cp's OEIS Frontend

The next term (a(7)) has 202 digits. - Harvey P. Dale, Aug 20 2015

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.

a(9) > 19. See A240766 for more information.

a(n) is also the n-values such that A240766(n) is nonzero.

It is known that a(n) must be == 3 mod 4 or 0 mod 4 (except a(1) = 2) due to the parity of the sum. If an n-value is congruent to 1 mod 4 or 2 mod 4, the sum will always be even and thus, not prime.

It is known that 31, 36, 40, 43, 47, 56, 67, 83, and 171 are members of this sequence.

If n-1 is not squarefree, then n is not a member of this sequence.