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A245373
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G.f.: Sum_{n>=0} x^n / ( (1+x)^(n+1) * (1 - 2*(n+1)*x) ).
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5
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1, 2, 6, 20, 80, 368, 1904, 10880, 67904, 459008, 3336704, 25925120, 214175744, 1873092608, 17276401664, 167504076800, 1702214549504, 18084149854208, 200388963958784, 2311212530401280, 27693720143396864, 344157474490155008, 4428851361694613504, 58933575269230837760
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OFFSET
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0,2
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COMMENTS
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A generalization of Peter Bala's formula in A229046 is as follows:
if F(x,y) = Sum_{n>=0} x^n / ( (1+x)^(n+1) * (1 - (n+1)*x*y) ) then
F(x,y) = Sum_{n>=0} n! * (x*y)^n * (1+x)^n / Product_{k=1..n} (1 + k*x*y);
further, F(x,y) = Sum_{n>=0} b(n,y)*x^n where b(n,y) is given by
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).
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LINKS
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FORMULA
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G.f.: Sum_{n>=0} n! * (2*x)^n * (1+x)^n / Product_{k=1..n} (1 + 2*k*x).
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).
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EXAMPLE
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G.f.: A(x) = 1 + 2*x + 6*x^2 + 20*x^3 + 80*x^4 + 368*x^5 + 1904*x^6 +...
where we have the following series identity:
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)) +...
is equal to
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)) +...
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PROG
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(PARI) {a(n)=polcoeff( sum(m=0, n, x^m/((1+x)^(m+1)*(1 - 2*(m+1)*x) +x*O(x^n))), n)}
for(n=0, 30, print1(a(n), ", "))
(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)}
for(n=0, 30, print1(a(n), ", "))
(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)))}
for(n=0, 30, print1(a(n), ", "))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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