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We conjecture that the sequence satisfies the stronger congruences supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. Examples of these congruences are given below.
More generally, for each integer m, we conjecture that the sequence a_m(n) := the n-th order Taylor polynomial of c(x)^(m*n) evaluated at x = 1 satisfies the same congruencessupercongruences. For cases see A099837 (m = -2), A100219 (m = -1), A000012 (m = 0), A333093 (m = 1), A333094 (m = 2), A333095 (m = 3), A333097 (m = 5).
a(n) = Sum_{k = 0..n} 4*n/(4*n+k)*binomial(4*n+2*k-1, k) for n >= 1.
a(n) = Sum_{k = 0..n} 4*n/(4*n+2*k)*binomial(4*n+2*k, k) for n >= 1. - Peter Bala, May 03 2024
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We conjecture that the sequence satisfies the stronger supercongruences congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. Examples of these congruences are given below.
More generally, for each integer m, we conjecture that the sequence a_m(n) := the n-th order Taylor polynomial of c(x)^(m*n) evaluated at x = 1 satisfies the same supercongruencescongruences. For cases see A099837 (m = -2), A100219 (m = -1), A000012 (m = 0), A333093 (m = 1), A333094 (m = 2), A333095 (m = 3), A333097 (m = 5).
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Examples of supercongruencescongruences:
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a(n) ~ 2^(6*n + 3) * 3^(6*n + 3/2) / (31 * sqrt(Pi*n) * 5^(5*n + 1/2)). - Vaclav Kotesovec, Mar 28 2020
Join[{1}, Table[4*Binomial[6*n-1, n] * HypergeometricPFQ[{1, -5*n, -n}, {1/2 - 3*n, 1 - 3*n}, 1/4]/5, {n, 1, 20}]] (* Vaclav Kotesovec, Mar 28 2020 *)
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