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a(n) = (8*n)!*(5*n)!*(3*n)! / ( (6*n)!*(4*n)!^2*n!^2 ).
2

%I #26 Oct 31 2024 01:35:25

%S 1,70,17550,5567380,1960044750,732012601320,283986961467300,

%T 113142133870180800,45969979122504907470,18961650930856541865100,

%U 7915377251895103264073800,3336455614603881320759754000,1417729131896719482585245182500,606517077508008639090614765297280

%N a(n) = (8*n)!*(5*n)!*(3*n)! / ( (6*n)!*(4*n)!^2*n!^2 ).

%H Paolo Xausa, <a href="/A364305/b364305.txt">Table of n, a(n) for n = 0..300</a>

%H R. Meštrović, <a href="http://arxiv.org/abs/1111.3057">Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011)</a>, arXiv:1111.3057 [math.NT], 2011.

%F a(n) = Sum_{k = 0..n} binomial(8*n, n-k)^2 * binomial(6*n+k-1, k).

%F a(n) = [x^n] (1 - x)^(2*n) * P(8*n,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial.

%F a(n) = (5/12)*(5*n-1)*(5*n-2)*(5*n-3)*(5*n-4)*(8*n-1)*(8*n-3)*(8*n-5)*(8*n-7)/((4*n-1)*(4*n-3)*(6*n-1)*(6*n-5)*n^2*(2*n-1)^2) * a(n-1) with a(0) = 1.

%F a(n) ~ c^n * sqrt(5)/(4*Pi*n), where c = (2^2)*(5^5)/(3^3).

%F a(n) = binomial(8*n,2*n)*binomial(5*n,n)*binomial(2*n,n)/binomial(4*n,n) = A001449(n) * A211421(n).

%F a(p) == a(1) (mod p^3) for all primes p >= 5.

%F Conjecture: the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and all positive integers n and k [added 16 Oct 2024: the conjecture follows from Meštrović, equation 39, since a(n) = binomial(8*n, 2*n)*binomial(5*n, n)* binomial(2*n, n)/binomial(4*n, n)].

%F a(n) = [x^n] G(x)^(10*n), where the power series G(x) = 1 + 7*x + 412*x^2 + 55524*x^3 + 10088066*x^4 + 2146473322*x^5 + 503731865112*x^6 + ... appears to have integer coefficients.

%F exp(Sum_{n >= 1} a(n)*x^n/n) = F(x)^10, where the power series F(x) = 1 + 7*x + 902*x^2 + 191779*x^3 + 50706776*x^4 + 15153397742*x^5 + 4898289306180*x^6 + ... appears to have integer coefficients.

%F From _Peter Bala_, Oct 16 2024: (Start)

%F For integer r and positive integer s, define sequences u(n) = { [x^(s*n)] G(x)^(r*n) : n >= 0 } and v(n) = { [x^(s*n)] F(x)^(r*n) : n >= 0 }, with the power series F(x) and G(x) as defined above. We conjecture that both sequences {u(n)} and {v(n)} satisfy the above supercongruences mod p^(3*k). (End)

%p seq( (8*n)!*(5*n)!*(3*n)! / ( (6*n)!*(4*n)!^2*n!^2 ), n = 0..13);

%t A364305[n_]:=(8n)!(5n)!(3n)!/((6n)!(4n)!^2n!^2);Array[A364305,15,0] (* _Paolo Xausa_, Oct 06 2023 *)

%Y Row 8 of A364303.

%Y Cf. A001449, A211421.

%K nonn,easy,changed

%O 0,2

%A _Peter Bala_, Jul 21 2023