OFFSET
1,1
COMMENTS
Here we allow the arithmetic progressions to contain one or more terms.
The first 1000 odd primes can be covered with 221 arithmetic progressions of primes (see Links).
Finding the smallest n for a given k is a set covering problem with a binary variable for each arithmetic progression and a constraint for each of the first k odd primes. - Rob Pratt, Aug 26 2019
LINKS
dxdy forum, Covering of primes with arithmetic progressions of primes (in Russian)
Dmitry Kamenetsky, Covering of the first 1000 odd primes
EXAMPLE
1 arithmetic progression of primes is needed to cover the first 3 odd primes: (3,5,7). So a(1) = 3. Note that we cannot cover the first 4 odd primes with 1 arithmetic progression.
2 arithmetic progressions of primes are needed to cover the first 5 odd primes: (3,7,11), (5,13). So a(2) = 5.
3 arithmetic progressions of primes are needed to cover the first 10 odd primes: (3,17,31), (5,11,17,23,29), (7,13,19). So a(3) = 10.
4 arithmetic progressions of primes are needed to cover the first 13 odd primes: (3,13,23), (5,17,29,41), (7,19,31,43), (11,37). So a(4) = 13.
5 arithmetic progressions of primes are needed to cover the first 18 odd primes: (5,11,17,23,29), (7,19,31,43), (41,47,53,59), (13,37,61), (3,67). So a(5) = 18.
CROSSREFS
KEYWORD
nonn,more,hard
AUTHOR
Dmitry Kamenetsky, Jul 20 2019
EXTENSIONS
a(27)-a(50) from Rob Pratt, Aug 26 2019
STATUS
approved