A231725 - OEIS

%I #47 Aug 05 2019 09:26:29

%S 0,1,0,1,2,3,2,1,4,1,2,3,2,1,10,1,2,3,6,1,4,5,2,5,2,1,4,1,8,3,2,3,4,1,

%T 2,3,2,1,4,1,2,3,6,1,12,5,2,3,8,1,4,5,2,11,2,1,6,1,4,3,2,3,4,1,2,5,2,

%U 1,4,1,22,3,2,57,10,1,2,3,6,1,4,11,2,11,8,1,4,7,4,3,2,3,4,1,2,3,2,1,16,1

%N Least positive integer k < n such that n + k + 2^k is prime, or 0 if such an integer k does not exist.

%C This was motivated by A231201 and A231557.

%C Conjecture: a(n) > 0 for all n > 3. We have verified this for n up to 2*10^6;  for example, we find the following relatively large values of a(n): a(65958) = 37055, a(299591) = 51116, a(295975) = 13128, a(657671) = 25724,  a(797083) = 44940, a(1278071) = 24146, a(1299037) = 34502, a(1351668) = 25121, a(1607237) = 34606, a(1710792) = 11187, a(1712889) = 18438.

%C I conjecture the opposite. In particular I expect that a(n) = 0 for infinitely many values of n. - _Charles R Greathouse IV_, Nov 13 2013

%H Zhi-Wei Sun, <a href="/A231725/b231725.txt">Table of n, a(n) for n = 1..10000</a>

%H Z.-W. Sun, <a href="http://arxiv.org/abs/1312.1166">On a^n+ bn modulo m</a>, arXiv preprint arXiv:1312.1166 [math.NT], 2013-2014.

%e a(3) = 0 since 3 + 1 + 2^1 = 6 and 3 + 2 + 2^2 = 9 are both composite.

%e a(5) = 2 since 5 + 1 + 2^1 = 8 is not prime, but 5 + 2 + 2^2 = 11 is prime.

%t Do[Do[If[PrimeQ[n+k+2^k],Print[n," ",k];Goto[aa]],{k,1,n-1}];

%t Print[n," ",0];Label[aa];Continue,{n,1,100}]

%o (PARI) a(n)=for(k=1,n-1,if(ispseudoprime(n+k+2^k),return(k)));0 \\ _Charles R Greathouse IV_, Nov 13 2013

%Y Cf. A000040, A000079, A231201, A231516, A231557, A231561, A231631, A231776.

%K nonn

%O 1,5

%A _Zhi-Wei Sun_, Nov 12 2013