A247339 a(n) is the least number k such that the greatest prime divisor of k^2+1 is the smallest prime divisor of n^2+1.

1, 2, 1, 4, 1, 6, 1, 2, 1, 10, 1, 2, 1, 14, 1, 16, 1, 2, 1, 20, 1, 2, 1, 24, 1, 26, 1, 2, 1, 4, 1, 2, 1, 5, 1, 36, 1, 2, 1, 40, 1, 2, 1, 5, 1, 12, 1, 2, 1, 9, 1, 2, 1, 54, 1, 56, 1, 2, 1, 5, 1, 2, 1, 4, 1, 66, 1, 2, 1, 5, 1, 2, 1, 74, 1, 23, 1, 2, 1, 6, 1, 2

Offset

1,2

Comments

a(n)=n if n^2+1 is prime and a(n)=1 if n is odd.

Conjecture: for all integer n, there exists at least an integer m <= n such that the smallest prime factor of n^2+1 is also the greatest prime factor of m^2+1. - Michel Lagneau, Sep 27 2015

Links

Michel Lagneau, Table of n, a(n) for n = 1..10000

Example

a(34)=5 because the greatest prime divisor of 5^2+1 = 2*13 is the smallest prime divisor of 34^2+1 =13*89.

Maple

with(numtheory):nn:=2000:T:=array(1..nn):U:=array(1..nn):
  for i from 1 to nn do:
    x:=factorset(i^2+1):T[i]:=x[1]:U[i]:=i:
  od:
    for n from 1 to 100 do:
     ii:=0:
      for k from 1 to 50000 while(ii=0) do:
       y:=factorset(k^2+1):n0:=nops(y):q:=y[n0]:
        if q=T[n]
         then
         ii:=1: printf(`%d, `, k):
         else
        fi:
     od:
   od:

Mathematica

Table[k = 1; While[FactorInteger[k^2 + 1][[-1, 1]] != FactorInteger[n^2 + 1][[1, 1]], k++]; k, {n, 82}] (* Michael De Vlieger, Sep 27 2015 *)

Prog

(PARI) a(n) = {f = factor(n^2+1)[1, 1]; k = 1; while (! ((g=factor(k^2+1)) && (g[#g~, 1] == f)), k++); k; } \\ Michel Marcus, Sep 14 2014

Crossrefs

Cf. A002522, A089120, A014442.

Sequence in context: A329642 A214052 A276094 * A281071 A256908 A346466

Adjacent sequences: A247336 A247337 A247338 * A247340 A247341 A247342

Keyword

nonn

Author

Michel Lagneau, Sep 14 2014

Status

approved