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URL: https://oeis.org/A359504

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A359504
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.
2
2, 3, 7, 5, 11, 23, 31, 17, 43, 47, 13, 67, 71, 79, 29, 59, 19, 103, 53, 107, 37, 131, 139, 73, 83, 167, 89, 179, 61, 41, 191, 101, 211, 109, 223, 239, 127, 263, 137, 283, 97, 151, 311, 331, 173, 347, 359, 367, 383, 193, 197, 419, 431, 439, 443, 149, 113, 227, 463
OFFSET
1,1
COMMENTS
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).
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.
Conjecture: P+1 has at most one prime factor not already in the sequence, so the requirement of taking "the smallest such" is unnecessary.
LINKS
EXAMPLE
For n=1, the sole product P is the empty product P=1, and P+1 = 2 is itself prime so a(1) = 2.
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.
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.
CROSSREFS
Sequence in context: A387076 A386364 A333086 * A064011 A050367 A354201
KEYWORD
nonn
AUTHOR
Joel Brennan, Jan 03 2023
EXTENSIONS
More terms from Kevin Ryde, Jan 10 2023
STATUS
approved