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
Keywords
Examples
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.
Links
- Diophante, A1880. NP en PA (in French).
Formula
a(2k+1) = 2.
Comments