I'm fairly convinced that for positive integer n > 0, if you choose the smallest k > 0 such that neither k-1 nor k+n divides X = LCM(k, k+1, ..., k+n-1), then A181063(n) = X. If this is so, it gives a pretty quick algorithm for extending A181063, supposing you have an large integer arithmetic package to compute the LCM.
-----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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