A072507
Dean Hickerson
dean at math.ucdavis.edu
Thu Jul 27 05:16:06 CEST 2006
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?
...
> 10 2 to 7
The maximum is 3, as in this example:
7939375 = 5^4 12703
7939376 = 2^4 496211
7939377 = 3^4 98017
If there were 4, then 2 of them, say n and n+2, must be even, and hence equal
to either 2^9, 2^4 p, or 2 p^4 for some odd prime p. The first doesn't
work, since neither 510 nor 514 has the right form. Since one of the 2
numbers is divisible by 4 and the other is not, we either have
n = 2 p^4 and n+2 = 2^4 q.
or
n = 2^4 p and n+2 = 2 q^4
The first case implies that p^4 == 7 (mod 8) and the second implies (since
p is odd) that q^4 == 9 (mod 16). Both of these are impossible.
Dean Hickerson
dean at math.ucdavis.edu
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