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%I #46 Jul 21 2023 10:07:35
%S 1,4,32,328,3840,48504,641984,8765712,122370048,1736921560,
%T 24975268032,362872728816,5317470233088,78479369810352,
%U 1165299414952320,17393306836535328,260791399517110272,3925811865435871896,59305018671515758784
%N a(n) = Sum_{k = 0..n} 4^(n-k)*binomial(n,k)*binomial(n-1,k)*binomial(2*k,k).
%C 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 the following modification of the right-hand side of the identity: a(n) = Sum_{k = 0..n} (-4)^(n-k)*binomial(n,k)*binomial(-n+k,k)* binomial(2*k,k), multipled by a factor (-1)^n to give positive terms.
%C 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.
%C From _Peter Bala_, Jul 07 2023 (Start):
%C 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).
%C 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).
%C We conjecture that the present sequence satisfies the same supercongruences. (End)
%H Seiichi Manyama, <a href="/A362676/b362676.txt">Table of n, a(n) for n = 0..833</a>
%H Robert Osburn and Brundaban Sahu, <a href="https://arxiv.org/abs/1201.6195">A supercongruence for generalized Domb numbers</a>, arXiv:1201.6195v2 [math.NT], Functiones et Approximatio. Comment. Math, Vol. 48, No 1, March 2013, 29-36.
%F a(n) = 4^n * hypergeom ([-n, 1 - n, 1/2], [1, 1], 1).
%F From _Vaclav Kotesovec_, Jul 04 2023: (Start)
%F Recurrence: (n-1)*n^2*(3*n^2 - 9*n + 7)*a(n) = 4*(n-1)*(15*n^4 - 60*n^3 + 80*n^2 - 40*n + 8)*a(n-1) - 4*(n-2)*(4*n - 7)*(4*n - 5)*(3*n^2 - 3*n + 1)*a(n-2).
%F a(n) ~ 2^(4*n - 1/2) / (Pi*n). (End)
%F a(n) = Sum_{k = 0..n} (-1)^k * binomial(-n,k) * binomial(2*n-2*k,n-k) * binomial(2*k,k). Cf. A081085. _Peter Bala_, Jul 07 2023
%F a(n) = binomial(2*n,n)*hypergeom([-n, n, 1/2], [1, 1/2 - n], 1). - _Peter Bala_, Jul 07 2023
%p seq(add(4^(n-k)*binomial(n,k)*binomial(n-1,k)*binomial(2*k,k), k = 0..n), n = 0..20);
%p # alternative faster program for large n
%p seq(simplify(4^n*hypergeom([-n, 1 - n, 1/2], [1, 1], 1)), n = 0..20);
%p # alternative (Peter Bala Jul 07 2023)
%p seq(add(binomial(n+k-1,k) * binomial(2*n-2*k,n-k) * binomial(2*k,k), k = 0..n), n = 0..20);
%t Table[4^n * HypergeometricPFQ[{-n, 1 - n, 1/2}, {1, 1}, 1], {n, 0, 20}] (* _Vaclav Kotesovec_, Jul 04 2023 *)
%o (Python)
%o from sympy import hyper, hyperexpand, S
%o def A362676(n): return int(hyperexpand(hyper((-n, 1-n, S.Half), [1,1], 1))*(1<<(n<<1))) # _Chai Wah Wu_, Jul 10 2023
%Y Cf. A000172, A363985 - A363990, A002895, A081085, A364111.
%K nonn,easy
%O 0,2
%A _Peter Bala_, Jul 03 2023
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