A380075 -id:A380075 - 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)

A380076 Indices of records in A379899.

Original entry on oeis.org

1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114

Offset: 1

Paolo Xausa, Jan 11 2025

nonn

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Cf. A379899, A380075 (records).

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

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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