A174162 -id:A174162 - cp's OEIS frontend

A379652 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.

2, 5, 11, 23, 47, 19, 3, 7, 31, 127, 17, 71, 13, 37, 151, 101, 29, 59, 239, 479, 137, 1103, 2207, 883, 2357, 41, 83, 167, 67, 271, 181, 727, 97, 1567, 6271, 113, 227, 911, 1823, 521, 149, 599, 109, 73, 197, 79, 53, 107, 43, 563, 347, 139, 373, 997, 307, 1231

Offset: 1

Robert C. Lyons, Dec 28 2024

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 17 primes.
- The first 10^3 terms include the first 131 primes.
- The first 10^4 terms include the first 808 primes.
- The first 10^5 terms include the first 4397 primes.
- The first 10^6 terms include the first 35801 primes.
- The first 10^7 terms include the first 253682 primes.
Conjecture: this sequence is a permutation of the primes.
If we start with 1 instead of 2 we get A379727. - N. J. A. Sloane, Dec 31 2024

			a(6) is 19 because the prime factors of c=2*a(5)+1 (i.e., 95) are 5 and 19, and 5 already appears in the sequence as a(2).
a(9) is 31 because the prime factors of c=2*a(8)+1 (i.e., 15) are 3 and 5 which already appear in the sequence as a(7) and a(2). The next value of c (i.e., 2*c+1) is 31, which is prime and does not already appear in the sequence.

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

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

c[_] := True; j = 2; c[2] = False;
{j}~Join~Reap[Do[m = 2*j + 1;
  While[Set[k, SelectFirst[FactorInteger[m][[All, 1]], c] ];
    !IntegerQ[k],
    m = 2*m + 1];
  c[k] = False; j = Sow[k], {120}] ][[-1, 1]] (* Michael De Vlieger, Dec 29 2024 *)

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)

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.

Original entry on oeis.org

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)

A174161 a(1) = 2. Let k >= 1 be the minimal integer such that 2ka(n-1) - 1 has at least one prime divisor which is not already in the sequence. Then a(n) is the smallest such divisor.

Original entry on oeis.org

2, 3, 5, 19, 37, 73, 29, 23, 7, 13, 17, 11, 43, 257, 79, 157, 313, 139, 277, 41, 163, 31, 61, 487, 59, 47, 281, 1123, 449, 359, 239, 53, 211, 421, 101, 67, 89, 71, 283, 113, 677, 2707, 5413, 433, 173, 691, 1381, 251, 167, 1669, 5563, 7417, 44501, 431, 1723, 2297

Offset: 1

Vladimir Shevelev, Mar 10 2010

nonn

Conjectures: 1) The sequence is a permutation of prime numbers; 2) k = k(n) runs all positive integers.

Ivan Neretin, Table of n, a(n) for n = 1..10000

Cf. A005132, A063733, A078943, A174162.

a = {2}; Do[k = 1; While[(d = Complement[FactorInteger[2 k a[[-1]] - 1][[All, 1]], a]) == {}, k++]; AppendTo[a, Min@d], {n, 50}]; a (* Ivan Neretin, Dec 04 2018 *)

Terms from a(22) onwards corrected by Ivan Neretin, Dec 04 2018

cp's OEIS Frontend

A379652 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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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=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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A174161 a(1) = 2. Let k >= 1 be the minimal integer such that 2*k*a(n-1) - 1 has at least one prime divisor which is not already in the sequence. Then a(n) is the smallest such divisor.

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A379652 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.

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.

A174161 a(1) = 2. Let k >= 1 be the minimal integer such that 2ka(n-1) - 1 has at least one prime divisor which is not already in the sequence. Then a(n) is the smallest such divisor.