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.

%I A359505 #12 Feb 13 2023 08:48:25
%S A359505 2,3,5,7,13,29,17,11,19,37,41,23,73,43,47,31,61,101,109,113,59,137,71,
%T A359505 173,181,97,193,67,53,79,157,107,229,127,257,281,149,151,103,313,317,
%U A359505 163,167,83,353,89,373,389,197,131,199,397,401,409,139,277,433,223
%N 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.
%C A359505 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).
%C A359505 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.
%C A359505 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).
%H A359505 Joel Brennan, <a href="/A359505/b359505.txt">Table of n, a(n) for n = 1..5000</a>
%e A359505 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.
%e A359505 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.
%Y A359505 Cf. A000945, A000946, A359504
%K A359505 nonn
%O A359505 1,1
%A A359505 _Joel Brennan_, Jan 24 2023