[seqfan] Re: Mobile phone security!

[Rob Arthan]

Neil,

OK. Thanks for the encouragement. Will do!

By the way, there is a blooper in my description below: “acyclic eulerian paths” should just read “acyclic paths”.

Regards,

Rob.

On 20 Sep 2014, at 21:30, Neil Sloane <njasloane at gmail.com> wrote:

> of course - please submit it!
> Neil
> 
> On Sat, Sep 20, 2014 at 4:17 PM, Rob Arthan <rda at lemma-one.com> wrote:
> 
>> I have been playing with a two-dimensional sequence inspired by the
>> security patterns that you can use like passwords on Android phones.
>> See:
>> 
>> 
>> http://math.stackexchange.com/questions/37167/combination-of-smartphones-pattern-password
>> 
>> But I don’t think any of the answers there except (possibly) the one I
>> added this
>> afternoon can be right.
>> 
>> Ignoring the minor detail that Android requires the patterns to have at
>> least
>> four points, this suggests a two-dimensional sequence a(m, n) defined as
>> follows.
>> Let G(m, n) be the graph whose vertices are the integer lattice points (p,
>> q)
>> with 0 <= p < m and 0 <= q < n. The graph has an edge between v
>> and w iff the line segment [v, w] does not contain any other
>> integer lattice points (equivalently, iff v - w = (i, j) with i and j
>> coprime).
>> a(m, n) is the number of acyclic eulerian paths in G(m, n).
>> 
>> I implemented a brute force search and got the following values of:
>> a(m, n) for 1 <= m <= 3 and 1 <= n <= 4:
>> 
>> 0, 2, 6, 12
>> 2, 60, 1058, 25080
>> 6, 1058, 140240, 58673472
>> 
>> I have only been able to verify these results independently
>> in the cases when one of m or n is 1 (which is easy because
>> the paths are uniquely determined by their end-points in that
>> case and in the case m = n = 2 (which is easy because G(2, 2)
>> is the complete graph on 4 vertices).  I would be very grateful
>> for confirmation of (or corrections to) my results and for any
>> thoughts on an efficient way of calculating the sequence.
>> 
>> My OEIS searches haven’t come up with anything like this.
>> Do people think this sequence is worth submitting to OEIS?
>> 
>> Regards,
>> 
>> Rob.
>> 
>> 
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>> 
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>> 
> 
> 
> 
> -- 
> Dear Friends, I have now retired from AT&T. New coordinates:
> 
> Neil J. A. Sloane, President, OEIS Foundation
> 11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
> Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
> Phone: 732 828 6098; home page: http://NeilSloane.com
> Email: njasloane at gmail.com
> 
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