A097040 a(n) = 2*sum(C(n,2k+1)*F(2k), k=0..floor((n-1)/2)), where F(n) are Fibonacci numbers A000045.

0, 0, 2, 8, 26, 76, 212, 576, 1542, 4092, 10802, 28424, 74648, 195808, 513242, 1344672, 3521994, 9223284, 24151052, 63235040, 165562430, 433465780, 1134856802, 2971140048, 7778620656, 20364814656, 53315973362, 139583348216

Offset

1,3

Comments

Create a triangle with first column T(n,1)=A000045(n) for n=0,1,2... The remaining terms T(r,c)=T(r,c-1)+T(r-1,c-1). The sum of all terms for the first n+1 rows of this triangle=a(n+2). The sum of the terms in row(n+1)= 0, 2, 6, 18, 50, 136, 364...with partial sums of these sums duplicating this sequence 0, 2, 8, 26, 76, 212, 576... - J. M. Bergot, Dec 19 2012

Links

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (4,-3,-2,1).

Formula

a(n) = F(2n-1)-F(n+1) = 2*A056014(n).

G.f. -2*x^3 / ( (x^2-3*x+1)*(x^2+x-1) ). - R. J. Mathar, Jan 08 2013

Mathematica

f[n_] := f[n] = f[n - 1] + f[n - 2]; f[0] = 0; f[1] = 1; Table[2 Sum[Binomial[n, 2k + 1]f[2k], {k, 0, Floor[(n - 1)/2]}], {n, 1, 30}]
Table[Fibonacci[2n-1]-Fibonacci[n+1], {n, 30}] (* Harvey P. Dale, Oct 05 2011 *)
LinearRecurrence[{4, -3, -2, 1}, {0, 0, 2, 8}, 29] (* Robert G. Wilson v, Dec 26 2012 *)

Crossrefs

Sequence in context: A167826 A301995 A325926 * A302237 A224289 A124721

Adjacent sequences: A097037 A097038 A097039 * A097041 A097042 A097043

Keyword

easy,nonn

Author

Mario Catalani (mario.catalani(AT)unito.it), Jul 22 2004

Status

approved