mersenneforum.org - k*b^n+/-1, Bases 271 and 11971

[QUOTE=robert44444uk;358380]I have given a little (I mean very little) thought to a set that is defined in terms of the following:

Set member z is the prime p which is the first instance of primes such that none of the primes q up to q(z-1) have a multiplicative order p-1 base q.

The set (I think) goes as follows - the first 21 members:

7,11,11,59,131,131,181,181,271,271,271,271,271,1531,2791,11971,11971,11971,11971,11971,11971...

corresponding to the primes 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73...

An example:
271 is 135order2, 30order3, 27order5, 135order7, 135 order11, 18order13, 135 order17, 30 order19, 18 order23, 6 order29, 45 order 31, 135 order37, 45 order41, but 270 order 43.

The list is not on OEIS so happy for someone to put it up if they like.

I'm trying to think of a use.

If the p is a base in k.p^n+/-1, then it should be possible to define k much smaller than q(z-1) primorial, that provides a relatively prime series with integer n increasing, as all members of the series cannot have factors smaller than q(z-1). The efficient k values are found by applying CRM.

This is a bit akin to the Payam series, y.M(x)*2^n+/-1 with M(x) the multiple of primes that are p-1 order2, but in this new instance M(x)=1 given that all the primes to q(z-1) are considered as part of the CRM.

It might be interesting to find a few efficient k and find some primes with the bases 271 and 11971

Comments/ observations/ continuation/ efficient k/ subsequent primes welcome[/QUOTE]

The sequence (when duplicated term removed) should be [URL="https://oeis.org/A029932"]A029932[/URL], your sequence is wrong.

the right sequence is 3, 7, 23, 41, 109, 191, 191, 191, 271, 271, 271, 271, 271, 271, 2791, 2791, 11971, 11971, 11971, 11971, 11971, 11971, 31771, 31771, 31771, 31771, 31771, 190321, 190321

271 and 11971 are really in the sequence, but your sequence is wrong.