A259732 Numbers n > 1 that divide ((p-1)/2)^3 + 2 for some odd prime p.

2, 3, 5, 10, 11, 17, 22, 23, 29, 31, 34, 41, 43, 46, 47, 53, 58, 59, 62, 71, 82, 83, 86, 89, 94, 101, 106, 107, 109, 113, 118, 121, 127, 131, 137, 142, 149, 157, 166, 167, 173, 178, 179, 187, 191, 197, 202

Offset

1,1

Comments

n = 3,5,10 works only once, for p=3 (3-1)/2=1, then 1^3 + 2 = 3 and for p=5 (5-1)/2=2, then 2^3+2 = 10.

This sequence is a subset of A057760, where all elements that are multiples of 3 and 5 are excluded, except the three above (3,5,10).

"Mirror sequence" of this one, when n divides ((p+1)/2)^3 - 2, p = prime, produces a sequence very close to this one, the only differences being 10 (excluded), 25 (included for p=5 (p+1)/2=3 then 3^3-2 = 25) and 6 (included for p=3 (p+1)/2=2 then 2^3-2 = 6).

Analyzing ((p-1)/2)^3 + 2 = (p^3 - 3(p(p-1)-5))/8, every composite x (mod 3) trying to divide this one will fail.

To prove 5 can't divide ((p-1)/2)^3 + 2 = (p^3 - 3p^2 + 3p + 15)/8 we use the last digit of p, which can be 1,3,7 or 9. This leads the last digit of the formula to be (1,9,7 or 3) + 15, so it cannot be divided by 5, unless the last digit of p is 5. This happens just for the only prime divisible by 5, i.e., 5 itself, which occurs only once.

A179871 looks very similar to this sequence.

Links

Table of n, a(n) for n=1..47.

Crossrefs

Cf. A057760, A040028.

Sequence in context: A076681 A047604 A104427 * A192229 A233696 A002263

Adjacent sequences: A259729 A259730 A259731 * A259733 A259734 A259735

Keyword

nonn

Author

Miguel Angel Velilla Mula, Jul 04 2015

Status

approved