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A337247
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a(n) = (Sum_{k=0..n-1} (-1)^k * (4k+1) * 160^(n-1-k) * C(2k,k) * Sum_{j=0..k} C(k,j) * C(k+2j,2j) * C(2j,j) * (-20)^(k-j)) / (n * C(2n,n)).
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2
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25, 809, 23020, 730325, 27867142, 1117643720, 42658771456, 1558395721085, 57260792702050, 2179584653311070, 84835851591609400, 3292250198848240760, 126379831667243976400, 4841030410501144484000, 186842197443136622824960, 7269291788529191112814925, 283472902036823148786161530
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OFFSET
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2,1
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COMMENTS
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Conjecture 1: a(n) is a positive integer for each n > 1. Moreover, a(n) is odd if and only if n = 2^k + 1 for some nonnegative integer k.
Conjecture 2: The infinite series Sum_{k>=0} (4*k+1)/(-160)^k * C(2k,k) * Sum_{j=0..k} C(k,j) * C(k+2j,2j) * C(2j,j) * (-20)^(k-j) has the value sqrt(30)/(5*Pi)*(5+c^(1/3))/c^(1/6), where c = 145 + 30*sqrt(6).
Conjecture 3. Let p be an odd prime different from 5, and let S(p) denote the sum Sum_{k=0..p-1} C(2k,k)/(-160)^k * Sum_{j=0..k} C(k,j) * C(k+2j,2j) * C2j,j) * (-20)^(k-j).
(i) If p == 1,3 (mod 8) and p = x^2 + 2*y^2 with x and y integers, then S(p) == (5/p)*(4x^2 - 2p) (mod p^2), where (5/p) is the Legendre symbol.
(ii) If p == 5,7 (mod 8), then S(p) == 0 (mod p^2).
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LINKS
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Zhi-Wei Sun, Table of n, a(n) for n = 2..100
Zhi-Wei Sun, Two curious series for 1/Pi, Question 369569 at MathOverflow, August 19-20, 2020.
Zhi-Wei Sun, New series for powers of Pi and related congruences, Electron. Res. Arch. 28(2020), no. 3, 1273-1342.
Zhi-Wei Sun, Some new series for 1/Pi motivated by congruences, arXiv:2009.04379 [math.NT], 2020.
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EXAMPLE
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a(2) = (160 - (4 + 1)*C(2,1)*(-20 + C(3,2)*C(2,1)))/(2*C(4,2)) = 300/12 = 25.
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MATHEMATICA
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a[n_]:=a[n]=Sum[(4k+1)(-1)^k*160^(n-1-k)*Binomial[2k, k]*Sum[Binomial[k, j]Binomial[k+2j, 2j]Binomial[2j, j](-20)^(k-j), {j, 0, k}], {k, 0, n-1}]/(n*Binomial[2n, n])
Table[a[n], {n, 2, 18}]
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CROSSREFS
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Cf. A000796, A000984, A336981, A336982.
Sequence in context: A277584 A223228 A132540 * A142998 A218479 A183879
Adjacent sequences: A337244 A337245 A337246 * A337248 A337249 A337250
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KEYWORD
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nonn
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AUTHOR
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Zhi-Wei Sun, Aug 20 2020
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STATUS
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approved
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