OFFSET
0,2
COMMENTS
Compare with the identity Sum_{k = 0..n} binomial(n, k)^2 * binomial(n+k, k) * binomial(2*n+k, n) = binomial(2*n, n)^3 = A002897(n).
It is easy to see that for odd prime p, binomial(2*n, n)^3 is divisible by p^3 for integer n in the interval [(p + 1)/2, p - 1]. A similar property appears to hold for the present sequence. We conjecture that for prime p >= 5, a(n) is divisible by p^3 for integer n in the interval [ceiling((2*p + 1)/3), p - 1] (checked up to p = 101).
More generally, for m >= 2, a similar divisibility property appears to hold for the sequence whose n-th term is equal to Sum_{k = 0..n} binomial(n, k)^2* binomial(n+k, k)*binomial((m + 1)*n + m*k, n).
FORMULA
a(n) = binomial(3*n, n)*hypergeom([-n, -n, (3*n+1)/2, (3*n+2)/2], [1, 1, n+1/2], 1).
P-recursive: 16*n^3*(5616*n^4 - 30888*n^3 + 63459*n^2 - 57709*n + 19600)*(4*n - 1)^2*(4*n - 3)^2*a(n) = 36*(72783360*n^11 - 655050240*n^10 + 2595613248*n^9 - 5966404272*n^8 + 8824615470*n^7 - 8803399545*n^6 + 6034085115*n^5 - 2836309905*n^4 + 893904075*n^3 - 179376410*n^2 + 20562360*n - 1019200)*a(n-1) + 27*n*(5616*n^4 - 8424*n^3 + 4491*n^2 - 991*n + 78)*(3*n - 4)^3*(3*n - 5)^3*a(n-2) with a(0) = 1, a(1) = 13.
a(n) ~ 3^(9*n/2) * (1 + sqrt(3))^(6*n + 3) / (Pi^(3/2) * n^(3/2) * 2^(9*n + 9/2)). - Vaclav Kotesovec, Jul 22 2024
EXAMPLE
Factorization of a(8) thru a(10) showing divisibility by 11^3:
a(8) = (3^6)*11^3*10667*18773
a(9) = (5^2)*7*(11^3)*(13^3)*3607*10103
a(10) = (11^3)*(13^4)*31*22699*68099.
MAPLE
seq(add(binomial(n, k)^2*binomial(n+k, k)*binomial(3*n+2*k, n), k = 0..n), n = 0..20);
# faster program for large n
seq(simplify(binomial(3*n, n)*hypergeom([-n, -n, (3*n+1)/2, (3*n+2)/2], [1, 1, n+1/2], 1)), n = 0..20);
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Jul 20 2024
STATUS
approved