Well, that's interesting. I would have guessed that A181063(n) is the smallest x = LCM(k..k+n-1) for the smallest k where neither k-1 nor k+n divides x (in fact, I did guess that). However, n = 14 proves me wrong. The smallest admissible k is 17, giving LCM(17..30) = 1164544781400. However, k =20 gives the yet smaller LCM(20..33) = 447069823200, which was already in the sequence. When I went to submit my (incorrect) b-file, the submission was rejected because of the disagreement at a(14). It's good to be schooled now and then.
-----Original Message----- From: math-fun [mailto:math-fun- bounces+davidwwilson=comcast.net@mailman.xmission.com] On Behalf Of W. Edwin Clark Sent: Thursday, November 20, 2014 10:13 PM To: math-fun Subject: Re: [math-fun] Divisor runs
These sequences seem to be relevant to some of your questions:
On Thu, Nov 20, 2014 at 7:43 PM, David Wilson <davidwwilson@comcast.net> wrote:
The number 6 has divisors (1, 2, 3, 6).
Thus it has a run of 3 contiguous divisors (1, 2, 3).
What is the smallest number N with
- A run of at least 5 divisors?
- A run of exactly 5 divisors?
- A maximum run of exactly 5 divisors?
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