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%I #7 Jul 19 2014 22:05:50
%S 1,2,6,20,80,368,1904,10880,67904,459008,3336704,25925120,214175744,
%T 1873092608,17276401664,167504076800,1702214549504,18084149854208,
%U 200388963958784,2311212530401280,27693720143396864,344157474490155008,4428851361694613504,58933575269230837760
%N G.f.: Sum_{n>=0} x^n / ( (1+x)^(n+1) * (1 - 2*(n+1)*x) ).
%C A generalization of Peter Bala's formula in A229046 is as follows:
%C if F(x,y) = Sum_{n>=0} x^n / ( (1+x)^(n+1) * (1 - (n+1)*x*y) ) then
%C F(x,y) = Sum_{n>=0} n! * (x*y)^n * (1+x)^n / Product_{k=1..n} (1 + k*x*y);
%C further, F(x,y) = Sum_{n>=0} b(n,y)*x^n where b(n,y) is given by
%C b(n,y) = Sum_{k=0..floor(n/2)} Sum_{i=0..k} (-1)^i * C(k,i) * (k-i+1)^(n-k) * y^(n-k).
%F G.f.: Sum_{n>=0} n! * (2*x)^n * (1+x)^n / Product_{k=1..n} (1 + 2*k*x).
%F a(n) = Sum_{k=0..floor(n/2)} Sum_{i=0..k} (-1)^i * binomial(k,i) * 2^(n-k) * (k-i+1)^(n-k).
%e G.f.: A(x) = 1 + 2*x + 6*x^2 + 20*x^3 + 80*x^4 + 368*x^5 + 1904*x^6 +...
%e where we have the following series identity:
%e A(x) = 1/((1+x)*(1-2*x)) + x/((1+x)^2*(1-4*x)) + x^2/((1+x)^3*(1-6*x))+ x^3/((1+x)^4*(1-8*x))+ x^4/((1+x)^5*(1-10*x)) + x^5/((1+x)^6*(1-12*x)) +...
%e is equal to
%e A(x) = 1 + 2*x*(1+x)/(1+2*x) + 2!*(2*x)^2*(1+x)^2/((1+2*x)*(1+4*x)) + 3!*(2*x)^3*(1+x)^3/((1+2*x)*(1+4*x)*(1+6*x)) + 4!*(2*x)^4*(1+x)^4/((1+2*x)*(1+4*x)*(1+6*x)*(1+8*x)) + 5!*(2*x)^5*(1+x)^5/((1+2*x)*(1+4*x)*(1+6*x)*(1+8*x)*(1+10*x)) +...
%o (PARI) {a(n)=polcoeff( sum(m=0, n, x^m/((1+x)^(m+1)*(1 - 2*(m+1)*x) +x*O(x^n))), n)}
%o for(n=0, 30, print1(a(n), ", "))
%o (PARI) {a(n)=polcoeff( sum(m=0, n, 2^m*m!*x^m*(1+x)^m/prod(k=1, m, 1+2*k*x +x*O(x^n))), n)}
%o for(n=0, 30, print1(a(n), ", "))
%o (PARI) {a(n)=sum(k=0, floor(n/2), sum(i=0, k, (-1)^i*binomial(k, i)*(k-i+1)^(n-k)*2^(n-k)))}
%o for(n=0, 30, print1(a(n), ", "))
%Y Cf. A229046, A245374, A245375, A245376.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Jul 19 2014
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