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A127358
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a(n) = Sum_{k=0..n} binomial(n, floor(k/2))*2^(n-k).
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8
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1, 3, 8, 21, 54, 138, 350, 885, 2230, 5610, 14088, 35346, 88596, 221952, 555738, 1391061, 3480870, 8708610, 21783680, 54483510, 136254964, 340729788, 852000828, 2130354786, 5326563004
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OFFSET
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0,2
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COMMENTS
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Hankel transform is (-1)^n. In general, given r >= 0, the sequence given by Sum_{k=0..n} binomial(n, floor(k/2))*r^(n-k)} has Hankel transform (1-r)^n. The sequence is the image of the sequence with g.f. (1+x)/(1-2x) under the Chebyshev mapping g(x) -> (1/sqrt(1-4x^2))g(xc(x^2)), where c(x) is the g.f. of the Catalan numbers A000108.
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LINKS
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FORMULA
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G.f.: (1/sqrt(1 - 4x^2))(1 + x*c(x^2))/(1 - 2*x*c(x^2)).
a(n) is the sum of top row terms of M^n, M is an infinite square production matrix as follows:
2, 1, 0, 0, 0, ...
1, 0, 1, 0, 0, ...
0, 1, 0, 1, 0, ...
0, 0, 1, 0, 1, ...
0, 0, 0, 1, 0, ...
... (End)
Conjecture: 2*n*a(n) + (-5*n-4)*a(n-1) + 2*(-4*n+13)*a(n-2) + 20*(n-2)*a(n-3) = 0. - R. J. Mathar, Nov 30 2012
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EXAMPLE
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a(3) = 21 = (12 + 6 + 2 + 1), where the top row of M^3 = (12, 6, 2, 1).
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MATHEMATICA
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Table[Sum[Binomial[n, Floor[k/2]]2^(n-k), {k, 0, n}], {n, 0, 30}] (* Harvey P. Dale, Jun 03 2012 *)
CoefficientList[Series[(1 + 2*x - Sqrt[1 - 4*x^2])/(2*Sqrt[1 - 4*x^2]*(x - 1 + Sqrt[1 - 4*x^2])), {x, 0, 50}], x] (* G. C. Greubel, May 22 2017 *)
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PROG
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(PARI) x='x+O('x^50); Vec((1 + 2*x - sqrt(1 - 4*x^2))/(2*sqrt(1 - 4*x^2)*(x - 1 + sqrt(1 - 4*x^2)))) \\ G. C. Greubel, May 22 2017
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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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STATUS
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approved
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