
COMMENTS

a(n) is definitely 0 for n == 3 mod 4 or 0 mod 4 (except a(4) = 2). This is the case for n = 3; 7,8; 11,12; 15,16; 19,20; ...
Further, if n is not squarefree, then a(n) = 0. Thus, if n is in A013929, then a(n) = 0. This is the case for n = 8, 9, 12, 16, 18, 20, 24, ...
Let S = n^k  (n1)^k  ...  3^k  2^k  1. Then S is divisible by gpf(n) when k is not a multiple of gpf(n)1, where gpf(x) denotes the greatest prime factor of x. This means that if a(n) is not 0, then a(n) must be a multiple of gpf(n)1. Note that this holds with the previous findings.
For n <= 200, if n = {21, 26, 29, 34, 38, 58, 61, 62, 65, 70, 74, 86, 89, 97, 101, 102, 106, 110, 133, 137, 142, 146, 157, 173, 178, 181, 182, 185, 190}, there is a pattern with the factorization of S when k is a multiple of gpf(n)1. Thus, a(n) = 0 is definite for these nvalues.
For other nvalues <= 200, there is not a concrete pattern when k is a multiple of gpf(n)1. If n = 14, a(n) > 30000, if n = {33, 37, 42, 46, 53, 57, 69, 73, 77, 78, 82, 85, 93, 94}, a(n) > 8000, and if n = {105, 109, 113, 114, 118, 122, 129, 130, 134, 138, 141, 145, 149, 154, 158, 161, 165, 166, 170, 174, 177, 186, 193, 197}, a(n) > 6000. Here, a(n) could still be nonzero.
For n <= 200, it is known that a(17) = 16, a(22) = 3190, a(30) = 124, a(41) = 520, a(66) = 260, and a(194) = 288.
