A104507 Row sums of triangle A104505, which is equal to the right-hand side of the triangle A084610 of coefficients in (1+x-x^2)^n.
1, 0, -2, -3, 2, 15, 19, -28, -134, -129, 353, 1254, 791, -4238, -11818, -3123, 49162, 110007, -17783, -554458, -996323, 690932, 6096792, 8624747, -12287153, -65419110, -69285296, 178655307, 684550946, 483569751, -2354830741, -6970706252, -2324044054, 29195280375, 68793790705
Offset: 0
Programs
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Mathematica
CoefficientList[Series[(x/((1 - x)) + 1/((-Sqrt[5 x^2 - 2 x + 1] + x + 1)) x (1 - (5 x - 1)/(Sqrt[5 x^2 - 2 x + 1]))), {x, 0, 40}], x] (* Vincenzo Librandi, Oct 05 2015 *) -
Maxima
a(n):=sum((-1)^j*binomial(n,j)*binomial(n-j-1,n-2*j),j,0,n/2); /* Vladimir Kruchinin, Oct 04 2015 */
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PARI
a(n)=sum(k=0,n,polcoeff((1+x-x^2)^n,n+k))
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PARI
a(n) = sum(k=0, n/2, (-1)^k*binomial(n,k)*binomial(n-k-1,n-2*k)); vector(40, n, a(n-1)) \\ Altug Alkan, Oct 04 2015
Formula
G.f.: (x/((1-x))+1/((-sqrt(5*x^2-2*x+1)+x+1))*x*(1-(5*x-1)/(sqrt(5*x^2-2*x+1)))). - Vladimir Kruchinin, Oct 04 2015
a(n) = Sum_{j=0..n/2}((-1)^j*binomial(n,j)*binomial(n-j-1,n-2*j)). - Vladimir Kruchinin, Oct 04 2015
From Peter Bala, Jul 24 2025: (Start)
a(n) = [x^n] (1 - x^2/(1 - x))^n. Cf. A246437.
The Gauss congruences hold: a(n*p^k) == a(n*p^(k-1)) (mod p^k) for all primes p and all positive integers n and k.
exp(Sum_{n >= 1} a(n)*x^n/n) = 1 - x^2 - x^3 + x^4 + 4*x^5 + ... is the g.f. of A108623.(End)