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A185657 Numbers n such that (n^2 + n + 41) / 43 is prime. 2

%I #27 Jul 13 2017 22:28:03

%S 41,44,84,87,127,130,170,173,213,216,256,259,299,302,342,345,385,388,

%T 428,471,474,517,557,560,600,643,646,689,729,732,772,815,818,944,947,

%U 987,990,1030,1033,1073,1076,1119,1159,1202,1205,1248,1288,1291,1331,1377,1463,1503,1546

%N Numbers n such that (n^2 + n + 41) / 43 is prime.

%C There exists an infinity of integers of the form (m^2 + m + 41) / 43 if we consider the numbers m of the form m = 1 + 43k => (m^2 + m + 41)/43 = k(43k+3)+1. It's highly probable that a(n) is infinite.

%C From _Robert Israel_, Jul 09 2017: (Start)

%C All terms == 1 or 41 (mod 43).

%C Bunyakovsky's conjecture implies there are infinitely many terms in both these classes. (End)

%H G. C. Greubel, <a href="/A185657/b185657.txt">Table of n, a(n) for n = 1..1000</a>

%p select(t -> isprime((t^2+t+41)/43), [seq(seq(43*j+k, k = [1,41]),j=0..100)]); # _Robert Israel_, Jul 09 2017

%t lst={}; Do[If[PrimeQ[(n^2+n+41)/43], AppendTo[lst, n]], {n, 2000}]; lst

%t Select[Range[1600],PrimeQ[(#^2+#+41)/43]&] (* _Harvey P. Dale_, Feb 12 2011 *)

%o (PARI) is(n)=isprime((n^2+n+41)/43) \\ _Charles R Greathouse IV_, Jun 13 2017

%Y Cf. A185658.

%K nonn,easy

%O 1,1

%A _Michel Lagneau_, Feb 08 2011

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