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%I A055623 #27 Oct 30 2022 18:19:59
%S A055623 5,13,89,389,2593,12401,77069,262897,11593,373649,766261,3358169,
%T A055623 12204889,18256561,23048897,12270077,297387757,310523021,297779117,
%U A055623 3670889597,5344989829,1481666377,2572421893,1113443017,121117598053,84676452781,790457451349,3498519134533,689101181569,3289884073409
%N A055623 First occurrence of run of primes congruent to 1 mod 4 of exactly length n.
%C A055623 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.
%C A055623 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.
%H A055623 J. K. Andersen, <a href="http://primerecords.dk/congruent-primes.htm">Consecutive Congruent Primes</a>.
%H A055623 Carlos Rivera's Prime Puzzles and Problems Connection, <a href="http://www.primepuzzles.net/puzzles/puzz_256.htm">Puzzle 256, Jack Brennen old records</a>
%F A055623 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.
%e A055623 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.
%e A055623 n=3: 83, 89, 97, 101, 103 are congruent to 3, 1, 1, 1, 3 modulo 4. So a(3) = 89.
%e A055623 a(33) = 3278744415797. - _Jens Kruse Andersen_, May 29 2006
%t A055623 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 *)
%Y A055623 Cf. A055624, A055626.
%K A055623 nonn
%O A055623 1,1
%A A055623 _Labos Elemer_, Jun 05 2000
%E A055623 Corrected and extended by _Reiner Martin_, Jul 18 2001
%E A055623 More terms from _Jens Kruse Andersen_, May 29 2006
%E A055623 Edited by _N. J. A. Sloane_, Jun 01 2006

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