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a(n) = Sum_{k = 0..n-1} (-1)^k*binomial(n,k)*binomial(n-1,k)^2.
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%I #23 Jul 27 2023 08:13:02

%S 0,1,-1,-8,15,126,-280,-2400,5775,50050,-126126,-1100736,2858856,

%T 25069968,-66512160,-585307008,1577585295,13919870250,-37978905250,

%U -335813478000,925166131890,8194328596740,-22754499243840,-201822515032320,564121960420200,5009403008531376

%N a(n) = Sum_{k = 0..n-1} (-1)^k*binomial(n,k)*binomial(n-1,k)^2.

%C Conjecture: the supercongruence a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) holds for all primes p >= 5 and positive integers n and k.

%C Compare with A005258(n-1) = Sum_{k = 0..n-1} (-1)^k*binomial(-n,k)*binomial(n-1,k)^2.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Dixon%27s_identity">Dixon's identity</a>.

%F a(n) = (1/n^2) * Sum_{k = 0..n} (-1)^(n+k) * k^2 * binomial(n,k)^3 for n >= 1.

%F a(n) = (1/(3*n)) * Sum_{k = 0..n} (-1)^(n+k+1) * (n - 3*k) * binomial(n,k)^3 for n >= 1.

%F a(2*n) = (-1)^n * (1/6) * (3*n)!/n!^3 for n >= 1; a(2*n+1) = (-1)^n * (3*n+1)/(2*n+1) * (3*n)!/n!^3.

%F a(2*n) = (1/3)*A361716(2*n); a(2*n+1) = A361711(2*n+1) = A361716(2*n+1).

%F a(2*n) = (1/6)*A245086(2*n) = (1/6)*(-1)^n*A006480(n) for n >= 1.

%F a(n) = hypergeom([-n, 1 - n, 1 - n], [1, 1], 1);

%F P-recursive: n^2*(n - 1)*(6*n^2 - 16*n + 11)*a(n) = - 6*(n - 1)*(3*n^2 - 6*n + 2)*a(n-1) - (3*n - 4)*(3*n - 5)*(3*n - 6)*(6*n^2 - 4*n + 1)*a(n-2) with a(0) = 0 and a(1) = 1.

%F a(n) = Sum_{k = 0..n-1} (-1)^k*binomial(n,k)*binomial(n+k-1,k)*binomial(2*n-k-1,n). - _Peter Bala_, Jul 01 2023

%p seq( add((-1)^k*binomial(n,k)*binomial(n-1,k)^2, k = 0..n-1), n = 0..25);

%o (PARI) a(n) = sum(k = 0, n-1, (-1)^k*binomial(n,k)*binomial(n-1,k)^2); \\ _Michel Marcus_, Mar 26 2023

%Y Cf. A005258, A006480, A245086, A361711, A361716.

%K sign,easy

%O 0,4

%A _Peter Bala_, Mar 21 2023