A253596 Numbers k such that A002313(m) is the greatest prime divisor of k^2 + 1 and A002313(m+1) is the greatest prime divisor of (k+1)^2 + 1 for some m.

1, 7, 31, 293, 1936, 2244, 4158, 5744, 11573, 25242, 285202, 339354

Offset: 1

Author

Michel Lagneau, Jan 05 2015

Keywords

• nonn
• more

Comments

A002313 contains the primes congruent to 1 or 2 (mod 4).

The corresponding indices m in A002313 are 1, 2, 6, 13, 69, 65, 322, 199, 130, 46, 1471, 866, ...

The corresponding primes A002313(m) are 2, 5, 37, 101, 809, 761, 4877, 2777, 1709, 509, 26821, 14957, ...

Examples

			31 is in the sequence because 31^2 + 1 = 2*13*37 and 32^2 + 1 = 5*5*41 with the property that 37 = A002313(6) and 41 = A002313(7).
		

Crossrefs

Cf. A002313, A014442.

Programs

Maple

with(numtheory): nn:=500000:print(1):
for n from 1 to nn do:
   p:=n^2+1:x:=factorset(p):n0:=nops(x):p1:=x[n0]:
   q:=(n+1)^2+1:y:=factorset(q):n1:=nops(y):p2:=y[n1]:ii:=0:
     for j from 2 by 2 to 1000 while(ii=0) do:
      pp:=p1+j:
      if type(pp,prime)=true and irem(pp,4)=1
      then
      p3:=pp:ii:=1:
      else
      fi:
    od:
    if p3=p2
    then
    print(n):
     else
     fi:
    od:

Mathematica

lst={};Do[If[Mod[Prime[i],4]==1||Mod[Prime[i],4]==2,AppendTo[lst,Prime[i]]],{i,1,1000}];Do[Do[If[FactorInteger[n^2+1][[-1]][[1]]==Part[lst,j]&&FactorInteger[(n+1)^2+1][[-1]][[1]]==Part[lst,j+1],Print[n]],{n,1,20000}],{j,1,999}]