A379776 -id:A379776 - cp's OEIS frontend

A379899 a(1) = 2. For n > 1, a(n) = smallest prime factor of c=a(n-1)+4 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 c+4; and so on.

2, 3, 7, 11, 5, 13, 17, 29, 37, 41, 53, 19, 23, 31, 43, 47, 59, 67, 71, 79, 83, 103, 107, 127, 131, 139, 151, 163, 167, 179, 61, 73, 89, 97, 101, 109, 113, 137, 149, 157, 173, 181, 193, 197, 229, 233, 241, 257, 269, 277, 281, 293, 313, 317, 337, 349, 353, 373

Offset: 1

Robert C. Lyons, Jan 05 2025

nonn

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 95 primes.
- The first 10^3 terms include the first 697 primes.
- The first 10^4 terms include the first 6783 primes.
- The first 10^5 terms include the first 98563 primes.
Conjecture: this sequence is a permutation of the primes.

			a(2) is 3 because the prime factors of c=a(1)+4 (i.e., 6) are 2 and 3, and 2 already appears in the sequence as a(1).
a(6) is 13 because the only prime factor of c=a(5)+4 (i.e., 9) is 3 which already appears in the sequence as a(2). The next value of c (i.e., c+4) is 13, which is prime and does not already appear in the sequence.

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

Cf. A031439, A072268, A131200, A174162, A379652, A379648, A379775, A379776, A379783.

Cf. A379784, A379900, A380075, A380076.

b:= proc(n) option remember; `if`(n<1, {}, b(n-1) union {a(n)}) end:
a:= proc(n) option remember; local c, p; p:= infinity;
      for c from a(n-1)+4 by 4 while p=infinity do
        p:= min(numtheory[factorset](c) minus b(n-1)) od; p
    end: a(1):=2:
seq(a(n), n=1..200);  # Alois P. Heinz, Jan 11 2025

nn = 120; c[_] := True; j = 2; s = 4; c[2] = False;
Reap[Do[m = j + s;
  While[k = SelectFirst[FactorInteger[m][[All, 1]], c];
    ! IntegerQ[k], m += s];
c[k] = False; j = Sow[k], {nn}] ][[-1, 1]] (* Michael De Vlieger, Jan 11 2025 *)

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 = c + 4
        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)

A379775 a(1) = 2. For n > 1, a(n) = smallest prime factor of c=2a(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 2c-1; and so on.

Original entry on oeis.org

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

Robert C. Lyons, Jan 02 2025

nonn

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.

			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.

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

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

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)

cp's OEIS Frontend

A379899 a(1) = 2. For n > 1, a(n) = smallest prime factor of c=a(n-1)+4 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 c+4; and so on.

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A379775 a(1) = 2. For n > 1, a(n) = smallest prime factor of c=2a(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 2c-1; and so on.

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cp's OEIS Frontend

A379899 a(1) = 2. For n > 1, a(n) = smallest prime factor of c=a(n-1)+4 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 c+4; and so on.

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Maple

Mathematica

Python

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.

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Programs

Python

A379775 a(1) = 2. For n > 1, a(n) = smallest prime factor of c=2a(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 2c-1; and so on.