[seqfan] Divisor chains
Chuck Seggelin
seqfan at plastereddragon.com
Tue May 4 05:25:00 CEST 2004
Sorry for the double-post... for n=25 there are 1589 chains. First 25 terms
are therefore:
1, 1, 1, 1, 1, 1, 1, 5, 4, 3, 2, 8, 4, 6, 47, 44, 6, 37, 6, 166, 462, 232,
372, 2130, 1589
-- Chuck
----- Original Message -----
From: "Chuck Seggelin" <seqfan at plastereddragon.com>
To: "Richard Guy" <rkg at cpsc.ucalgary.ca>; "Math Fun"
<math-fun at mailman.xmission.com>; <seqfan at ext.jussieu.fr>
Cc: "Vaderlind, Paul -- Paul Vaderlind" <paul at matematik.su.se>; "Paul
Vaderlind" <paul at math.su.se>; "Loren Larson" <lllarsson at earthlink.net>
Sent: Monday, May 03, 2004 11:13 PM
Subject: Re: [seqfan] Divisor chains
> Wrote a litte recursive Maple code to solve this problem. My output
agrees
> with Richards for n from 1 to 9 and 11. For n=10 I find 3 chains
(Richard's
> + 2), for n=12 I find 8 chains (Richard's + 3), for n=13 I find 4
(Richard's
> 4'th n=13 chain includes "3" twice), for n=14 I find 6 chains (Richard's +
> 2). Other than the errant chain for 13, the chains Richard provides
appear
> to be correct.
>
> This makes the sequence up to 14 look like this:
>
> 1, 1, 1, 1, 1, 1, 1, 5, 4, 3, 2, 8, 4, 6
>
> From there on the values start getting much larger (here is n=15 to n=24):
>
> 47, 44, 6, 37, 6, 166, 462, 232, 372, 2130
>
> If anyone is interested I'm happy to share the code. I'm waiting on n=25
> right now. The trend is generally upward so I suspect there are MANY
chains
> for n=25.
>
> Given how many chains there are for these numbers, I would expect the
number
> of chains for n=37 to be pretty high, although for some primes (such as 17
> and 19) there are small numbers of chains. Richard do you suggest n=37
> because you believe there to be very few chains?
>
> -- Chuck
>
> ----- Original Message -----
> From: "Richard Guy" <rkg at cpsc.ucalgary.ca>
> To: "Math Fun" <math-fun at mailman.xmission.com>; <seqfan at ext.jussieu.fr>
> Cc: "Vaderlind, Paul -- Paul Vaderlind" <paul at matematik.su.se>; "Paul
> Vaderlind" <paul at math.su.se>; "Loren Larson" <lllarsson at earthlink.net>
> Sent: Monday, May 03, 2004 4:20 PM
> Subject: [seqfan] Divisor chains
>
>
> > The following might be an acceptable sequence
> > for Neil Sloane's OEIS, but someone needs to
> > do some work
> >
> > 1 2 3 4 5 6 7 8 9 10 11 12 13 14 ...
> > 1 1 1 1 1 1 1 5 4 1 2 5 5 4 ...
> >
> > It arises from a problem I got recently from
> > Paul Vaderlind.
> >
> > If the sequence of numbers from 1 to 37,
> > is arranged so that each term is a divisor
> > of the sum of preceding ones, starting
> > 37, 1, ...
> > what is the next term?
> >
> > The answer is either 2 or 19. P'r'aps
> > I won't spoil your fun by pointing out
> > which. But I will spoil it by asking:
> > How do you know that there is such a
> > `divisor chain' ?
> >
> > The sequence is (my present state of knowledge
> > of) the number of divisor chains of length n.
> > Here are the ones I found
> >
> > 1 2 1 3 1 2 4 2 3 1
> >
> > 5 1 2 4 3 6 2 4 3 5 1 7 1 2 5 3 6 4
> >
> > 8 2 5 3 6 4 7 1
> > 8 4 2 7 3 1 5 6
> > 8 4 2 7 3 6 5 1
> > 8 4 3 5 1 7 2 6
> > 8 4 6 3 7 2 5 1
> >
> > 9 1 2 4 8 6 5 7 3
> > 9 1 2 6 3 7 4 8 5
> > 9 3 4 8 6 5 7 2 1
> > 9 3 6 2 1 7 4 8 5
> >
> > 10 2 4 8 3 9 6 7 1 5
> >
> > 11 1 2 7 3 8 4 9 5 10 6
> > 11 1 4 8 6 10 5 9 2 7 3
> >
> > 12 2 1 5 10 3 11 4 8 7 9 6
> > 12 2 7 3 8 4 9 5 10 6 11 1
> > 12 3 5 10 2 8 4 11 1 7 9 6
> > 12 4 8 3 9 6 2 11 5 10 7 1
> > 12 4 8 6 10 5 9 2 7 3 11 1
> >
> > 13 1 2 8 3 9 4 10 5 11 6 12 7
> > 13 1 2 8 6 10 4 11 5 12 9 3 7
> > 13 1 2 8 6 10 4 11 5 3 9 12 7
> > 13 1 2 8 12 4 10 5 11 3 9 3 7
> > 13 1 2 8 12 4 10 5 11 6 9 3 7
> >
> > 14 2 8 6 10 4 11 5 3 9 12 7 13 1
> > 14 2 8 6 10 4 11 5 12 9 3 7 13 1
> > 14 2 8 12 4 10 5 11 6 9 3 7 13 1
> > 14 2 8 12 9 3 4 13 1 11 7 6 10 5
> >
> > Enough sins of omission, to say nothing of
> > commission, for today. Please check and extend.
> > Any ideas for proving anything? R.
> >
> >
>
>
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