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A361885
a(n) = (1/n) * Sum_{k = 0..2*n} (n+2*k) * binomial(n+k-1,k)^3.
3
9, 979, 165816, 33372819, 7380882509, 1732912534168, 424032181044264, 106952563532680339, 27609695174536836075, 7259294757681340436979, 1937215339689731617386000, 523352118643145676922317336, 142854011885066484369862826496, 39337931825265398967484384872560
OFFSET
0,1
COMMENTS
Compare with the closed form evaluation of the binomial sums (1/n) * Sum_{k = 0..2*n} (-1)^k * (n + 2*k) * binomial(n+k-1,k) = binomial(3*n,n) and (1/n) * Sum_{k = 0..2*n} (n + 2*k) * binomial(n+k-1,k)^2 = binomial(3*n,n)^2.
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.
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.
FORMULA
a(n) ~ 5 * 3^(9*n + 3/2) / (19 * Pi^(3/2) * n^(3/2) * 2^(6*n + 3)). - Vaclav Kotesovec, Mar 29 2023
MAPLE
seq( (1/n)*add((n + 2*k) * binomial(n+k-1, k)^3, k = 0..2*n), n = 1..20);
MATHEMATICA
Table[Sum[(n+2*k) * Binomial[n+k-1, k]^3, {k, 0, 2*n}]/n, {n, 1, 20}] (* Vaclav Kotesovec, Mar 29 2023 *)
PROG
(PARI) a(n) = (1/n) * sum(k = 0, 2*n, (n+2*k) * binomial(n+k-1, k)^3); \\ Michel Marcus, Mar 30 2023
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Mar 28 2023
STATUS
approved