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.
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
Keywords
Examples
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.
Links
- Joel Brennan, Table of n, a(n) for n = 1..5000
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