|
%I #16 Mar 30 2023 05:08:35
%S 3,435,79464,16551315,3732732003,887492378136,219081875199120,
%T 55618197870142611,14429522546341842225,3808899907812064500435,
%U 1019705941257612879722400,276212555234100323977483800,75563424471884688135891640224
%N a(n) = (1/n) * Sum_{k = 0..2*n} (-1)^k * (n+2*k) * binomial(n+k-1,k)^3.
%C Compare with the closed form evaluation of the binomial sum (1/n) * Sum_{k = 0..2*n} (-1)^k * (n + 2*k) * binomial(n+k-1,k) = binomial(3*n,n).
%C The binomial coefficients u(n) := binomial(3*n,n) = A005809(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..2*n} (n + 2*k) * binomial(n+k-1,k)^m and c_m(n) = (1/n) * Sum_{k = 0..2*n} (-1)^k * (n + 2*k) * binomial(n+k-1,k)^m may satisfy the same congruences.
%F a(n) ~ 3^(9*n + 3/2) / (7 * Pi^(3/2) * n^(3/2) * 2^(6*n + 3)). - _Vaclav Kotesovec_, Mar 29 2023
%p seq( (1/n)*add( (-1)^k * (n + 2*k) * binomial(n+k-1,k)^3, k = 0..2*n), n = 1..20);
%t Table[Sum[(-1)^k * (n+2*k) * Binomial[n+k-1,k]^3, {k,0,2*n}]/n, {n,1,20}] (* _Vaclav Kotesovec_, Mar 29 2023 *)
%o (PARI) a(n) = (1/n) * sum(k = 0, 2*n, (-1)^k * (n+2*k) * binomial(n+k-1,k)^3); \\ _Michel Marcus_, Mar 30 2023
%Y Cf. A005809, A361883, A361884, A361885.
%K nonn,easy
%O 1,1
%A _Peter Bala_, Mar 28 2023
|
