A072507

[hv at crypt.org]

Dean Hickerson <dean at math.ucdavis.edu> wrote:
:Mostly to Franklin T. Adams-Watters:
:
:> The next question is, what is the length, for a given n, of the longest 
:> sequence of integers with exactly n divisors?
:...
:> 6 4 or 5
:...
:> If the case n=6 can be resolved, I will go ahead and submit the sequence.
:
:The longest sequence has length 5:
:
:    10093613546512321 = 7^2 205992113194129
:    10093613546512322 = 2 71040881^2
:    10093613546512323 = 3^2 1121512616279147
:    10093613546512324 = 2^2 2523403386628081
:    10093613546512325 = 5^2 403744541860493
:
:I don't know if this is the smallest example; I only looked for ones in which
:the last 4 numbers have the forms  2 p^2,  3^2 q,  2^2 r,  and  5^2 s,  with
:primes p, q, r, and s.

Assuming my code is correct, it confirms that there is no smaller solution.

#!/usr/bin/perl -w
use strict;
BEGIN { $Math::Pari::initprimes = 72e6 }
use Math::Pari qw/ prime divisors /;

my($count, $s) = (0, "");
for (my($pi, $p) = (1, 2); $p < 71040881; $p = prime(++$pi)) {
  my($min, $max) = map 2 * $p * $p, 1 .. 2;
  while (--$min) { last unless @{ divisors($min) } == 6 };
  while (++$max) { last unless @{ divisors($max) } == 6 };
  print("\n"), die $min + 1 if $max - $min == 6;
}

Hugo