A359504 -id:A359504 - cp's OEIS frontend

A359505 a(1)=2, a(2)=3, and for n >= 3, a(n) is calculated by considering in ascending order all products P of (distinct) terms from {a(1..n-1)} until finding one where P-1 has a prime factor not in {a(1..n-1)}, in which case a(n) is the smallest such prime factor.

2, 3, 5, 7, 13, 29, 17, 11, 19, 37, 41, 23, 73, 43, 47, 31, 61, 101, 109, 113, 59, 137, 71, 173, 181, 97, 193, 67, 53, 79, 157, 107, 229, 127, 257, 281, 149, 151, 103, 313, 317, 163, 167, 83, 353, 89, 373, 389, 197, 131, 199, 397, 401, 409, 139, 277, 433, 223

Offset: 1

Joel Brennan, Jan 24 2023

nonn

A new prime is always found since at worst P can be the product of all primes {a(1..n-1)} and then P-1 certainly has a prime factor not among a(1..n-1).
Taking products P in ascending order generally results in smaller quantities to consider than always taking the product of all primes as done in A084598.
Conjecture: P-1 has at most one prime factor not already in the sequence, so the requirement of taking "the smallest such" is unnecessary (verified up to n=10000).

			For n=3, the primes so far are 2 and 3 but products P=2 or P=3 have P-1 = 1 or 2 which have no new prime factor. Product P = 2*3 = 6 has P-1 = 5 which is a new prime so a(3) = 5.
For n=4, the smallest product P with a new prime in P-1 is P = 3*5 = 15 for which P-1 = 14 and a(4) = 7 is its smallest new prime factor.

Joel Brennan, Table of n, a(n) for n = 1..5000

Cf. A000945, A000946, A359504

cp's OEIS Frontend

A359505 a(1)=2, a(2)=3, and for n >= 3, a(n) is calculated by considering in ascending order all products P of (distinct) terms from {a(1..n-1)} until finding one where P-1 has a prime factor not in {a(1..n-1)}, in which case a(n) is the smallest such prime factor.

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