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A364111
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a(n) = Sum_{k = 0..n} binomial(n+k-1,k)^2 * binomial(2*n-2*k,n-k) * binomial(2*k,k).
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2
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1, 4, 76, 2560, 106060, 4864504, 237354880, 12079462560, 633885607500, 34050190896040, 1863047125801576, 103465470769890112, 5817117095161011328, 330450303019252600240, 18937657945720403830240, 1093557503049551583194560, 63566414131528881235953228, 3716526456851323626808570632
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OFFSET
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0,2
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COMMENTS
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Compare with the Domb numbers A002895, which are defined by A002895(n) = Sum_{k = 0..n} binomial(n,k)^2 * binomial(2*n-2*k,n-k) * binomial(2*k,k).
The supercongruences A002895(n*p^r) == A002895(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and positive integers n and r (see Osburn and Sahu).
We conjecture that the present sequence satisfies the same supercongruences.
More generally, let A >= 2, B and C be positive integers. Then we conjecture that the sequence whose terms are given by Sum_{k = 0..n} binomial(n+k-1,k)^A * binomial(2*n-2*k,n-k)^B * binomial(2*k,k)^C also satisfies the same supercongruences.
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LINKS
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FORMULA
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a(n) = Sum_{k = 0..n} binomial(-n,k)^2 * binomial(2*n-2*k,n-k) * binomial(2*k,k).
a(n) = binomial(2*n,n)*hypergeom([-n, n, n, 1/2], [1, 1, 1/2 - n], 1).
a(n) ~ 2^(6*n-1) / (sqrt(3) * Pi^(3/2) * n^(3/2)). - Vaclav Kotesovec, Jul 09 2023
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MAPLE
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seq(add(binomial(n+k-1, k)^2 * binomial(2*n-2*k, n-k) * binomial(2*k, k)), n = 0..20);
# faster program for large n
seq(simplify(binomial(2*n, n)*hypergeom([-n, n, n, 1/2], [1, 1, 1/2 - n], 1)), n = 0..20);
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MATHEMATICA
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Table[Binomial[2*n, n] * HypergeometricPFQ[{-n, n, n, 1/2}, {1, 1, 1/2 - n}, 1], {n, 0, 20}] (* Vaclav Kotesovec, Jul 09 2023 *)
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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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