A258840 a(n) is the least integer k such that there are n values of i <= k for which gpf(i^2 + 1) = gpf(k^2 + 1), where gpf(x) is the greatest prime factor of x.

1, 3, 7, 38, 47, 157, 302, 327, 515, 616, 697, 798, 818, 1303, 2818, 3141, 3323, 5648, 6962, 9193, 9872, 13213, 13747, 15445, 16271, 17149, 18263, 20491, 20727, 24389, 26915, 29078, 31867, 37848, 38007, 40182, 41508, 43328, 46349, 55025, 62258, 63133, 66893

Offset: 1

Author

Michel Lagneau, Jun 12 2015

Keywords

• nonn

Comments

A014442(n) gives the largest prime factor of n^2 + 1.

The primes of the sequence are 3, 7, 47, 157, 1303, 3323, 46349, ...

The corresponding sequence Gpf(a(n)^2+1) is 2, 5, 5, 17, 17, 29, 37, 37, 101, 101, 101, 101, 101, 101, 101, 101, 101, 97, 97, 97, 97, 401, 349, 389, 557, 557, 557, 557, 557, 421, 421, 421, 557, ... and it is interesting to observe the frequency of repetitions for the numbers 5, 17, 37, 97, 101, 557, ...

Examples

			a(3) = 7 because gpf(7^2 + 1) = gpf(3^2 + 1) = gpf(2^2 + 1) = 5 => 3 occurrences.
a(4) = 38 because gpf(38^2 + 1) = gpf(21^2 + 1) = gpf(13^2 + 1) = gpf(4^2 + 1) = 17 => 4 occurrences.
		

Crossrefs

Cf. A014442, A242012.

Programs

Maple

with(numtheory):nn:=70000:T:=array(1..nn):k:=0:kk:=1:
for m from 1 to nn do:
x:=factorset(m^2+1):n1:=nops(x):p:=x[n1]:k:=k+1:T[k]:=p:
od:
for n from 1 to 43 do:jj:=0:for k from kk to nn while(jj=0) do:
  q:=T[k]:ii:=0:jj:=0:
    for i from 1 to k do:
      if T[i]=q then ii:=ii+1:
      else
      fi:
    od:if ii=n then jj:=1:kk:=k:
    printf ( "%d %d \n",n,k):else fi:
  od:od:

PARI

gpf(n) = my(f=factor(n^2+1)); f[#f~,1];
nboc(k) = my(gpfk = gpf(k)); sum(i=1, k, gpf(i) == gpfk);
a(n) = my(k = 1); while (nbo(k) != n, k++); k; \\ Michel Marcus, Jun 12 2015