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%I #22 Jun 18 2014 04:21:59
%S 1,4,32,814,86600,39560554,75654970772,594996059517934,
%T 19035905851947436400,2460857798358946973785234,
%U 1280109151917797032199865564812,2672783800502564772495577135824089014,22366199286781599568269093307412768076442280
%N a(n) = Sum_{k=0..n} binomial(n,k) * (1 + 2^k)^k.
%F a(n) = Sum_{k=0..n} binomial(n,k) * (1 + 2^k)^(n-k) * 2^(k^2).
%F O.g.f.: Sum_{n>=0} (1 + 2^n)^n * x^n / (1-x)^(n+1).
%F O.g.f.: Sum_{n>=0} 2^(n^2) * x^n / (1 - (1+2^n)*x)^(n+1).
%F E.g.f.: exp(x) * Sum_{n>=0} (1 + 2^n)^n * x^n / n!.
%F a(n) ~ 2^(n^2). - _Vaclav Kotesovec_, Jun 18 2014
%e O.g.f.: A(x) = 1 + 4*x + 32*x^2 + 814*x^3 + 86600*x^4 + 39560554*x^5 +...
%e where the g.f. may be expressed by the series identity:
%e A(x) = 1/(1-x) + 3*x/(1-x)^2 + 5^2*x^2/(1-x)^3 + 9^3*x^3/(1-x)^4 + 17^4*x^4/(1-x)^5 + 33^5*x^5/(1-x)^6 + 65^6*x^6/(1-x)^7 +...
%e A(x) = 1/(1-2*x) + 2*x/(1-3*x)^2 + 2^4*x^2/(1-5*x)^3 + 2^9*x^3/(1-9*x)^4 + 2^16*x^4/(1-17*x)^5 + 2^25*x^5/(1-33*x)^6 + 2^36*x^6/(1-65*x)^7 +...
%e Illustration of initial terms:
%e a(0) = 1;
%e a(1) = 1 + (1+2);
%e a(2) = 1 + 2*(1+2) + (1+2^2)^2;
%e a(3) = 1 + 3*(1+2) + 3*(1+2^2)^2 + (1+2^3)^3;
%e a(4) = 1 + 4*(1+2) + 6*(1+2^2)^2 + 4*(1+2^3)^3 + (1+2^4)^4;
%e a(5) = 1 + 5*(1+2) + 10*(1+2^2)^2 + 10*(1+2^3)^3 + 5*(1+2^4)^4 + (1+2^5)^5; ...
%e Also, by a binomial identity we have
%e a(0) = 1;
%e a(1) = 2 + 2;
%e a(2) = 2^2 + 2*(1+2)*2 + 2^4;
%e a(3) = 2^3 + 3*(1+2)^2*2 + 3*(1+2^2)*2^4 + 2^9;
%e a(4) = 2^4 + 4*(1+2)^3*2 + 6*(1+2^2)^2*2^4 + 4*(1+2^3)*2^9 + 2^16;
%e a(5) = 2^5 + 5*(1+2)^4*2 + 10*(1+2^2)^3*2^4 + 10*(1+2^3)^2*2^9 + 5*(1+2^4)*2^16 + 2^25; ...
%t Table[Sum[Binomial[n,k]*(1+2^k)^k,{k,0,n}],{n,0,20}] (* _Vaclav Kotesovec_, Jun 18 2014 *)
%o (PARI) {a(n)=sum(k=0, n, binomial(n, k)*(1+2^k)^k)}
%o for(n=0, 20, print1(a(n), ", "))
%o (PARI) {a(n)=sum(k=0, n, binomial(n, k)*(1+2^k)^(n-k)*2^(k^2))}
%o for(n=0, 20, print1(a(n), ", "))
%Y Cf. A244004, A136516.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Jun 17 2014
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