cp's OEIS Frontend

A158013 100*a(n)+13 and 100*a(n)+27 are consecutive primes, i.e., a prime gap 14.

%I A158013 #13 Aug 01 2015 12:16:28
%S A158013 1,106,133,154,184,217,220,307,334,436,454,496,505,574,580,604,616,
%T A158013 631,805,892,1009,1015,1045,1132,1174,1189,1198,1204,1360,1408,1444,
%U A158013 1504,1510,1627,1702,1708,1771,1954,1984,2101,2182,2218,2221,2245,2260,2281
%N A158013 100*a(n)+13 and 100*a(n)+27 are consecutive primes, i.e., a prime gap 14.
%C A158013 Notes:
%C A158013 1) Necessarily a(n)=3k+1: a(n)=3k => 100*3k+27= 3*(100k+9), divisible by 3 a(n)= 3k+2 => 100*(3k+2)+13=3*(100k+71), divisible by 3.
%C A158013 2) It is conjectured that sequence is infinite.
%C A158013 3) Each sequence 100*b(n)+13 and 100*c(n)+27 includes an infinite number of primes (because of DIRICHLET's theorem).
%C A158013 4) Analogous sequences for investigation of prime gaps are obvious and useful.
%D A158013 N. G. Tchudakoff, On the difference between two neighboring prime numbers, Math. Sb. 1, (1936), 799-814.
%D A158013 R. K. Guy, Unsolved problems in number theory
%H A158013 A. E. Ingham, <a href="http://dx.doi.org/10.1093/qmath/os-8.1.255">On the difference between consecutive primes</a>, Quart. J. Math. Oxford 8 (1937), 255-266.
%F A158013 p(k+1)=100*a(n)+27 and p(k)=100*a(n)+13 where p(k) is the k-th prime => prime gap p(k+1)-p(k)=14.
%e A158013 1) 113=P(30) and 127=P(31) => a(1)=1.
%e A158013 2) 1613=P(255) and 1627=P(258) prime too but 1619=P(256), 1621=P(257) => 1613 and 1627 are not consecutive primes.
%e A158013 3) next: 10613=P(1295), 10627 = P(1296) => a(2)=106.
%t A158013 fQ[n_] := PrimeQ[ Range[100 n + 13, 100 n + 27, 2]] == {True, False, False, False, False, False, False, True}; Select[ Range@ 2295, fQ@# &] (* _Robert G. Wilson v_, Mar 13 2009 *)
%Y A158013 Cf. A157772 (primes ending with "13" ordered in natural growing size).
%K A158013 nonn
%O A158013 1,2
%A A158013 Ulrich Krug (leuchtfeuer37(AT)gmx.de), Mar 11 2009
%E A158013 a(31)-a(46) from _Robert G. Wilson v_, Mar 13 2009