A342309 When A123556(n) is the number of primes (not necessarily consecutive) in the longest arithmetic progression of primes with common difference n, a(n) is the smallest term of the first such arithmetic progression.

2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 5, 2, 3, 2, 7, 2, 59, 2, 3, 2, 3, 2, 7, 2, 5, 2, 3, 2, 31, 2, 3, 2, 3, 2, 5, 2, 3, 2, 7, 2, 5, 2, 3, 2, 7, 2, 5, 2, 3, 2, 3, 2, 11, 2, 5, 2, 3, 2, 31, 2, 3, 2, 3, 2, 7, 2, 5, 2, 3, 2, 23, 2, 3, 2, 7, 2, 5, 2, 3, 2, 13, 2, 13

Offset

1,1

Comments

Inspired by problem A1880 in Diophante (see link).

a(n) is the last term of the n-th row of A124064.

About the k-tuples conjecture in A123556: let p = A053669(n), if the arithmetic progression of p elements starting at p with difference n consists of primes, then A123556(n) = p, otherwise A123556(n) = p-1. The first 21 terms of this sequence are precisely the same as A053669, then a(22) = 7 (corresponding to the arithmetic progression (7,29)) while A053669(22) = 3.

Links

Table of n, a(n) for n=1..90.

Diophante, A1880. NP en PA (in French).

Formula

a(2k+1) = 2.

Example

There are only two consecutive primes (p,p+1) = (2,3), hence a(1) = 2.
(5,29,53) is the smallest arithmetic progression of 3 primes with common difference of 24, but (59,83,107,131) is the smallest arithmetic progression of 4 primes with common difference of 24 and there does not exist such an arithmetic progression with length > 4; hence, a(24) = 59.
(3,29), (5,31), (11,37) are the first three arithmetic progressions of primes with common difference of 26, the smallest term of the first arithmetic progression (3,29) is 3, and there does not exist such an arithmetic progression with length > 2, hence a(26) = 3.

Crossrefs

Cf. A053669, A123556, A124064.

Sequence in context: A284017 A236454 A053669 * A112047 A112048 A060395

Adjacent sequences: A342306 A342307 A342308 * A342310 A342311 A342312

Keyword

nonn

Author

Bernard Schott, Mar 08 2021

Status

approved