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%I #27 Nov 25 2020 16:50:06
%S 1,4,25,188,1671,17190,201125,2638984,38390179,613363466,10678267425,
%T 201215691660,4080450217247,88609322165902,2051573162708125,
%U 50450534991347216,1313219083705400475,36072797094375866898,1042811362801447763225,31647646914322017237652,1006032342980535954429463
%N A binomial recursion : a(n) = p(n) (see formula).
%H Vaclav Kotesovec, <a href="/A135147/b135147.txt">Table of n, a(n) for n = 1..400</a>
%F 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 Lim n-->oo p(n)/q(n) = (3 - 2*log(2))/(2*log(2) - 1) = 4.17739889912417966161076...
%F a(n) ~ (3 - 2*log(2)) * n * n! / (8 * log(2)^(n+2)). - _Vaclav Kotesovec_, Nov 25 2020
%F E.g.f.: (1 - exp(x)) * (2*x - 1 - exp(x)) / (2*(2 - exp(x))^2). - _Vaclav Kotesovec_, Nov 25 2020
%t z[1]:= x; z[n_] := 1 + Sum[(2 + Binomial[n, k])*z[k], {k, 1, n - 1}]; Table[ Coefficient[z[n], x], {n, 1, 20}] (* _G. C. Greubel_, Sep 28 2016 *)
%t 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], {n, 1, 30}] (* _Vaclav Kotesovec_, Nov 25 2020 *)
%t nmax = 30; Rest[CoefficientList[Series[(1 - E^x)*(-1 - E^x + 2*x)/(2*(2 - E^x)^2), {x, 0, nmax}], x] * Range[0, nmax]!] (* _Vaclav Kotesovec_, Nov 25 2020 *)
%o (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 Cf. A135148, A135149, A135150, A135074, A135075.
%K nonn
%O 1,2
%A _Benoit Cloitre_, Nov 20 2007
%E More terms from _Amiram Eldar_, Nov 25 2020
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