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, 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

Author

Joel Brennan, Jan 03 2023

Keywords

• nonn

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.

Examples

			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.
		

Links

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

Crossrefs

Cf. A000945, A000946.

Extensions

More terms from Kevin Ryde, Jan 10 2023