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%I #19 Mar 30 2023 05:08:24
%S 2,66,2540,110530,5197752,257490156,13238524728,699822144450,
%T 37800431926400,2077184897317816,115757876008359312,
%U 6526739641107783916,371641758587326581200,21341134886976332825400,1234474507620634579565040
%N a(n) = (1/n) * Sum_{k = 0..n} (-1)^(n+k) * (n + 2*k) * binomial(n+k-1,k)^3.
%C Compare with the closed form evaluation of the binomial sums (1/n) * Sum_{k = 0..n} (-1)^(n+k) * (n + 2*k) * binomial(n+k-1,k) = binomial(2*n,n) and (1/n) * Sum_{k = 0..n} (n + 2*k) * binomial(n+k-1,k)^2 = binomial(2*n,n)^2.
%C The central binomial coefficients u(n) := binomial(2*n,n) = A000984(n) satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) for positive integers n and r and all primes p >= 5. We conjecture that the present sequence satisfies the same congruences.
%C More generally, for m >= 3, the sequences {b_m(n) : n >= 1} and {c_m(n) : n >= 1} defined by b_m(n) = (1/n) * Sum_{k = 0..n} (n + 2*k) * binomial(n+k-1,k)^m and c_m(n) = (1/n) * Sum_{k = 0..n} (-1)^k * (n + 2*k) * binomial(n+k-1,k)^m may both satisfy the same congruences.
%F a(n) ~ 2^(6*n) / (3 * Pi^(3/2) * n^(3/2)). - _Vaclav Kotesovec_, Mar 29 2023
%p seq( (1/n)*add((-1)^(n+k) * (n + 2*k) * binomial(n+k-1, n-1)^3, k = 0..n), n = 1..20);
%t Table[Sum[(-1)^(n+k) * (n + 2*k) * Binomial[n+k-1,k]^3, {k,0,n}]/n, {n,1,20}] (* _Vaclav Kotesovec_, Mar 29 2023 *)
%o (PARI) a(n) = (1/n) * sum(k = 0, n, (-1)^(n+k) * (n + 2*k) * binomial(n+k-1,k)^3); \\ _Michel Marcus_, Mar 30 2023
%Y Cf. A000984, A002894, A361883, A361885, A361886.
%K nonn,easy
%O 1,1
%A _Peter Bala_, Mar 28 2023
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