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
KEYWORD
easy,nonn
AUTHOR
Mario Catalani (mario.catalani(AT)unito.it), Jul 22 2004
STATUS
approved