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A135148 A binomial recursion: a(n) = q(n) (see formula).

%I A135148 #35 Jan 29 2025 07:56:35
%S A135148 0,1,6,45,400,4115,48146,631729,9189972,146829039,2556200086,
%T A135148 48167698733,976792093784,21211601837803,491112582793626,
%U A135148 12077021182230057,314362864408454236,8635229233659916007,249631741661080132766,7575921686807827601701,240827454421807200901728
%N A135148 A binomial recursion: a(n) = q(n) (see formula).
%H A135148 Vaclav Kotesovec, <a href="/A135148/b135148.txt">Table of n, a(n) for n = 1..400</a>
%F A135148 Let z(1) = x and z(n) = 1 + Sum_{k=1..n-1} (2 + binomial(n,k))*z(k), then z(n) = p(n)*x + q(n).
%F A135148 Limit_{n->oo} p(n)/q(n) = (3 - 2*log(2))/(2*log(2) - 1) = 4.177398899124179661610768...
%F A135148 a(n) ~ (2*log(2) - 1) * n * n! / (8 * log(2)^(n+2)). - _Vaclav Kotesovec_, Nov 25 2020
%F A135148 E.g.f.: (1 - exp(x)) * (exp(x) - 2*x - 1) / (2*(2 - exp(x))^2). - _Vaclav Kotesovec_, Nov 25 2020
%F A135148 a(n+1) = Sum_{k = 1..n} Stirling2(n, k)*A142980(k). - _Peter Bala_, Dec 10 2024
%t A135148 z[1] := x; z[n_] := 1 + Sum[(2 + Binomial[n, k])*z[k], {k, 1, n - 1}]; Table[ Coefficient[z[n], x, 0], {n, 1, 20}] (* _G. C. Greubel_, Sep 28 2016 *)
%t A135148 z[1] := x; z[n_] := z[n] = Expand[1 + Sum[(2 + Binomial[n, k])*z[k], {k, 1, n-1}]]; Table[Coefficient[z[n], x, 0], {n, 1, 30}] (* _Vaclav Kotesovec_, Nov 25 2020 *)
%t A135148 nmax = 30; Rest[CoefficientList[Series[(1 - E^x)*(E^x - 2*x - 1)/(2*(2 - E^x)^2), {x, 0, nmax}], x] * Range[0, nmax]!] (* _Vaclav Kotesovec_, Nov 25 2020 *)
%o A135148 (PARI) r=1; s=2; v=vector(120, j, x); for(n=2, 120, g=r+sum(k=1, n-1, (s+binomial(n, k))*v[k]); v[n]=g); z(n)=v[n]; p(n)=polcoeff(z(n), 1); q(n)=polcoeff(z(n), 0); a(n)=p(n);
%Y A135148 Cf. A135147, A135149, A135150, A135074, A135075, A142980.
%K A135148 nonn,easy
%O A135148 1,3
%A A135148 _Benoit Cloitre_, Nov 20 2007