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A295112
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a(n) = Sum_{k=0..n} binomial(n,2*k)*binomial(2*k,k)/(2*k-1).
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3
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-1, -1, 1, 5, 13, 29, 63, 139, 317, 749, 1827, 4575, 11699, 30419, 80161, 213573, 574253, 1556077, 4244835, 11647151, 32122231, 88995879, 247573565, 691246369, 1936445619, 5441165699, 15331341373, 43308322049, 122624939677, 347957102909, 989335822559, 2818200111867
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OFFSET
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0,4
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COMMENTS
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As binomial(2*k,k) = 2*(2*k-1)*A000108(k-1) for all k = 1,2,..., we see that a(n) is always an odd integer. Clearly, a(n) > 0 for all n > 1. a(n) can be viewed as an analog of Motzkin numbers, which should have some combinatorial interpretations.
Conjecture: The sequence a(n+1)/a(n) (n = 5,6,...) is strictly increasing with limit 3, and the sequence a(n+1)^(1/(n+1))/a(n)^(1/n) (n = 9,10,...) is strictly decreasing to the limit 1.
See also A295113 for a conjecture involving the current sequence.
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LINKS
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Zhi-Wei Sun, Conjectures involving arithmetical sequences, arXiv:1208.2683 [math.CO], 2012-2013; in: Number Theory: Arithmetic in Shangri-La (eds., S. Kanemitsu, H. Li and J. Liu), Proc. 6th China-Japan Seminar (Shanghai, August 15-17, 2011), World Sci., Singapore, 2013, pp. 244-258.
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FORMULA
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Via the Zeilberger algorithm we have the recurrence: (n+3)*a(n+3) = (3n+7)*a(n+2) + (n-5)*a(n+1) - 3*(n+1)*a(n) for any nonnegative integer n.
a(n) = -hypergeom([-1/2, 1/2 - n/2, -n/2], [1/2, 1], 4). - Peter Luschny, Nov 15 2017
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EXAMPLE
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a(3) = 5 since binomial(3,2*0)*binomial(2*0,0)/(2*0-1) + binomial(3,2*1)*binomial(2*1,1)/(2*1-1) = -1 + 3*2 = 5.
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MAPLE
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a := n -> -hypergeom([-1/2, 1/2 - n/2, -n/2], [1/2, 1], 4):
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MATHEMATICA
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W[n_]:=W[n]=Sum[Binomial[n, 2k]Binomial[2k, k]/(2k-1), {k, 0, n/2}]; Table[W[n], {n, 0, 35}]
a[n_] := -AppellF1[-n, -1/2, -1/2, 1, 2, -2]; Table[a[n], {n, 0, 31}] (* Peter Luschny, Nov 15 2017 *)
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CROSSREFS
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KEYWORD
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sign,easy
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AUTHOR
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STATUS
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approved
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