%I
%S 5,3,5,7,13,3,13,3,5,3,3,11,0,7,5,19,37,11,5,7,5,7,37,11,5,31,53,31,
%T 13,23,5,7,5,7,13,23,13,19,5,7,421,47,5,7,5,11,13,11,5,43,5,11,61,23,
%U 5,19,5,7,5,5,53,7,17,7,13,11,13,7,113,7,373,11,17,7
%N Smallest prime p such that 2n+1 = p^3  2q for some odd prime q, or 0 if no such prime exists.
%C a(n) = 0 for n = 13, 171, 364, 1098, 2456, 3429, 6083, 7812, 9841, 12194, 14895, 17968,... and 2n+1 = 27, 343, 729,... is a class of cubes.
%C The corresponding primes q are in A224730.
%C Conjecture: The odd numbers different from a cube are of the form m = p^3  2q where p and q are prime numbers.
%C Remark: Its converse is false: there exists cubes m = c^3 that are in the sequence with the form c^3 = p^3  2q, where pc = 2, and q of the form x^2 +x*y+y^2 (see A007645). For example: 5^3 = 7^3  2*109.
%H Michel Lagneau, <a href="/A185046/b185046.txt">Table of n, a(n) for n = 1..10000</a>
%e a(4) = 7 because, for (p, q) = (7, 167) => 2*4+1 = 9 = 7^3  2*167 = 343  334 = 9.
%p for n from 3 by 2 to 200 do:
%p jj:=0:
%p for j from 1 to 10000 while (jj=0) do:
%p p:=ithprime(j):q:=(p^3n)/2:
%p if q> 0 and type(q,prime)=true
%p then
%p jj:=1:printf(`%d, `,p):
%p else
%p fi:
%p od:
%p if jj=0 then
%p printf(`%d, `,0):
%p else
%p fi:
%p od:
%Y Cf. A007645, A224730.
%K nonn
%O 1,1
%A _Michel Lagneau_, Apr 17 2013
