A031439 -id:A031439 - cp's OEIS frontend

A072268 a(0)=1; a(n+1) = 1 + f(a(n))^2, where f(x) is the largest prime factor of x (A006530).

1, 2, 5, 26, 170, 290, 842, 177242, 160802, 2810, 78962, 9223370, 5033760602, 2935496262242, 2154284576409188208716642, 1379590379356276893461978662419832989306970202, 10320758390549056348725939119133160378521185060950774444682

Offset: 0

Reinhard Zumkeller, Jul 08 2002

nonn

Is the sequence bounded?

Essentially the same as A031439; a(n) = A031439(n-1)^2 + 1. - Charles R Greathouse IV, May 08 2009

			Given a(5)=290: a(6) = 1 + lpf(a(5))^2 = 1 + lpf(290)^2 = 1 + 29^2 = 842.

Cf. A031439.

with(numtheory): a[0]:=1: a[1]:=2: for n from 2 to 20 do b:=factorset(a[n-1]): a[n]:=1+op(nops(b),b)^2: od: seq(a[n],n=0..20); # Emeric Deutsch, Feb 05 2006

NestList[1+FactorInteger[#][[-1,1]]^2&,1,17] (* Harvey P. Dale, Feb 01 2022 *)

More terms from Emeric Deutsch, Feb 05 2006

a(16) corrected by T. D. Noe, Nov 26 2007

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.

Original entry on oeis.org

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)

A056650 a(n) is the greatest prime factor of a(n-1)^2+a(n-1)+1.

Original entry on oeis.org

2, 7, 19, 127, 5419, 1009, 9181, 1423, 96493, 163350799, 25249969, 212520319916977, 784949209969, 145542538757017, 147660435988297, 2508855873622663, 3565137918692593, 6521735641, 11273204227, 16141059763, 125679599753438821, 240780337980146570229319, 7282590063606707136764017

Offset: 0

Henry Bottomley, Aug 09 2000

nonn

Except for the first term the initial (ten at least) terms are one more than a multiple of 6, with the result that a(n)^2+a(n)+1 is in these cases an odd multiple of 3 and appears to be 3 times the product of primes all of which are one more than a multiple of 6.

Dennis Langdeau, Table of n, a(n) for n = 0..31

Cf. A031439, A045375.

More terms from Vladeta Jovovic, Nov 26 2001

A125256 Smallest odd prime divisor of n^2 + 1.

Original entry on oeis.org

5, 5, 17, 13, 37, 5, 5, 41, 101, 61, 5, 5, 197, 113, 257, 5, 5, 181, 401, 13, 5, 5, 577, 313, 677, 5, 5, 421, 17, 13, 5, 5, 13, 613, 1297, 5, 5, 761, 1601, 29, 5, 5, 13, 1013, 29, 5, 5, 1201, 41, 1301, 5, 5, 2917, 17, 3137, 5, 5, 1741, 13, 1861, 5, 5, 17, 2113, 4357, 5, 5

Offset: 2

Nick Hobson, Nov 26 2006

Any odd prime divisor of n^2+1 is congruent to 1 modulo 4.
n^2+1 is never a power of 2 for n > 1; hence a prime divisor congruent to 1 modulo 4 always exists.
a(n) = 5 if and only if n is congruent to 2 or -2 modulo 5.
If the map "x -> smallest odd prime divisor of n^2+1" is iterated, does it always terminate in the 2-cycle (5 <-> 13)? - Zoran Sunic, Oct 25 2017

			The prime divisors of 8^2 + 1 = 65 are 5 and 13, so a(7) = 5.

D. M. Burton, Elementary Number Theory, McGraw-Hill, Sixth Edition (2007), p. 191.

Ray Chandler, Table of n, a(n) for n = 2..20001 (first 999 terms from Nick Hobson)

Cf. A002522, A002496, A014442, A057207, A031439.
For bisections see A256970, A293958.
See also A125257, A125258, A294656, A294657, A294658.

with(numtheory, factorset);
A125256 := proc(n) local t1,t2;
if n <= 1 then return(-1); fi;
if (n mod 5) = 2 or (n mod 5) = 3 then return(5); fi;
t1 := numtheory[factorset](n^2+1);
t2:=sort(convert(t1,list));
if (n mod 2) = 1 then return(t2[2]); fi;
t2[1];
end;
[seq(A125256(n),n=1..40)]; # N. J. A. Sloane, Nov 04 2017

Table[Select[First/@FactorInteger[n^2+1],OddQ][[1]],{n,2,68}] (* James C. McMahon, Dec 16 2024 *)

vector(68, n, if(n<2, "-", factor(n^2+1)[1+(n%2),1]))

A125256(n)=factor(n^2+1)[1+bittest(n,0),1] \\ M. F. Hasler, Nov 06 2017

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)

A166134 a(n+1) is the smallest divisor of a(n)^2+1 that does not yet appear in the sequence, with a(1) = 1.

