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%I #15 Jan 17 2019 15:54:56
%S 8,24,32,40,48,72,120,144,152,176,216,264,320,400,520,592,600,824,856,
%T 872,936,992,1064,1072,1112,1336,1392,1408,1584,1720,2080
%N Record (maximal) gaps between primes of the form 8k + 3.
%C Dirichlet's theorem on arithmetic progressions suggests that average gaps between primes of the form 8k + 3 below x are about phi(8)*log(x). This sequence shows that the record gap ending at p grows almost as fast as phi(8)*log^2(p). Here phi(n) is A000010, Euler's totient function; phi(8)=4.
%C Conjecture: a(n) < phi(8)*log^2(A269422(n)) almost always.
%C A269421 lists the primes preceding the maximal gaps.
%C A269422 lists the corresponding primes at the end of the maximal gaps.
%H Alexei Kourbatov, <a href="http://arxiv.org/abs/1610.03340">On the distribution of maximal gaps between primes in residue classes</a>, arXiv:1610.03340 [math.NT], 2016.
%H Alexei Kourbatov, <a href="https://arxiv.org/abs/1709.05508">On the nth record gap between primes in an arithmetic progression</a>, arXiv:1709.05508 [math.NT], 2017; <a href="https://doi.org/10.12988/imf.2018.712103">Int. Math. Forum, 13 (2018), 65-78</a>.
%H Alexei Kourbatov and Marek Wolf, <a href="https://arxiv.org/abs/1901.03785">Predicting maximal gaps in sets of primes</a>, arXiv preprint arXiv:1901.03785 [math.NT], 2019.
%e The first two primes of the form 8k + 3 are 3 and 11, so a(1)=11-3=8. The next prime of this form is 19; the gap 19-11 is not a record so nothing is added to the sequence. The next prime of this form is 43 and the gap 43-19=24 is a new record, so a(2)=24.
%o (PARI) re=0; s=3; forprime(p=11, 1e8, if(p%8!=3, next); g=p-s; if(g>re, re=g; print1(g", ")); s=p)
%Y Cf. A007520, A269421, A269422.
%K nonn
%O 1,1
%A _Alexei Kourbatov_, Feb 25 2016