

A231725


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


5



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
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OFFSET

1,5


COMMENTS

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


LINKS

ZhiWei 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], 20132014.


EXAMPLE

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.


MATHEMATICA

Do[Do[If[PrimeQ[n+k+2^k], Print[n, " ", k]; Goto[aa]], {k, 1, n1}];
Print[n, " ", 0]; Label[aa]; Continue, {n, 1, 100}]


PROG

(PARI) a(n)=for(k=1, n1, if(ispseudoprime(n+k+2^k), return(k))); 0 \\ Charles R Greathouse IV, Nov 13 2013


CROSSREFS

Cf. A000040, A000079, A231201, A231516, A231557, A231561, A231631, A231776.
Sequence in context: A060475 A168069 A280929 * A106559 A280047 A106377
Adjacent sequences: A231722 A231723 A231724 * A231726 A231727 A231728


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Nov 12 2013


STATUS

approved



