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

1,2

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.

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

Table of n, a(n) for n=1..46.

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

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.

Example

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.

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 *)

Crossrefs

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

Sequence in context: A095645 A039554 A186458 * A250643 A163625 A070796

Adjacent sequences: A158010 A158011 A158012 * A158014 A158015 A158016

Keyword

nonn

Author

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

Extensions

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

Status

approved