[seqfan] (pseudo)prime generating recursion

[Tomasz Ordowski]

Dear SeqFans!

For a fixed integer base b > 1,

a(n) is the smallest k > a(n-1) such that b^(k-1) == 1 (mod a(n-1)*k), with
a(0) = 1.

For n > 0, a(n) is prime or pseudoprime (a Fermat pseudoprime to base b).

If for a base b, a(n) is a prime for every n > 0, then a(n) is the smallest
prime p > a(n-1) that does not divide b such that b^(p-1) == 1 (mod
a(n-1)), with a(0) = 1.

For b = 2, a(n) = A306826(n) : https://oeis.org/A306826

For b = 3, a(n) = 1, 2, 5, 13, 19, 37, 73, 97, 193, 241, 601, 751, 2251,
3001, 4001, 16001, 96001, 160001, 1120001, 4480001, 13440001, 20160001,
23385601, 29232001, 36540001, 38628001, 115884001, 231768001, 579420001,
1448550001, ...
All the above terms are prime [data from Amiram Eldar].

For b = 4, a(n) = 1, 3, 5, 7, 13, 19, 37, 73, 91, 97, 193, 277, 461, 691,
1151, 14951, ...
Note that a(8) = 91 = 7*13 is composite (a Fermat pseudoprime to base 4).

For b = 5, a(n) = 1, 2, 3, 7, 13, 17, 97, 193, 577, 1153, 3457, 10369, ...
So far, the above sequence is without composite numbers. Are further terms
also prime numbers?

Conjecture: For any integer base b > 1, a(n) is prime for almost all n.
Seems that at most finitely many terms are composite.

It is worth searching for composite terms in further such sequences.

Best regards,

Thomas Ordowski
_______________
For b = n > 1, a(1) = A053669(n):
Smallest prime p not dividing n.
See https://oeis.org/A053669