A247339 - OEIS

%I #17 Sep 27 2015 17:50:44

%S 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,

%T 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,

%U 4,1,66,1,2,1,5,1,2,1,74,1,23,1,2,1,6,1,2

%N 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.

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

%C 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

%H Michel Lagneau, <a href="/A247339/b247339.txt">Table of n, a(n) for n = 1..10000</a>

%e 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.

%p with(numtheory):nn:=2000:T:=array(1..nn):U:=array(1..nn):

%p   for i from 1 to nn do:

%p     x:=factorset(i^2+1):T[i]:=x[1]:U[i]:=i:

%p   od:

%p     for n from 1 to 100 do:

%p      ii:=0:

%p       for k from 1 to 50000 while(ii=0) do:

%p        y:=factorset(k^2+1):n0:=nops(y):q:=y[n0]:

%p         if q=T[n]

%p          then

%p          ii:=1: printf(`%d, `,k):

%p          else

%p         fi:

%p      od:

%p    od:

%t 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 *)

%o (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

%Y Cf. A002522, A089120, A014442.

%K nonn

%O 1,2

%A _Michel Lagneau_, Sep 14 2014