

A259732


Numbers n > 1 that divide ((p1)/2)^3 + 2 for some odd prime p.


0



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
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OFFSET

1,1


COMMENTS

n = 3,5,10 works only once, for p=3 (31)/2=1, then 1^3 + 2 = 3 and for p=5 (51)/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 3^32 = 25) and 6 (included for p=3 (p+1)/2=2 then 2^32 = 6).
Analyzing ((p1)/2)^3 + 2 = (p^3  3(p(p1)5))/8, every composite x (mod 3) trying to divide this one will fail.
To prove 5 can't divide ((p1)/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



