A293379 - OEIS

%I #26 Jun 12 2020 11:14:25

%S 1,1,2,8,60,708,11508,238356,5997924,177615460,6050014228,

%T 233010132788,10011985592516,474815000195844,24635182609510260,

%U 1388071373846060244,84406416179170495524,5509502700599270439588,384233531128293605883092,28513258202704586092168820,2243347226931600749380032388,186528883284172684793892938308,16343425532977997699916585695796

%N G.f.: exp( Sum_{n>=1} A052886(n) * x^n/n ).

%C The e.g.f. of A052886 equals Series_Reversion( log(1 + x - x^2) ), which also equals (1 - sqrt(5 - 4*exp(x)) )/2.

%C This sequence is motivated by the following conjectures:

%C (C1) Given integer series G(x) such that G(0) = G'(0) = 1, define L(n) by

%C Sum_{n>=1} L(n) * x^n/n! = Series_Reversion( log(G(x)) )

%C then exp( Sum_{n>=1} L(n) * x^n/n ) is also an integer series;

%C (C2) Given G(x) = 1 + x*G(x)^m, define L(n) by

%C Sum_{n>=1} L(n) * x^n/n! = Series_Reversion( log(G(x)) )

%C then exp( Sum_{n>=1} L(n) * x^n/n ) = (1 + m*x)/(1 + (m-1)*x).

%H Paul D. Hanna, <a href="/A293379/b293379.txt">Table of n, a(n) for n = 0..300</a>

%F a(n) ~ sqrt(5) * n^(n-2) / (2^(3/2) * exp(n) * (log(5/4))^(n-1/2)). - _Vaclav Kotesovec_, Oct 26 2017

%F a(n) = (1/n)*Sum_{m=1..n} m! * C(m-1) * Sum_{s=m..n} Stirling2(s,m) * a(n-s) for n >= 1 with a(0) = 1, where C(n) = A000108(n). - _Petros Hadjicostas_, Jun 12 2020

%e G.f.: A(x) = 1 + x + 2*x^2 + 8*x^3 + 60*x^4 + 708*x^5 + 11508*x^6 + 238356*x^7 + 5997924*x^8 + 177615460*x^9 + 6050014228*x^10 +...

%e such that

%e log(A(x)) = x + 3*x^2/2 + 19*x^3/3 + 207*x^4/4 + 3211*x^5/5 + 64383*x^6/6 + 1581259*x^7/7 + 45948927*x^8/8 + 1541641771*x^9/9 +...+ A052886(n)*x^n/n +...

%e Note that the e.g.f. of A052886 equals Series_Reversion( log(1 + x - x^2) ) = x + 3*x^2/2! + 19*x^3/3! + 207*x^4/4! + 3211*x^5/5! + 64383*x^6/6! + 1581259*x^7/7! + 45948927*x^8/8! +...

%t nmax = 20; A052886 = CoefficientList[Series[(1 - Sqrt[5 - 4*E^x])/2, {x, 0, nmax}], x] * Range[0, nmax]!; Clear[a]; a[n_]:= a[n] = Sum[A052886[[n-k+1]]*a[k], {k, 0, n-1}] / n; a[0] = 1; Table[a[n], {n, 0, nmax}] (* _Vaclav Kotesovec_, Oct 26 2017 *)

%o (PARI) {a(n) = my(A,L); L = x*serlaplace( 1/x*serreverse( log(1+x-x^2 +O(x^(n+2))) ) ); A = exp(L); polcoeff(A,n)}

%o for(n=0,30,print1(a(n),", "))

%o (PARI) c(n) = (1/(n+1))*binomial(2*n,n); /* A000108 */

%o lista(nn)={my(a=vector(nn)); a[1]=1; for(n=2, nn, a[n] = (1/(n-1))*sum(m=1, n-1, m!*c(m-1)*sum(s=m,n-1, stirling(s,m,2)*a[n-s]))); a;} \\ _Petros Hadjicostas_, Jun 12 2020

%Y Cf. A000108, A052886, A294115.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Oct 22 2017