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A357513
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a(n) = numerator of Sum_{k = 1..n} (1/k^3) * binomial(n,k)^2 * binomial(n+k,k)^2 for n >= 1 with a(0) = 0
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6
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0, 4, 81, 14651, 956875, 1335793103, 697621869, 3929170277787, 573290332967211, 8235727724024089939, 172296487023049395523, 5032311952710217004416313, 114828404520381550476341513, 5947240175728534283432460589661, 144126887537331651710781931325261
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OFFSET
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0,2
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COMMENTS
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Let m be a nonnegative integer and set u(n) = the numerator of Sum_{k = 0..n} 1/k^(2*m+1) * binomial(n,k)^2 * binomial(n+k,k)^2. We conjecture that u(p-1) == 0 (mod p^4) for all primes p, with a finite number of exceptions that depend on m.
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LINKS
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FORMULA
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Conjecture: a(p-1) == 0 (mod p^4) for all primes p >= 3 except p = 7 (checked up to p = 499).
Note: the Apery numbers A(n) = A005259(n) = Sum_{k = 0..n} binomial(n,k)^2 * binomial(n+k,k)^2 satisfy the supercongruence A(p-1) == 1 (mod p^3) for all primes p >= 5 (see, for example, Straub, Introduction).
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EXAMPLE
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a(11 - 1) = 172296487023049395523 = (11^4)*(43^2)*6163*1032705769 == 0 (mod 11^4).
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MAPLE
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seq(numer(add( (1/k^3) * binomial(n, k)^2 * binomial(n+k, k)^2, k = 1..n )), n = 0..20);
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PROG
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(PARI) a(n) = if (n, numerator(sum(k=1, n, binomial(n, k)^2*binomial(n+k, k)^2/k^3)), 0); \\ Michel Marcus, Oct 04 2022
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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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STATUS
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approved
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