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A361884 a(n) = (1/n) * Sum_{k = 0..n} (-1)^(n+k) * (n + 2*k) * binomial(n+k-1,k)^3. 3
2, 66, 2540, 110530, 5197752, 257490156, 13238524728, 699822144450, 37800431926400, 2077184897317816, 115757876008359312, 6526739641107783916, 371641758587326581200, 21341134886976332825400, 1234474507620634579565040 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
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.
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.
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.
LINKS
FORMULA
a(n) ~ 2^(6*n) / (3 * Pi^(3/2) * n^(3/2)). - Vaclav Kotesovec, Mar 29 2023
MAPLE
seq( (1/n)*add((-1)^(n+k) * (n + 2*k) * binomial(n+k-1, n-1)^3, k = 0..n), n = 1..20);
MATHEMATICA
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 *)
PROG
(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
CROSSREFS
Sequence in context: A218433 A092884 A230735 * A261784 A350730 A255435
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Mar 28 2023
STATUS
approved

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Last modified June 19 19:59 EDT 2024. Contains 373507 sequences. (Running on oeis4.)