[seqfan] Re: Puzzle of David Wilson

[Matthijs Coster]

Hello Bob and all of them who are interested in the Puzzle,

I started an experiment with r = 25/27. I decided to have a very small 
set S, but then I continued with all fractions in S until the fractions 
exceeded all the norm (i.e. the size of the denominator) which was fixed 
at 10^80. As long the size of succeeders were smaller than the norm the 
fractions were added to S. The result was that 8.2 % of the succeeder's 
had smaller norms than the originals. Let No and Ns be the norms of 
originals and succeeder respectively. I made a table of the quotients 
floor(Ns/No) (increasing case) and floor(No/Ns) (decreasing case):

Explanation: First column n; second column increasing case; third column 
decreasing case
1 72575 39547
2 69859 13382
3 60794 6670
4 56907 4023
5 55422 2628
6 53012 1847
7 51051 1410
8 49402 1146
9 43799 911
10 32041 793
11 31461 595
12 30447 495
13 29612 466
14 28794 361
15 27919 331
16 27269 316
17 26087 275
18 25993 260
19 25461 192
20 24963 175
21 24071 160
22 23798 184
23 23064 150
24 22626 133
25 22248 118
26 21514 124
27 21248 111
28 20895 90
29 3302 93
30 0 2674

Very remarkable are the enormous large numbers which appear in 
floor(No/Ns):

21043, 68249, 22535 and 139697

It is rather comparable to the convergents of continued fractions.
By the way, I didn't found a Wilson--chain.

Best regards,

--Matthijs Coster


Bob Selcoe schreef op 28-1-2015 om 23:47:
> Hi Matthijs and Seqfans,
>
> Thanks for writing the paper and citing me, Matthijs!   I will need 
> time to review.
>
> One thing first - I'd be surprised if all real positive rational 
> numbers < 1 didn't create Wilson-chains.
>
> Would you mind trying my "modular approach" with, say, 9/11?  It may 
> require a somewhat different program - one that (I think) would 
> eliminate many of the iterations required with the current program.
>
> Here, the first term (11/2) would be the last term in your current 
> program, then just make a "modular chain" until you reach 2/11.
>
> So the first few steps are:
>
> R_(1-9/11) = 11/2 = 121/22.   9/11 = 18/22.   121 mod 18 = 15.
>
> R_(15/22) = 22/15 = 242/165.   9/11 = 135/165.   242 mod 135 = 107.
>
> R_(107/165) = 165/107 = 1815/1177.  9/11 = 963/1177.   1815 mod 963 = 
> 852.
> etc.
>
> So generally, when a/b > 1/2, the process is first find b^2 mod 
> a(b-a), = a' and b(b-a) = b', then continue the process with b*b' mod 
> a*a' = a'' and b*b' = b'', etc., until reaching (b-a)/b.
>
> (Of course, all fractions - and thus the a-primes and b-primes - may 
> be in reduced form).
>
> Certainly, this doesn't preclude other possible Wilson-chains existing 
> for 9/11 (as you've shown occurs with 7/9); but I think this 
> particular approach can lead to more economical paths.  While it might 
> take many steps, each step really is only one iteration; so I would 
> think it wouldn't be too intense for the computer.  I would do this 
> myself but I don't know the first thing about computer programming!
>
> Best,
> Bob S.
>
>
>
> --------------------------------------------------
> From: "Matthijs Coster" <seqfan at matcos.nl>
> Sent: Wednesday, January 28, 2015 7:59 AM
> To: "_Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
> Subject: [seqfan] Puzzle of David Wilson
>
>> LS,
>>
>> For everybody who is interested in the puzzle of David Wilson (z -> z 
>> + kr, or z -> 1/z).
>> I wrote a paper with my results.
>> See: http://www.matcos.nl/sequences/SeqFanPuzzle.pdf.
>>
>> Please send me your commands!
>>
>> Best regards,
>>
>> --Matthijs
>>
>>
>>
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