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A181553
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Coefficient of x^n in (x^2 + 98*x + 1)^n.
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3
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1, 98, 9606, 941780, 92352070, 9058034748, 888610349724, 87192397723368, 8557276143987270, 840005101192014380, 82474083957903064756, 8099197733721011526168, 795527368821049695145756, 78154959591300863484042200, 7679729103551077344613236600, 754784236214755050742369782480, 74197094919316919158188333048390
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OFFSET
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0,2
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COMMENTS
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On Jan 29 2011, Zhi-Wei Sun conjectured that Sum_{k>=0} (40k + 3)*a(k)*binomial(4k, 2k)*binomial(2k, k)/112^(2k) = 70*sqrt(21)/(9*Pi).
He also conjectured that 2n(2n + 1)*binomial(2n, n) divides Sum_{k=0..n-1} (40k + 3)*a(k)*binomial(4k, 2k)*binomial(2k, k)112^{2(n - 1 - k)} for each n = 2, 3, 4, ... and that for any prime p different from 2 and 7 we have the congruence Sum_{k=0..p-1} (40k + 3)*a(k)*binomial(4k, 2k)*binomial(2k, k)/112^(2k) == p(-21/p)(5 - 2(-2/p)) (mod p^2).
Another conjecture of his states that for any prime p == 1, 3 (mod 8) with p = x^2 + 2y^2 (x, y integers) we have Sum_{k=0..p-1} a(k)*binomial(4k, 2k)*binomial(2k, k)/112^(2k) == (7/p)(4x^2 - 2p) (mod p^2). By a result of Sun, Sum_{k=0..p-1} (40k + 3)*a(k)*binomial(4k, 2k)*binomial(2k, k)/112^(2k) == 0 (mod p^2) for any prime p = 5,7 (mod 8).
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LINKS
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FORMULA
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a(n) = Sum_{k = 0..floor(n/2)} binomial(n, 2*k)*binomial(2*k, k) *(98)^{n - 2*k}.
G.f.: (9600*x^2 - 196*x + 1)^(-1/2).
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EXAMPLE
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For n = 2 we have a(2) = coefficient of x^2 in (x^2 + 98x + 1)^2 = 9606.
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MATHEMATICA
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A[n_] := If[n > 0, Coefficient[(x^2 + 98x + 1)^n, x^n], 1]; Table[A[n], {n, 0, 20}]
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PROG
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(PARI) x='x+O('x^20); Vec((9600*x^2 - 196*x + 1)^(-1/2)) \\ G. C. Greubel, Mar 06 2017
(Magma) m:=20; R<x>:=PowerSeriesRing(Rationals(), m); Coefficients(R!( 1/Sqrt(9600*x^2 - 196*x + 1) )); // G. C. Greubel, Nov 10 2018
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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