A224795 -id:A224795 - cp's OEIS frontend

A224794 Smallest prime p such that 2n+1 = 2p - q^3 for some odd prime q, or 0 if no such prime exists.

173, 2459, 17, 67, 19, 113497, 179, 71, 23, 73, 677, 25339, 0, 74453, 29, 79, 31, 1117, 191, 83, 193, 25349, 37, 1123, 197, 89, 41, 3594557, 43, 1129, 3461, 12227, 47, 97, 701, 647551, 3467, 101, 53, 103, 1124087, 647557, 709, 107, 59, 109, 61, 113539, 6133

Offset: 1

Michel Lagneau, Apr 18 2013

nonn

See the odd numbers of the form p^3 - 2q where p and q are primes (A224730, A185046).
a(n) = 0 for n = 13, 62, 171, 364, 665, 1098, ... and 2n+1 = 27, 125, 343, 729, ... are the odd cubes 3^3, 5^3, 7^3, ...
The corresponding primes q are in A224795.
Conjecture: The odd numbers different from a cube are of the form m = 2p - q^3 where p and q are prime numbers.
Remark: if m = c^3 for any odd integer c, then c^3 = 2p - q^3 is impossible because c^3 + q^3 = 2p => (c+q)(c^2 - cq + q^2) with c+q even of the form c+q = 2a => p = a(c^2 - cq + q^2) absurd because p is prime.

			a(4) = 67 because, for (p, q) = (67, 5), 2*4 + 1 = 9 = 2*67 - 5^3 = 134 - 125 = 9.

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

Cf. A185046, A224730, A224795.

for n from 3 by 2 to 200 do:jj:=0:for j from 1 to 50000 while (jj=0) do:q:=ithprime(j):p:=(q^3+n)/2:if type(p,prime)=true  then jj:=1: printf(`%d, `,p):else fi:od:if jj=0 then printf(`%d, `,0):else fi:od:

cp's OEIS Frontend

A224794 Smallest prime p such that 2n+1 = 2p - q^3 for some odd prime q, or 0 if no such prime exists.

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