A072507
hv at crypt.org
hv at crypt.org
Fri Jul 28 20:25:29 CEST 2006
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
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