[seqfan] Divisor chains

[Chuck Seggelin]

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.
>
>