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A251183 a(n) = Sum_{k=0..n} binomial(n,k) * (-1)^(n-k) * (2^k + 1)^k.

%I A251183 #12 Jan 25 2015 07:43:28
%S A251183 1,2,20,662,80744,38724842,75185316860,593940189201662,
%T A251183 19026394370724334544,2460515237682283759296722,
%U A251183 1280059938187721964276628432100,2672755638642586172344574379667166822,22366135140308331991861876017552822964355384,749476887752252949738787898479359344347670538150842
%N A251183 a(n) = Sum_{k=0..n} binomial(n,k) * (-1)^(n-k) * (2^k + 1)^k.
%F A251183 G.f.: Sum_{n>=0} (2^n + 1)^n * x^n / (1+x)^(n+1).
%F A251183 G.f.: Sum_{n>=0} 2^(n^2) * x^n / (1+x - 2^n*x)^(n+1).
%F A251183 a(n) ~ 2^(n^2). - _Vaclav Kotesovec_, Jan 25 2015
%e A251183 G.f.: A(x) = 1 + 2*x + 20*x^2 + 662*x^3 + 80744*x^4 + 38724842*x^5 +...
%e A251183 where we have the identity:
%e A251183 (1) A(x) = 1/(1-x) + (2+1)*x/(1+x)^2 + (2^2+1)^2*x^2/(1+x)^3 + (2^3+1)^3*x^3/(1+x)^4 + (2^4+1)^4*x^4/(1+x)^5 + (2^5+1)^5*x^5/(1+x)^6 +...
%e A251183 (2) A(x) = 1 + 2*x/(1+x - 2*x)^2 + 2^4*x^2/(1+x - 2^2*x)^3 + 2^9*x^3/(1+x - 2^3*x)^4 + 2^16*x^4/(1+x - 2^4*x)^5 + 2^25*x^5/(1+x - 2^5*x)^6 +...
%e A251183 Illustration of initial terms:
%e A251183 a(0) = 1;
%e A251183 a(1) = -1 + (2+1) = 2;
%e A251183 a(2) = 1 - 2*(2+1) + (2^2+1)^2 = 20;
%e A251183 a(3) = -1 + 3*(2+1) - 3*(2^2+1)^2 + (2^3+1)^3 = 662;
%e A251183 a(4) = 1 - 4*(2+1) + 6*(2^2+1)^2 - 4*(2^3+1)^3 + (2^4+1)^4 = 80744;
%e A251183 a(5) = -1 + 5*(2+1) - 10*(2^2+1)^2 + 10*(2^3+1)^3 - 5*(2^4+1)^4 + (2^5+1)^5 = 38724842; ...
%t A251183 Table[Sum[Binomial[n,k] * (-1)^(n-k) * (2^k + 1)^k,{k,0,n}],{n,0,20}] (* _Vaclav Kotesovec_, Jan 25 2015 *)
%o A251183 (PARI) {a(n)=sum(k=0, n, binomial(n, k) * (-1)^(n-k) * (2^k + 1)^k )}
%o A251183 for(n=0, 15, print1(a(n), ", "))
%o A251183 (PARI) {a(n)=local(A=1); A=sum(m=0, n, (2^m + 1)^m * x^m / (1+x +x*O(x^n) )^(m+1) ); polcoeff(A, n)}
%o A251183 for(n=0, 15, print1(a(n), ", "))
%o A251183 (PARI) {a(n)=local(A=1); A=sum(m=0, n, 2^(m^2) * x^m / (1+x - 2^m*x +x*O(x^n) )^(m+1) ); polcoeff(A, n)}
%o A251183 for(n=0, 15, print1(a(n), ", "))
%Y A251183 Cf. A251182, A251184.
%K A251183 nonn
%O A251183 0,2
%A A251183 _Paul D. Hanna_, Jan 19 2015