The OEIS is supported by the many generous donors to the OEIS Foundation.

a(n) = largest prime q such that q | 2^p - 2 and p - 1 | q - 1, where p = prime(n).

0

`%I #32 Oct 18 2017 13:44:57
`

`%S 2,3,5,7,31,13,257,73,683,113,331,109,61681,5419,2796203,1613,3033169,
`

`%T 1321,599479,122921,38737,22366891,8831418697,2931542417,22253377,
`

`%U 268501,131071,28059810762433,279073,54410972897,77158673929,145295143558111,2879347902817,10052678938039,616318177,1133836730401,121369
`

`%N a(n) = largest prime q such that q | 2^p - 2 and p - 1 | q - 1, where p = prime(n).
`

`%C First conjecture: a(n) > prime(n) for all n > 6. _Robert Israel_ tested the author's conjecture up to prime(95) = 499. The prime factorizations of the numbers 2^(p-1)-1 for larger p can be checked in available tables, see A005420.
`

`%C Second conjecture: a(n) = gpf(2^prime(n) - 2) for almost all n, in the sense that the set of exceptions {10, 16, 37, 40, ...} has zero natural density.
`

`%C Primes p for which p - 1 does not divide gpf(2^p - 2) - 1 are 29, 53, 157, 173, ...
`

`%e For prime(5) = 11, 2^11-2 = 2*3*11*31 and 11-1 | 31-1, so a(5) = 31.
`

`%Y Cf. A000040, A005420.
`

`%K nonn
`

`%O 1,1
`

`%A _Thomas Ordowski_, Sep 01 2017
`