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%I #7 Aug 24 2017 08:29:53
%S 1,4,45,1494,167321,70339178,118127981277,818113700595166,
%T 23602509419592675345,2828610325004443707717522,
%U 1397057668479359172475738134221,2819825298850525709434498781240666534,23097546959835633409694123795378169234099369
%N a(n) = Sum_{k=0..n} C(n,k) * (n + 2^k)^k.
%F a(n) = Sum_{k=0..n} C(n,k) * (1 + n*2^k)^(n-k) * 2^(k^2).
%F a(n) ~ 2^(n^2). - _Vaclav Kotesovec_, Aug 24 2017
%e E.g.f.: A(x) = 1 + 4*x + 45*x^2/2! + 1494*x^3/3! + 167321*x^4/4! + 70339178*x^5/5! +...
%e ILLUSTRATION OF INITIAL TERMS:
%e a(1) = (1+2^0)^0 + (1+2^1)^1 = 4;
%e a(2) = (2+2^0)^0 + 2*(2+2^1)^1 + (2+2^2)^2 = 45;
%e a(3) = (3+2^0)^0 + 3*(3+2^1)^1 + 3*(3+2^2)^2 + (3+2^3)^3 = 1494;
%e a(4) = (4+2^0)^0 + 4*(4+2^1)^1 + 6*(4+2^2)^2 + 4*(4+2^3)^3 + (4+2^4)^4 = 167321; ...
%e where we have the binomial identity:
%e a(1) = (1+1*2^0)^1*2^0 + (1+2^1)^0*2^1 = 4;
%e a(2) = (1+2*2^0)^2*2^0 + 2*(1+2*2^1)^1*2^1 + (1+2*2^2)^0*2^4 = 45;
%e a(3) = (1+3*2^0)^3*2^0 + 3*(1+3*2^1)^2*2^1 + 3*(1+3*2^2)^1*2^4 + (1+3*2^3)^0*2^9 = 1494;
%e a(4) = (1+4*2^0)^4*2^0 + 4*(1+4*2^1)^3*2^1 + 6*(1+4*2^2)^2*2^4 + 4*(1+4*2^3)^1*2^9 + (1+4*2^4)^0*2^16 = 167321; ...
%t Table[Sum[Binomial[n,k](n+2^k)^k,{k,0,n}],{n,0,20}] (* _Harvey P. Dale_, Jun 15 2017 *)
%o (PARI) {a(n) = sum(k=0,n,binomial(n,k) * (n + 2^k)^k )}
%o for(n=0,15,print1(a(n),", "))
%o (PARI) {a(n) = sum(k=0,n,binomial(n,k) * (1 + n*2^k)^(n-k) * 2^(k^2) )}
%o for(n=0,15,print1(a(n),", "))
%K nonn
%O 0,2
%A _Paul D. Hanna_, Jul 05 2014
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