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A097040 a(n) = 2*sum(C(n,2k+1)*F(2k), k=0..floor((n-1)/2)), where F(n) are Fibonacci numbers A000045. 1
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 (list; graph; refs; listen; history; text; internal format)
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

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Last modified January 29 04:57 EST 2020. Contains 331335 sequences. (Running on oeis4.)