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A167871
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a(n) = 64^n * Sum_{k=0..n} binomial(2*k,k)^3 / 64^k.
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5
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1, 72, 4824, 316736, 20614104, 1335305664, 86248451520, 5560325134848, 357992555533272, 23026456586057408, 1479999826835627328, 95071036081670530560, 6104320340924619384256, 391801560518407856592384
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OFFSET
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0,2
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COMMENTS
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p^2 divides all a(n) from n = (p-1)/2 to n = p-1 for prime p of the form p = 4k+3, p = {3,7,11,19,23,31,43,47,59,...} = A002145.
p^2 divides all a(n) from n = (2p-1 - (p-1)/2) to n = 2p-1 for prime p of the form p = 4k+3.
p^2 divides all a(n) from n = (3p-1 - (p-1)/2) to n = 3p-1 for prime p of the form p = 4k+3.
p^2 divides all a(n) from n = (p^2-1)/2 to n = p^2-1 for prime p of the form p = 4k+3.
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REFERENCES
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W. Feller, An Introduction to Probability Theory and Its Applications, Vol. 1, 3rd ed., Wiley, 1968, p. 361.
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LINKS
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FORMULA
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a(n) = 64^n * Sum_{k=0..n} binomial(2*k,k)^3 / 64^k.
Recurrence: n^3*a(n) = 8*(4*n-1)*(4*n^2 - 2*n + 1)*a(n-1) - 512*(2*n-1)^3 *a(n-2). - Vaclav Kotesovec, Aug 14 2013
a(n) ~ 64^n*(Pi/GAMMA(3/4)^4 - 2/(Pi^(3/2)*sqrt(n))). - Vaclav Kotesovec, Aug 14 2013
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MATHEMATICA
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Table[64^n Sum[Binomial[2k, k]^3/64^k, {k, 0, n}], {n, 0, 20}] (* Vincenzo Librandi, Mar 26 2012 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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