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A135075 A binomial recursion : a(n) = q(n) (see formula).

%I A135075 #22 Nov 25 2020 04:23:12
%S A135075 0,1,5,33,265,2505,27261,335757,4617461,70138689,1166295457,
%T A135075 21072290241,411069239997,8611025176533,192788027607293,
%U A135075 4594027768539585,116093660372707273,3101080076109154137,87305805274735566669
%N A135075 A binomial recursion : a(n) = q(n) (see formula).
%H A135075 Robert Israel, <a href="/A135075/b135075.txt">Table of n, a(n) for n = 1..423</a>
%F A135075 Let z(1) = x and z(n) = 1 + Sum_{k=1,..,n-1} ( (1 + binomial(n,k))*z(k) ), then z(n) = p(n)*x + q(n). Lim n-->infinity p(n)/q(n) = (3*Pi - 14)/ (8 - 3*Pi) = 3.2111824896280692148...
%F A135075 E.g.f.: g(x) = ((-3*x-8)*exp(x)+6*x+4)/(9*exp(x)-18) -exp(3*x/2)*(-4*arctan(exp(x/2)/sqrt(2-exp(x)))+Pi+8/3)/(6*(2-exp(x))^(3/2))  satisfies (exp(x)-2) g'(x) + 3 g(x) + x = 0. - _Robert Israel_, Mar 06 2017
%F A135075 a(n) ~ (3*Pi - 8) * sqrt(n) * n! / (9 * sqrt(Pi) * log(2)^(n + 3/2)). - _Vaclav Kotesovec_, Nov 25 2020
%p A135075 A[1]:= 0:
%p A135075 for n from 2 to 50 do
%p A135075   A[n]:= 1 + add((1+binomial(n,k))*A[k],k=1..n-1)
%p A135075 od:
%p A135075 seq(A[i],i=1..50); # _Robert Israel_, Mar 06 2017
%t A135075 z[1] := x; z[n_] := 1 + Sum[(1 + Binomial[n, k])*z[k], {k, 1, n - 1}]; Table[Coefficient[z[n], x, 0], {n, 1, 10}] (* _G. C. Greubel_, Sep 22 2016 *)
%t A135075 z[1] := x; z[n_] := z[n] = Expand[1 + Sum[(1 + Binomial[n, k])*z[k], {k, 1, n-1}]]; Table[Coefficient[z[n], x, 0], {n, 1, 30}] (* _Vaclav Kotesovec_, Nov 25 2020 *)
%o A135075 (PARI) r=1; s=1; 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)=q(n);
%Y A135075 Cf. A135074.
%K A135075 nonn
%O A135075 1,3
%A A135075 _Benoit Cloitre_, Nov 17 2007