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

1, 106, 133, 154, 184, 217, 220, 307, 334, 436, 454, 496, 505, 574, 580, 604, 616, 631, 805, 892, 1009, 1015, 1045, 1132, 1174, 1189, 1198, 1204, 1360, 1408, 1444, 1504, 1510, 1627, 1702, 1708, 1771, 1954, 1984, 2101, 2182, 2218, 2221, 2245, 2260, 2281

Offset: 1

Author

Ulrich Krug (leuchtfeuer37(AT)gmx.de), Mar 11 2009

Keywords

• nonn

Comments

Notes:

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.

2) It is conjectured that sequence is infinite.

3) Each sequence 100*b(n)+13 and 100*c(n)+27 includes an infinite number of primes (because of DIRICHLET's theorem).

4) Analogous sequences for investigation of prime gaps are obvious and useful.

Examples

			1) 113=P(30) and 127=P(31) => a(1)=1.
2) 1613=P(255) and 1627=P(258) prime too but 1619=P(256), 1621=P(257) => 1613 and 1627 are not consecutive primes.
3) next: 10613=P(1295), 10627 = P(1296) => a(2)=106.
		

References

• N. G. Tchudakoff, On the difference between two neighboring prime numbers, Math. Sb. 1, (1936), 799-814.
• R. K. Guy, Unsolved problems in number theory

Links

• A. E. Ingham, On the difference between consecutive primes, Quart. J. Math. Oxford 8 (1937), 255-266.

Crossrefs

Cf. A157772 (primes ending with "13" ordered in natural growing size).

Programs

Mathematica

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 *)

Formula

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.

Extensions

a(31)-a(46) from Robert G. Wilson v, Mar 13 2009