A259732 - OEIS

%I #20 Aug 16 2015 23:42:42

%S 2,3,5,10,11,17,22,23,29,31,34,41,43,46,47,53,58,59,62,71,82,83,86,89,

%T 94,101,106,107,109,113,118,121,127,131,137,142,149,157,166,167,173,

%U 178,179,187,191,197,202

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

%C 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.

%C 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).

%C "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).

%C 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.

%C 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.

%C A179871 looks very similar to this sequence.

%Y Cf. A057760, A040028.

%K nonn

%O 1,1

%A _Miguel Angel Velilla Mula_, Jul 04 2015