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A363987
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a(n) = Sum_{k = 0..n} (-4)^(n-k)*binomial(n,k)*binomial(3*n+k,k)*binomial(2*k,k).
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1
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1, 4, 72, 1336, 27816, 609504, 13849776, 322761632, 7665078696, 184706717728, 4502875483072, 110816709352848, 2748733193029488, 68633934354206784, 1723482744564382272, 43491464193991134336, 1102200555724995901864, 28038534962615758665120, 715655490507796997136960, 18321118155905647835092032
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OFFSET
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0,2
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COMMENTS
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The sequence of Franel numbers A000172 satisfies the identity A000172(n) = Sum_{k = 0..n} (-4)^(n-k)*binomial(n,k)*binomial(n+2*k,2*k)*binomial(2*k,k). The present sequence comes from a modification of the right-hand side of the identity.
The Franel numbers satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and positive integers n and r. We conjecture that the present sequence satisfies the same supercongruences.
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LINKS
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FORMULA
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a(n) = (-4)^n*hypergeom(-n, 3*n+1, 1/2], [1, 1], 1).
P-recursive: n^2*(4*n - 1)*(3*n - 1)*(3*n - 2)*(4*n - 3)*P(n-1)*a(n) = 16*(57753*n^10 - 500526*n^9 + 1886256*n^8 - 4059105*n^7 + 5508054*n^6 - 4910492*n^5 + 2904096*n^4 - 1121483*n^3 + 269766*n^2 - 36410*n + 2100)*a(n-1) + 192*(3*n - 5)^2*(n - 1)^2*(3*n - 4)^2*P(n)*a(n-2), where P(n) = 279*n^4 - 465*n^3 + 276*n^2 - 68*n + 6 and a(0) = 1, a(1) = 4.
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MAPLE
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seq(add((-4)^(n-k)*binomial(n, k)*binomial(3*n+k, k)*binomial(2*k, k), k = 0..n), n = 0..20);
# alternative faster program for large n
seq(simplify((-4)^n*hypergeom([-n, 3*n+1, 1/2], [1, 1], 1)), n = 0..20);
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MATHEMATICA
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Table[(-4)^n*HypergeometricPFQ[{-n, 3*n+1, 1/2}, {1, 1}, 1], {n, 0, 20}] (* Vaclav Kotesovec, Jul 17 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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EXTENSIONS
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STATUS
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approved
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