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A167859
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a(n) = 4^n * Sum_{k=0..n} binomial(2*k, k)^2 / 4^k.
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7
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1, 8, 68, 672, 7588, 93856, 1229200, 16695424, 232418596, 3293578784, 47309094672, 686870685312, 10059942413584, 148412250014336, 2202990595617344, 32873407393419776, 492791264816231204
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OFFSET
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0,2
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COMMENTS
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Every a(n) from a((p-1)/2) to a(p-1) is divisible by prime p for p = {7, 47, 191, 383, 439, 1151, 1399, 2351, 2879, 3119, 3511, 3559, ...} = A167860, apparently a subset of primes of the form 8n+7 (A007522).
7^3 divides a(13) and 7^2 divides a(10)-a(13).
Every a(n) from a(kp-1 - (p-1)/2) to a(kp-1) is divisible by prime p from A167860.
Every a(n) from a((p^2-1)/2) to a(p^2-1) is divisible by prime p from A167860. For p=7 every a(n) from a((p^3-1)/2) to a(p^3-1) and from a((p^4-1)/2) to a(p^4-1)is divisible by p^2.
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LINKS
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FORMULA
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Recurrence: n^2*a(n) = 4*(5*n^2 - 4*n + 1)*a(n-1) - 16*(2*n - 1)^2*a(n-2). - Vaclav Kotesovec, Oct 20 2012
G.f.: 2*EllipticK(4*sqrt(x))/(Pi*(1-4*x)), where EllipticK is the complete elliptic integral of the first kind, using the Gradshteyn and Ryzhik convention, also used by Maple. In the convention of Abramowitz and Stegun, used by Mathematica, this would be written as 2*K(16*x)/(Pi*(1-4*x)). - Robert Israel, Sep 21 2016
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MAPLE
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add( (binomial(2*k, k)/2^k)^2, k=0..n) ;
4^n*% ;
end proc:
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MATHEMATICA
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Table[4^n*Sum[Binomial[2*k, k]^2/4^k, {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Oct 20 2012 *)
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PROG
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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