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%I #46 Feb 13 2023 08:48:10
%S 2,3,7,5,11,23,31,17,43,47,13,67,71,79,29,59,19,103,53,107,37,131,139,
%T 73,83,167,89,179,61,41,191,101,211,109,223,239,127,263,137,283,97,
%U 151,311,331,173,347,359,367,383,193,197,419,431,439,443,149,113,227,463
%N a(n) is calculated by considering in ascending order all products P of zero or more 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 A new prime is always found since at worst P can be the product of all primes {a(1..n-1)} and per Euclid's proof of the infinitude of primes, P+1 then certainly has a prime factor not among a(1..n-1).
%C Taking products P in ascending order generally results in smaller quantities to consider than always taking the product of all primes as done in A000945, the Euclid-Mullin sequence.
%C Conjecture: P+1 has at most one prime factor not already in the sequence, so the requirement of taking "the smallest such" is unnecessary.
%H Joel Brennan, <a href="/A359504/b359504.txt">Table of n, a(n) for n = 1..5000</a>
%e For n=1, the sole product P is the empty product P=1, and P+1 = 2 is itself prime so a(1) = 2.
%e For n=3, the primes so far are 2,3 but products P=2 or P=3 have P+1 = 3 or 4 which have no new prime factor. Product P = 2*3 = 6 has P+1 = 7 which is a new prime so a(3) = 7.
%e For n=4, the smallest product P with a new prime in P+1 is P = 2*7 = 14 for which P+1 = 15 and a(4) = 5 is its smallest new prime factor.
%Y Cf. A000945, A000946.
%K nonn
%O 1,1
%A _Joel Brennan_, Jan 03 2023
%E More terms from _Kevin Ryde_, Jan 10 2023
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