OFFSET
1,2
COMMENTS
All a(n) are either 1, semiprime or prime.
a(n) = 1 for n = 1 and n = {7, 13, 17, 19, 24, 25, 27, 31, 32, 34, 37, 38, 43, 45, 47, 49, ...} = A067656 = numbers n such that n!*B(2*n) is an integer, where the B(2*n)'s are the Bernoulli numbers.
p divides a(p-1) for prime p > 3. p divides a((p-1)/2) for prime p > 3.
a(p-1) = p*(2p-1) is a semiprime hexagonal number for prime p = {7, 19, 31, 37, 79, 97, 139, 157, 199, 211, 229, 271, 307, 331, 337, 367, 379, 439, 499, ...} = A005382(n) for n > 2, where A005382(n) are the numbers n such that n and 2*n-1 are primes.
a(p-1) = p for prime p = {5, 11, 13, 17, 23, 29, 41, 43, 47, 53, 59, 61, 67, 71, 73, 83, 89, ...} = primes that do not belong to A005382(n).
a((p-1)/2) = p for prime p = {5, 7, 11, 17, 19, 23, 29, 31, 41, 43, 47, 53, 59, 67, 71, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 163, 167, 173, 179, 181, 191, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 259, 271, 281, 283, 293, 307, 311, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 401, ...}, which is apparently the union of {5} and A034849(n).
FORMULA
a(n) = numerator((Sum_{k=1..n} k^2) / (Product_{k=1..n} k^2)).
a(n) = numerator(n*(n+1)*(2*n+1)/6/(n!)^2).
EXAMPLE
The first few fractions are 1, 5/4, 7/18, 5/96, 11/2880, 91/518400, 1/181440, 17/135475200, 19/8778792960, ... = A125294/A334735. - Petros Hadjicostas, May 09 2020
MATHEMATICA
Table[Numerator[n(n+1)(2n+1)/6/(n!)^2], {n, 1, 500}]
PROG
(PARI) a(n) = numerator(sum(k=1, n, k^2)/prod(k=1, n, k^2)); \\ Michel Marcus, May 09 2020
CROSSREFS
KEYWORD
nonn,frac
AUTHOR
Alexander Adamchuk, Jan 17 2007
STATUS
approved