Original entry on oeis.org

1, 2, 5, 13, 10, 101, 5101, 26, 677, 45833, 65, 2113, 446477, 130, 16901, 41, 29, 421, 17, 58, 673, 45293, 25, 313, 97, 941, 34057, 50, 61, 1861, 1229, 773, 59753, 89, 34, 1157, 82, 269, 194, 617, 38069, 55740337, 145, 10513, 11052317, 12215371106849

Offset: 1

Franklin T. Adams-Watters, Oct 07 2009

nonn

All members of the sequence can be represented as the sum of two relatively prime numbers (A008784). It appears that the sequence is infinite and that all such numbers are present.

			After a(4)=13, the divisors of 13^2+1=170 are 1,2, 5, 10, 17, 34, 85, 170. 1, 2, and 5 have already occurred, so a(5) = 10.

Ivan Neretin, Table of n, a(n) for n = 1..1000

Cf. A166133, A008784, A031439, A002522.

Nest[Append[#, Min[Complement[Divisors[#[[-1]]^2 + 1], #]]] &, {1}, 45] (* Ivan Neretin, Sep 03 2015 *)

invec(v,x,n)=for(i=1,n,if(v[i]==x,return(1)));0
bl(n)={local(v,d,ds);
v=vector(n,i,1);
for(i=2,n,
ds=divisors(v[i-1]^2+1);
for(k=2,#ds,d=ds[k];if(!invec(v,d,i-1),v[i]=d;break)));
v}

A083557 a(n) is the greatest prime factor of 3*a(n-1)+2.

Original entry on oeis.org

3, 11, 7, 23, 71, 43, 131, 79, 239, 719, 127, 383, 1151, 691, 83, 251, 151, 13, 41, 5, 17, 53, 23, 71, 43, 131, 79, 239, 719, 127, 383, 1151, 691, 83, 251, 151, 13, 41, 5, 17, 53, 23, 71, 43, 131, 79, 239, 719, 127, 383, 1151, 691, 83, 251, 151, 13, 41, 5, 17, 53, 23

Offset: 1

Yasutoshi Kohmoto, Jun 05 2003

nonn

Conjecture: if a(1)=m then the sequence becomes cyclic, for any m.

Conjecture verified up to 25000000 by Jud McCranie, Jun 11 2003

Cf. A031439, A031440, A031442, A082021, A082132, A084923.

f[n_] := Flatten[Table[ #[[1]], {1}] & /@ FactorInteger[ 3n + 2 ]][[ -1]]; NestWhileList[f, 3, UnsameQ, All]
NestList[FactorInteger[3#+2][[-1,1]]&,3,70] (* Harvey P. Dale, Feb 21 2013 *)

lista(nn) = {print1(a = 3, ", "); for (n=1, nn, a = vecmax(factor(3*a+2)[,1]); print1(a, ", "););} \\ Michel Marcus, Jul 15 2017

G.f.: x*(3 + 11*x + 7*x^2 + 23*x^3 + 71*x^4 + 43*x^5 + 131*x^6 + 79*x^7 + 239*x^8 + 719*x^9 + 127*x^10 + 383*x^11 + 1151*x^12 + 691*x^13 + 83*x^14 + 251*x^15 + 151*x^16 + 13*x^17 + 41*x^18 + 2*x^19 + 6*x^20 + 46*x^21) / (1 - x^19) (conjectured). - Colin Barker, Jul 15 2017

A279687 a(0) = 1, a(n) is the least prime factor of a(n-1)^2+1 that has not previously appeared in the sequence for n > 0.

Original entry on oeis.org

1, 2, 5, 13, 17, 29, 421, 401, 37, 1877, 41

Offset: 0

Franklin T. Adams-Watters, Dec 17 2016

			a(7) is a prime factor of a(6)^2+1 = 421^2 + 1 = 177242, which factors as 2*13*17*401. 2, 13, and 17 have already appeared in the sequence, so a(7) = 401.
a(10)^2+1 = 882 = 2 * 29^2. Both 2 and 29 have already appeared in the sequence, so it terminates.

Cf. A031439.

cp's OEIS Frontend

A072268 a(0)=1; a(n+1) = 1 + f(a(n))^2, where f(x) is the largest prime factor of x (A006530).

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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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A056650 a(n) is the greatest prime factor of a(n-1)^2+a(n-1)+1.

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A125256 Smallest odd prime divisor of n^2 + 1.

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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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A166134 a(n+1) is the smallest divisor of a(n)^2+1 that does not yet appear in the sequence, with a(1) = 1.

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A083557 a(n) is the greatest prime factor of 3*a(n-1)+2.

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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.