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A087522 a(0) = 2; for n>=1, a(n) = smallest prime p such that p+1 is divisible by an n-th power > 1. 15

%I #18 Apr 25 2019 03:36:03

%S 2,2,3,7,31,31,127,127,1279,3583,5119,6143,8191,8191,81919,131071,

%T 131071,131071,524287,524287,14680063,14680063,109051903,109051903,

%U 654311423,738197503,738197503,2147483647,2147483647,2147483647,2147483647

%N a(0) = 2; for n>=1, a(n) = smallest prime p such that p+1 is divisible by an n-th power > 1.

%C Trivially the n-th power under consideration is 2^n for n > 1.

%H John Mason, using Robert Israel's data for A127582, <a href="/A087522/b087522.txt">Table of n, a(n) for n = 0..3310</a>

%F a(n) << 37^n by Xylouris' improvement to Linnik's theorem. - _Charles R Greathouse IV_, Dec 10 2013

%e a(1) = 2 because 3^1|3.

%e a(2) = 3 because 2^2|4.

%e a(3) = 7 because 2^3|8.

%o (PARI) okdivs(pp1, n) = fordiv(pp1, d, if ((d>1) && ispower(d, n), return (1))); 0

%o a(n) = {if (n == 0, return (2)); p = 2; while (! okdivs(p+1, n), p = nextprime(p+1)); return (p);} \\ _Michel Marcus_, Sep 14 2013

%Y A127582 is identical except for a(1).

%K nonn

%O 0,1

%A _Amarnath Murthy_, Sep 11 2003

%E More terms from _Ray Chandler_, Sep 14 2003

%E Edited by _N. J. A. Sloane_, Jul 03 2008

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