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

0,3

Links

Table of n, a(n) for n=0..34.

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

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 *)

Prog

(PARI) a(n)=sum(k=0, n, polcoeff((1+x-x^2)^n, n+k))
(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 */
(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

Crossrefs

Cf. A104505.

Sequence in context: A174111 A367566 A164661 * A101033 A136454 A025522

Adjacent sequences: A104504 A104505 A104506 * A104508 A104509 A104510

Keyword

sign

Author

Paul D. Hanna, Mar 11 2005

Status

approved