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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.
1
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
1,1
COMMENTS
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).
LINKS
EXAMPLE
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.
CROSSREFS
KEYWORD
nonn
AUTHOR
Joel Brennan, Jan 24 2023
STATUS
approved