OFFSET
0,2
COMMENTS
A companion sequence to A374605.
Conjecture: for prime p, a(n) is divisible by p^3 for integer n in the interval [(1/3)*(2*p + (p/3)), p - 1], where (p/3) denotes the Legendre symbol.
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 + 1, n).
FORMULA
a(n) = binomial(3*n+1,n) * hypergeom([-n, -n, (3*n+2)/2, (3*n+3)/2], [1, 1, n+3/2], 1).
P-recursive: 16*(5616*n^4 - 22464*n^3 + 33183*n^2 - 21438*n + 5116)*(4*n - 1)^2*(4*n + 1)^2*n^3*a(n) = 36*(36391680*n^10 - 181958400*n^9 + 367112736*n^8 - 376700544*n^7 + 196483887*n^6 - 35225037*n^5 - 11205753*n^4 + 5552733*n^3 - 287502*n^2 - 163800*n + 20800)*(2*n - 1)*a(n-1) + 27*(n - 1)*(5616*n^4 - 513*n^2 + 13)*(3*n - 2)^3*(3*n - 4)^3*a(n-2) with a(0) = 1 and a(1) = 16.
a(n) ~ 3^((9*n + 3)/2) * (1 + sqrt(3))^(6*n+3) / (Pi^(3/2) * n^(3/2) * 2^(9*n + 13/2)). - Vaclav Kotesovec, Jul 16 2024
EXAMPLE
Factorization of a(9) thru a(12) showing divisibility by 13^3:
a(9) = (2^5)*5*(11^3)*(13^3)*1109*46351
a(10) = (11^3)*(13^3)*803962100633
a(11) = (2^8)*(3^3)*5*(13^3)*(17^3)*67*79*118057
a(12) = (2^6)*7*(13^3)*(17^3)*4836341163053.
MAPLE
seq( add(binomial(n, k)^2*binomial(n+k, k)*binomial(3*n+2*k+1, n), k = 0..n), n = 0..20);
# faster program for large n
seq(simplify( binomial(3*n+1, n) * hypergeom([-n, -n, (3*n+2)/2, (3*n+3)/2], [1, 1, n+3/2], 1)), n = 0..20);
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Jul 13 2024
STATUS
approved