cp's OEIS Frontend

A251182 a(n) = Sum_{k=0..n} binomial(n, k) * (2^k - 1)^k.

%I A251182 #10 Jan 25 2015 05:12:13
%S A251182 1,2,12,374,52056,28885802,62696043492,533314127677214,
%T A251182 17882368106959072176,2375841795610783881752402,
%U A251182 1255349217681407318492850586812,2644225793295900889867998778302561734,22235530372401312606563659670627119777739016,747102526634678016053120249315815798707672485729402
%N A251182 a(n) = Sum_{k=0..n} binomial(n, k) * (2^k - 1)^k.
%F A251182 G.f.: Sum_{n>=0} (2^n - 1)^n * x^n / (1-x)^(n+1).
%F A251182 G.f.: Sum_{n>=0} 2^(n^2) * x^n / (1-x + 2^n*x)^(n+1).
%F A251182 a(n) ~ 2^(n^2). - _Vaclav Kotesovec_, Jan 25 2015
%e A251182 G.f.: A(x) = 1 + 2*x + 12*x^2 + 374*x^3 + 52056*x^4 + 28885802*x^5 +...
%e A251182 where we have the identity:
%e A251182 (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 A251182 (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 + 2^36*x^6/(1-x + 2^6*x)^7 +...
%e A251182 Illustration of initial terms:
%e A251182 a(0) = 1;
%e A251182 a(1) = 1 + (2-1) = 2;
%e A251182 a(2) = 1 + 2*(2-1) + (2^2-1)^2 = 12;
%e A251182 a(3) = 1 + 3*(2-1) + 3*(2^2-1)^2 + (2^3-1)^3 = 374;
%e A251182 a(4) = 1 + 4*(2-1) + 6*(2^2-1)^2 + 4*(2^3-1)^3 + (2^4-1)^4 = 52056;
%e A251182 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 = 28885802; ...
%t A251182 Table[1 + Sum[Binomial[n,k] * (2^k-1)^k,{k,1,n}],{n,0,15}] (* _Vaclav Kotesovec_, Jan 25 2015 *)
%o A251182 (PARI) {a(n)=sum(k=0, n, binomial(n, k) * (2^k - 1)^k )}
%o A251182 for(n=0, 15, print1(a(n), ", "))
%o A251182 (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 A251182 for(n=0, 15, print1(a(n), ", "))
%o A251182 (PARI) {a(n)=local(A=1); A=sum(m=0, n, 2^(m^2) * x^m / (1-x + x*2^m +x*O(x^n))^(m+1) ); polcoeff(A, n)}
%o A251182 for(n=0, 15, print1(a(n), ", "))
%Y A251182 Cf. A251183, A251184.
%K A251182 nonn
%O A251182 0,2
%A A251182 _Paul D. Hanna_, Jan 19 2015