A231776 -id:A231776 - cp's OEIS frontend

A231725 Least positive integer k < n such that n + k + 2^k is prime, or 0 if such an integer k does not exist.

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, 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, 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

Offset: 1

Zhi-Wei Sun, Nov 12 2013

nonn

This was motivated by A231201 and A231557.
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.
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

			a(3) = 0 since 3 + 1 + 2^1 = 6 and 3 + 2 + 2^2 = 9 are both composite.
a(5) = 2 since 5 + 1 + 2^1 = 8 is not prime, but 5 + 2 + 2^2 = 11 is prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Z.-W. Sun, On a^n+ bn modulo m, arXiv preprint arXiv:1312.1166 [math.NT], 2013-2014.

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

Do[Do[If[PrimeQ[n+k+2^k],Print[n," ",k];Goto[aa]],{k,1,n-1}];
Print[n," ",0];Label[aa];Continue,{n,1,100}]

a(n)=for(k=1,n-1,if(ispseudoprime(n+k+2^k),return(k)));0 \\ Charles R Greathouse IV, Nov 13 2013

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A231725 Least positive integer k < n such that n + k + 2^k is prime, or 0 if such an integer k does not exist.

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