A185046 Smallest prime p such that 2n+1 = p^3 - 2q for some odd prime q, or 0 if no such prime exists.

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, 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, 5, 19, 5, 7, 5, 5, 53, 7, 17, 7, 13, 11, 13, 7, 113, 7, 373, 11, 17, 7

Offset: 1

Author

Michel Lagneau, Apr 17 2013

Keywords

• nonn

Comments

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.

The corresponding primes q are in A224730.

Conjecture: The odd numbers different from a cube are of the form m = p^3 - 2q where p and q are prime numbers.

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 p-c = 2, and q of the form x^2 +x*y+y^2 (see A007645). For example: 5^3 = 7^3 - 2*109.

Examples

			a(4) = 7 because, for (p, q) = (7, 167) => 2*4+1 = 9 = 7^3 - 2*167 = 343 - 334 = 9.
		

Links

• Michel Lagneau, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A007645, A224730.

Programs

Maple

for n from 3 by 2 to 200 do:
      jj:=0:
          for j from 1 to 10000 while (jj=0) do:
             p:=ithprime(j):q:=(p^3-n)/2:
             if q> 0 and type(q,prime)=true
             then
             jj:=1:printf(`%d, `,p):
             else
             fi:
         od:
            if jj=0 then
            printf(`%d, `,0):
            else
            fi:
     od: