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A097040
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a(n) = 2*sum(C(n,2k+1)*F(2k), k=0..floor((n-1)/2)), where F(n) are Fibonacci numbers A000045.
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1
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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
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OFFSET
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1,3
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COMMENTS
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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
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LINKS
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FORMULA
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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
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MATHEMATICA
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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 *)
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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Mario Catalani (mario.catalani(AT)unito.it), Jul 22 2004
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STATUS
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approved
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