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A363985
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a(n) = Sum_{k = 0..n} (-4)^(n-k)*binomial(n,k)*binomial(2*n+k,k)*binomial(2*k,k).
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7
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1, 2, 26, 272, 3418, 44252, 597104, 8249152, 116158298, 1659335492, 23979247276, 349798313152, 5142733169776, 76108788764192, 1132729444052288, 16940944956246272, 254449319912898394, 3836162994088105172, 58028561918702719604
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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.
More generally, define two families of sequences {u_m(n): n >= 0} and {v_m(n): n >= 0}, depending on an integer parameter m, by u_m(n) = Sum_{k = 0..n} (-4)^(n-k)*binomial(n,k)*binomial(m*n + k,k)*binomial(2*k,k) and v_m(n) = Sum_{k = 0..n} (-4)^(n-k)*binomial(n,k)*binomial(m*n + 2*k,2*k)*binomial(2*k,k). In this notation, the Franel numbers A000172 = v_1. Note that u_0(n) = v_0(n) = (-1)^n*binomial(2*n,n). We conjecture that, for all m in Z, the sequences u_m and v_m satisfy the above supercongruences.
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LINKS
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FORMULA
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a(n) = (-4)^n*hypergeom(-n, 2*n+1, 1/2], [1, 1], 1).
P-recursive: (20*n^2 - 55*n + 38)*n^2*(2*n - 1)^2*a(n) = (960*n^6 - 4560*n^5 + 8564*n^4 - 8107*n^3 + 4085*n^2 - 1044*n + 108)*a(n-1) + 64*(20*n^2 - 15*n + 3)*(n - 1)^2*(2*n - 3)^2*a(n-2) with a(0) = 1 and a(1) = 2.
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MAPLE
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seq(add((-4)^(n-k)*binomial(n, k)*binomial(2*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, 2*n+1, 1/2], [1, 1], 1)), n = 0..20);
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MATHEMATICA
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Table[(-4)^n*HypergeometricPFQ[{-n, 2*n+1, 1/2}, {1, 1}, 1], {n, 0, 20}] (* Vaclav Kotesovec, Jul 17 2023 *)
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PROG
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(Python)
from sympy import hyper, hyperexpand, S
def A363985(n): return int(hyperexpand(hyper((-n, (m:=n<<1)+1, S.Half), [1, 1], 1))*(-(1<<m) if n&1 else 1<<m)) # Chai Wah Wu, Jul 10 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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