A113158 Primes such that the sum of the predecessor and successor primes is divisible by 43.

521, 821, 859, 1069, 1459, 1549, 2203, 2411, 2539, 2837, 2969, 3011, 3089, 3359, 3613, 3823, 4259, 4339, 4561, 4643, 4783, 5503, 5557, 6067, 6619, 6967, 7481, 7699, 7741, 8263, 8779, 9419, 10103, 12041, 12379, 12641, 12899, 13417, 13721, 13759

Offset: 1

Author

Jonathan Vos Post, Jan 05 2006

Keywords

• easy
• nonn

Comments

A112681 is mod 3 analogy. A112794 is mod 5 analogy. A112731 is mod 7 analogy. A112789 is mod 11 analogy. A112795 is mod 13 analogy. A112796 is mod 17 analogy. A112804 is mod 19 analogy. A112847 is mod 23 analogy. A112859 is mod 29 analogy.

Examples

			a(1) = 521 since prevprime(521) + nextprime(521) = 509 + 523 = 1032 = 43 * 24.
a(2) = 821 since prevprime(821) + nextprime(821) = 811 + 823 = 1634 = 43 * 38.
a(3) = 859 since prevprime(859) + nextprime(859) = 857 + 863 = 1720 = 43 * 40.
a(4) = 1069 since prevprime(1069)+nextprime(1069) = 1063+1087 = 2150 = 43 * 50.
		

Crossrefs

Cf. A000040, A112681, A112794, A112731, A112789, A112795, A112796, A112804, A112847, A112859, A113155, A113156, A113157, A113158.

Programs

Mathematica

Prime@Select[Range[2, 1657], Mod[Prime[ # - 1] + Prime[ # + 1], 43] == 0 &] (* Robert G. Wilson v *)

Formula

a(n) = prime(i) is in this sequence iff prime(i-1)+prime(i+1) = 0 mod 43. a(n) = A000040(i) is in this sequence iff A000040(i-1)+A000040(i+1) = 0 mod 43.

Extensions

More terms from Robert G. Wilson v, Jan 11 2006