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Re: Finding pairs of numbers that meet a specific criteria
Posted:
Nov 5, 2011 9:06 AM
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hi All,
I'v been following this thread with keen interest.
Sequence 2,5,7,10,13,15,18 ... is known as http://oeis.org/A001950 "Upper
Wythoff sequence (a Beatty sequence): a(n) = floor(n*phi^2), where phi =
(1+sqrt(5))/2" if you drop the initial (0,0) pair.
Along with previous remarks, it all boils down to pairs (z(n)-n, z(n)) with
z(n)=Floor[n*phi^2]:
For those into Mathematica: Table[z=Floor[n*GoldenRatio^2] ;
{z-n,z},{n,100}]
{{1,2},{3,5},{4,7},{6,10},{8,13},{9,15},{11,18},{12,20},{14,23},{16,26}, ...
... , {153,248},{155,251},{156,253},{158,256},{160,259},{161,261}}
In the exploration phase, I naively explored the problem using a backtrack
algorithm:
base3=Table[{j,i+j},{i,1,12},{j,i,2i}]
{ {{1, 2}, {2, 3}},
{{2, 4}, {3, 5}, {4, 6}},
{{3, 6}, {4, 7}, {5, 8}, {6, 9}},
{{4, 8}, {5, 9}, {6, 10}, {7, 11}, {8, 12}}
...etc..
<<DiscreteMath`Combinatorica`;
queerQueens=Backtrack[base3,(And[UnsameQ@@Flatten[#],UnsameQ@@Subtract@@@#])
&,True&,One]
{{1, 2}, {3, 5}, {4, 7}, {6, 10}, {8, 13}, {9, 15}, {11, 18}, {12, 20}, {14,
23}, {16, 26}, {17, 28}, {19, 31}}
and that confirmed the structure "nth pair is (x, x+n-1)" reported earlier.
Wouter.
----- Original Message -----
From: "Walter Wallis" <wdwallis@siu.edu>
To: "Esther Brink" <bestes32@hotmail.com>
Cc: <discretemath@mathforum.org>; <benb@wcjc.edu>
Sent: Thursday, November 03, 2011 6:53 PM
Subject: Re: Finding pairs of numbers that meet a specific criteria
> Call the pairs (a_i, b_i) where a_i ? b_i. Then neither sequence, the
> a_i's or the b_i's, is in the handbok of integer sequences.
> Interesting.
>
>
> On Wed, Nov 2, 2011 at 7:23 PM, Esther Brink <bestes32@hotmail.com> wrote:
> > Musicgold:
> > Not so fast! It may be possible to work out a better formula!
> > As a first small step, if I'm not mistaken the "x" in her post need only
> > be the smallest positive integer not in the immediately preceding pair.
> > That would make the problem amenable to "recursive" methods.
> > Let's see if some of the experts can refine Dr. Richardson's analysis
> > further.
> > Thanks for a great problem.
> > Ben
> >> Date: Wed, 2 Nov 2011 10:34:03 -0400
> >> From: discussions@mathforum.org
> >> To: discretemath@mathforum.org
> >> Subject: Re: Finding pairs of numbers that meet a specific criteria
> >>
> >> Angela,
> >>
> >> Brilliant! Yes, it seems to be the solution. I did not occur to me.
> >> Thanks.
>
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