A379775 a(1) = 2. For n > 1, a(n) = smallest prime factor of c=2*a(n-1)-1 that is not in {a(1), ..., a(n-1)}; if all prime factors of c are in {a(1), ..., a(n-1)}, then we try the next value of c, which is 2*c-1; and so on.

2, 3, 5, 17, 11, 7, 13, 97, 193, 769, 29, 19, 37, 73, 577, 1153, 461, 307, 613, 31, 61, 241, 113, 449, 23, 89, 59, 233, 929, 619, 1237, 2473, 43, 337, 673, 269, 179, 211, 421, 41, 107, 71, 47, 67, 53, 139, 277, 79, 157, 313, 1249, 227, 151, 601, 1201, 4801

Offset: 1

Author

Robert C. Lyons, Jan 02 2025

Keywords

• nonn

Comments

The following are some statistics about how many terms of the sequence are required, so that the first k primes are included:

- The first 10^2 terms include the first 28 primes.

- The first 10^3 terms include the first 92 primes.

- The first 10^4 terms include the first 710 primes.

- The first 10^5 terms include the first 4848 primes.

- The first 10^6 terms include the first 29442 primes.

- The first 10^7 terms include the first 260324 primes.

Conjecture: this sequence is a permutation of the primes.

Examples

			a(6) is 7 because the prime factors of c=2*a(5)-1 (i.e., 21) are 3 and 7, and 3 already appears in the sequence as a(2).
a(8) is 97 because the only prime factor of c=2*a(7)-1 (i.e., 25) is 5 which already appears in the sequence as a(3). The next value of c (i.e., 2*c-1) is 49; its only prime factor is 7 which already appears in the sequence as a(6). The next value of c (i.e., 2*c-1) is 97, which is prime and does not already appear in the sequence.
		

Links

• Robert C. Lyons, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A031439, A072268, A131200, A174162, A379652, A379648, A379776.

Programs

Python

from sympy import primefactors
seq = [2]
seq_set = set(seq)
max_seq_len=100
while len(seq) <= max_seq_len:
    c = seq[-1]
    done = False
    while not done:
        c = 2*c-1
        factors = primefactors(c)
        for factor in factors:
            if factor not in seq_set:
                seq.append(factor)
                seq_set.add(factor)
                done = True
                break
print(seq)