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A185657

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

41, 44, 84, 87, 127, 130, 170, 173, 213, 216, 256, 259, 299, 302, 342, 345, 385, 388, 428, 471, 474, 517, 557, 560, 600, 643, 646, 689, 729, 732, 772, 815, 818, 944, 947, 987, 990, 1030, 1033, 1073, 1076, 1119, 1159, 1202, 1205, 1248, 1288, 1291, 1331, 1377, 1463, 1503, 1546

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OFFSET

1,1

COMMENTS

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.

From Robert Israel, Jul 09 2017: (Start)

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

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

LINKS

G. C. Greubel, Table of n, a(n) for n = 1..1000

MAPLE

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

MATHEMATICA

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

Select[Range[1600], PrimeQ[(#^2+#+41)/43]&] (* Harvey P. Dale, Feb 12 2011 *)

PROG

(PARI) is(n)=isprime((n^2+n+41)/43) \\ Charles R Greathouse IV, Jun 13 2017