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

1,2

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, ...

Links

Table of n, a(n) for n=1..43.

Example

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.

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:

Prog

(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

Crossrefs

Cf. A014442, A242012.

Sequence in context: A162926 A042895 A173561 * A209029 A373774 A169741

Adjacent sequences: A258837 A258838 A258839 * A258841 A258842 A258843

Keyword

nonn

Author

Michel Lagneau, Jun 12 2015

Status

approved