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%I #16 Jun 13 2015 00:51:23
%S 0,0,2,8,26,76,212,576,1542,4092,10802,28424,74648,195808,513242,
%T 1344672,3521994,9223284,24151052,63235040,165562430,433465780,
%U 1134856802,2971140048,7778620656,20364814656,53315973362,139583348216
%N a(n) = 2*sum(C(n,2k+1)*F(2k), k=0..floor((n-1)/2)), where F(n) are Fibonacci numbers A000045.
%C 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
%H Harvey P. Dale, <a href="/A097040/b097040.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-3,-2,1).
%F a(n) = F(2n-1)-F(n+1) = 2*A056014(n).
%F G.f. -2*x^3 / ( (x^2-3*x+1)*(x^2+x-1) ). - _R. J. Mathar_, Jan 08 2013
%t 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}]
%t Table[Fibonacci[2n-1]-Fibonacci[n+1],{n,30}] (* _Harvey P. Dale_, Oct 05 2011 *)
%t LinearRecurrence[{4, -3, -2, 1}, {0, 0, 2, 8}, 29] (* _Robert G. Wilson v_, Dec 26 2012 *)
%K easy,nonn
%O 1,3
%A Mario Catalani (mario.catalani(AT)unito.it), Jul 22 2004
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