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%I #8 Sep 09 2017 22:34:24
%S 173,2459,17,67,19,113497,179,71,23,73,677,25339,0,74453,29,79,31,
%T 1117,191,83,193,25349,37,1123,197,89,41,3594557,43,1129,3461,12227,
%U 47,97,701,647551,3467,101,53,103,1124087,647557,709,107,59,109,61,113539,6133
%N Smallest prime p such that 2n+1 = 2p - q^3 for some odd prime q, or 0 if no such prime exists.
%C See the odd numbers of the form p^3 - 2q where p and q are primes (A224730, A185046).
%C 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, ...
%C The corresponding primes q are in A224795.
%C Conjecture: The odd numbers different from a cube are of the form m = 2p - q^3 where p and q are prime numbers.
%C 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.
%H Michel Lagneau, <a href="/A224794/b224794.txt">Table of n, a(n) for n = 1..10000</a>
%e a(4) = 67 because, for (p, q) = (67, 5), 2*4 + 1 = 9 = 2*67 - 5^3 = 134 - 125 = 9.
%p 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:
%Y Cf. A185046, A224730, A224795.
%K nonn
%O 1,1
%A _Michel Lagneau_, Apr 18 2013
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