|
|
COMMENTS
|
a(n) is definitely 0 for n == 1 mod 4 or 2 mod 4 (except a(2) = 1). This is the case for n=5,6, 9,10, 13,14, 17,18, ...
Further, if n-1 is not squarefree, then a(n) = 0. Thus, if n-1 is in A013929, then a(n) = 0. This is the case for n = 5, 9, 10, 13, 17, 19, 21, ...
From the first two comments, we can conclude that there are an infinite number of 0 entries.
Let S = n^k + (n-1)^k + ... + 3^k + 2^k. Then, S is divisible by gpf(n-1) when k is not a multiple of gpf(n-1)-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)-1 for any n. Note that this holds with the previous findings.
For n <= 200, if n = {15, 23, 24, 32, 44, 59, 60, 68, 71, 75, 87, 88, 95, 96, 104, 107, 115, 120, 123, 131, 132, 140, 144, 151, 156, 159, 164, 167, 168, 184, 187, 188, 191, 195}, there is a pattern with the factorization of S when k is a multiple of gpf(n-1)-1. Thus, a(n) = 0 is definite for these n-values.
For other n-values <= 200, there is not a concrete pattern when k is a multiple of gpf(n-1)-1. If n = 20 or n = 72, a(n) > 10000, if n = {27, 35, 39, 48, 52, 63, 79, 80, 84, 92}, a(n) > 7500, and if n = {103, 108, 111, 112, 116, 119, 124, 128, 135, 139, 143, 147, 152, 155, 160, 175, 179, 180, 183, 192, 196, 200}, a(n) > 5000. Here, a(n) could still be nonzero.
For n < 200, it is known that a(31) = 2528, a(36) = 12, a(40) = 360, a(43) = 6, a(47) = 66, a(56) = 1580, a(67) = 390, a(83) = 80, and a(171) = 1984.
|
