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

Author

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

Keywords

• easy
• nonn

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).

Programs

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 *)

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