A055623 First occurrence of run of primes congruent to 1 mod 4 of exactly length n.

5, 13, 89, 389, 2593, 12401, 77069, 262897, 11593, 373649, 766261, 3358169, 12204889, 18256561, 23048897, 12270077, 297387757, 310523021, 297779117, 3670889597, 5344989829, 1481666377, 2572421893, 1113443017, 121117598053, 84676452781, 790457451349, 3498519134533, 689101181569, 3289884073409

Offset: 1

Author

Labos Elemer, Jun 05 2000

Keywords

• nonn

Comments

The term "exactly" means that before the first and after the last terms of the run, the next primes are not congruent to 1 modulo 4.

Carlos Rivera's Puzzle 256 includes Jack Brennen's a(29) starting at 689101181569 to 689101182437 and asks if anyone can break that 1999 record.

Examples

			a(3)=89 because here n=3 and 89 is the start of a run of exactly 3 consecutive primes congruent to 1 mod 4.
n=3: 83, 89, 97, 101, 103 are congruent to 3, 1, 1, 1, 3 modulo 4. So a(3) = 89.
a(33) = 3278744415797. - _Jens Kruse Andersen_, May 29 2006
		

Links

• J. K. Andersen, Consecutive Congruent Primes.
• Carlos Rivera's Prime Puzzles and Problems Connection, Puzzle 256, Jack Brennen old records

Crossrefs

Cf. A055624, A055626.

Programs

Mathematica

nn = 10; t = Table[0, {nn}]; found = 0; p = 1; cnt = 0; While[found < nn, p = NextPrime[p]; If[Mod[p, 4] == 1, cnt++, If[0 < cnt <= nn && t[[cnt]] == 0, t[[cnt]] = NextPrime[p, -cnt]; found++]; cnt = 0]]; t (* T. D. Noe, Jun 21 2013 *)

Formula

Compute sequence of primes congruent to 1 mod 4. When first occurrence of run of exactly length n is found, add first prime to sequence.

Extensions

Corrected and extended by Reiner Martin, Jul 18 2001

More terms from Jens Kruse Andersen, May 29 2006

Edited by N. J. A. Sloane, Jun 01 2006