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@cpsc.ucalgary.ca> To: "Math Fun" <math-fun@mailman.xmission.com>; <seqfan@ext.jussieu.fr> Cc: "Vaderlind, Paul -- Paul Vaderlind" <paul@matematik.su.se>; "Paul Vaderlind" <paul@math.su.se>; "Loren Larson" <lllarsson@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.