2!;
%e A352448 n=3: 0 = 1 + -9 + 42/2! + -78/3!;
%e A352448 n=4: 0 = 1 + -12 + 92/2! + -392/3! + 728/4!;
%e A352448 n=5: 0 = 1 + -15 + 160/2! + -1120/3! + 4600/4! + -8520/5!;
%e A352448 n=6: 0 = 1 + -18 + 246/2! + -2424/3! + 16104/4! + -64752/5! + 119952/6!;
%e A352448 ...
%t A352448 terms = 17; A[_] = 0; Do[A[x_] = Exp[2x*A[x]]/(1-x) + O[x]^terms // Normal, terms]; CoefficientList[A[x], x]Range[0,terms-1]! (* _Stefano Spezia_, Mar 24 2025 *)
%o A352448 (PARI) {a(n) = n!*polcoeff( (1/x)*serreverse( x/(exp(2*x +x^2*O(x^n)) + x) ),n)}
%o A352448 for(n=0,30,print1(a(n),", "))
%o A352448 (PARI) my(x='x+O('x^30)); Vec(serlaplace(lambertw(-2*x/(1-x))/(-2*x))) \\ _Michel Marcus_, Mar 17 2022
%o A352448 (PARI) a(n) = n!*sum(k=0, n, 2^k*(k+1)^(k-1)*binomial(n, k)/k!); \\ _Seiichi Manyama_, Mar 03 2023
%Y A352448 Cf. A352410, A352411, A352412.
%Y A352448 Cf. A361068.
%K A352448 nonn
%O A352448 0,2
%A A352448 _Paul D. Hanna_, Mar 16 2022