[math-fun] 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.
You seem to be considering only those chains for which the first element is the largest. Otherwise we have chains such as (2 1 3), (2 4 3 1), (2 4 3 1 5), (3 1 4 2 5), or (4 1 5 2 3). Gene --- Richard Guy <rkg@cpsc.ucalgary.ca> wrote:
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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OK, but that's another sequence. I would rule out your second and third examples on the grounds that 4 doesn't divide 2. R. On Mon, 3 May 2004, Eugene Salamin wrote:
You seem to be considering only those chains for which the first element is the largest. Otherwise we have chains such as (2 1 3), (2 4 3 1), (2 4 3 1 5), (3 1 4 2 5), or (4 1 5 2 3).
Gene
--- Richard Guy <rkg@cpsc.ucalgary.ca> wrote:
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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On Mon, 3 May 2004, Richard Guy wrote:
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 ...
I compute up to n = 19 the following number of divisor chains: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 1 1 1 1 1 1 5 4 3 2 8 4 6 47 44 6 37 6 which differs from your values at 10, 12, 13, 14. It's interesting that a(15) > a(16). One might think powers of 2 would give larger values. If anyone cares to check my chains from 10 to 16 here they are: 10 2 4 8 6 5 7 3 9 1 10 2 4 8 3 9 6 7 1 5 10 2 6 3 7 4 8 5 9 1 11 1 4 8 6 10 5 9 2 7 3 11 1 2 7 3 8 4 9 5 10 6 12 3 5 10 2 1 11 4 8 7 9 6 12 2 7 1 11 3 9 5 10 4 8 6 12 2 7 3 8 4 9 5 10 6 11 1 12 4 8 3 9 6 2 11 5 10 7 1 12 4 8 6 10 5 9 2 7 3 11 1 12 6 9 3 2 8 4 11 5 10 7 1 12 3 5 10 2 8 4 11 1 7 9 6 12 2 1 5 10 3 11 4 8 7 9 6 13 1 2 8 6 10 4 11 5 3 9 12 7 13 1 2 8 6 10 4 11 5 12 9 3 7 13 1 2 8 12 4 10 5 11 6 9 3 7 13 1 2 8 3 9 4 10 5 11 6 12 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 3 9 12 4 13 1 11 7 6 10 5 14 2 8 3 9 4 10 5 11 6 12 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 15 3 2 1 7 14 6 4 13 5 10 8 11 9 12 15 5 10 6 12 8 14 7 11 2 9 1 4 13 3 15 5 4 3 9 12 2 10 6 11 1 13 7 14 8 15 5 4 2 13 1 8 12 3 7 14 6 9 11 10 15 5 10 3 11 4 12 2 1 9 6 13 7 14 8 15 5 10 3 11 4 6 9 7 14 12 8 13 1 2 15 3 6 4 14 7 1 10 5 13 2 8 11 9 12 15 3 2 4 12 9 5 10 6 11 1 13 7 14 8 15 1 2 9 3 10 4 11 5 12 6 13 7 14 8 15 5 10 3 11 1 9 6 12 4 2 13 7 14 8 15 5 4 12 9 3 2 10 6 11 1 13 7 14 8 15 1 4 2 11 3 9 5 10 12 6 13 7 14 8 15 1 8 12 6 2 11 5 10 14 7 13 4 9 3 15 5 10 2 1 11 4 6 9 7 14 12 8 13 3 15 3 6 4 14 7 1 2 13 5 10 8 11 9 12 15 5 10 6 3 13 4 7 9 12 14 1 11 2 8 15 5 10 2 4 9 3 12 6 11 1 13 7 14 8 15 5 4 2 13 3 6 1 7 14 10 8 11 9 12 15 5 10 3 11 1 9 2 4 12 6 13 7 14 8 15 5 10 6 3 13 4 14 7 11 8 12 9 1 2 15 5 4 2 13 1 8 6 9 7 14 12 3 11 10 15 5 10 6 9 3 4 13 1 11 7 12 2 14 8 15 5 10 6 2 1 13 4 14 7 11 8 12 9 3 15 5 1 7 2 6 3 13 4 14 10 8 11 9 12 15 5 10 2 4 12 6 9 3 11 1 13 7 14 8 15 5 4 2 13 3 14 7 1 8 12 6 9 11 10 15 5 10 6 12 4 2 9 3 11 1 13 7 14 8 15 5 4 2 13 3 6 8 7 9 12 14 1 11 10 15 5 10 6 12 8 14 7 11 2 1 13 4 9 3 15 5 10 6 3 13 4 14 7 11 8 2 1 9 12 15 3 9 1 2 10 4 11 5 12 6 13 7 14 8 15 5 10 6 9 3 8 14 7 11 2 1 13 4 12 15 1 2 3 7 14 6 4 13 5 10 8 11 9 12 15 1 8 12 9 5 10 6 11 7 14 2 4 13 3 15 5 10 6 3 13 4 8 2 11 7 14 1 9 12 15 5 10 6 9 3 12 4 2 11 1 13 7 14 8 15 5 10 2 1 11 4 12 3 9 6 13 7 14 8 15 5 2 11 3 6 14 7 9 8 10 1 13 4 12 15 5 10 3 11 4 8 14 2 6 13 7 1 9 12 15 5 10 6 2 1 13 4 14 7 11 8 3 9 12 15 3 9 1 14 6 4 13 5 7 11 8 12 2 10 15 3 6 12 9 5 10 4 2 11 1 13 7 14 8 15 1 8 3 9 6 2 11 5 10 14 7 13 4 12 15 5 4 12 9 3 8 14 2 6 13 7 1 11 10 15 3 2 10 5 7 14 8 1 13 6 4 11 9 12 15 5 10 6 3 13 4 14 7 1 2 8 11 9 12 15 5 4 3 9 12 8 14 2 6 13 7 1 11 10 16 4 10 15 3 2 5 11 6 12 7 13 8 14 9 1 16 2 6 12 9 5 10 15 3 13 7 14 8 1 11 4 16 8 12 9 5 10 6 11 7 14 2 4 13 3 15 1 16 8 12 9 5 10 15 3 13 7 1 11 2 14 6 4 16 8 3 9 6 2 11 5 10 14 7 13 4 12 15 1 16 8 12 9 15 10 5 3 13 7 14 2 6 1 11 4 16 8 12 9 5 10 15 3 13 7 14 2 6 1 11 4 16 2 6 4 14 3 15 10 7 11 8 12 9 13 5 1 16 8 6 15 9 2 14 5 3 13 7 1 11 10 12 4 16 4 2 11 3 9 5 10 12 6 13 7 14 8 15 1 16 2 9 3 10 4 11 5 12 6 13 7 14 8 15 1 16 2 9 3 15 5 10 12 6 13 7 14 8 1 11 4 16 8 12 6 2 11 5 10 14 7 13 4 9 3 15 1 16 8 2 13 3 14 7 9 12 6 15 5 10 1 11 4 16 8 2 13 1 5 15 3 7 14 6 9 11 10 12 4 16 2 6 12 9 15 10 5 3 13 7 14 8 1 11 4 16 8 12 3 13 2 9 7 14 6 15 5 10 1 11 4 16 2 9 3 15 5 10 6 11 1 13 7 14 8 12 4 16 8 2 13 3 6 1 7 14 5 15 9 11 10 12 4 16 8 6 3 11 4 12 5 13 2 10 15 7 14 9 1 16 8 12 3 13 4 14 5 15 9 11 10 6 7 1 2 16 8 12 3 13 4 14 7 11 2 15 5 10 6 9 1 16 8 12 9 15 5 13 6 14 7 3 2 10 1 11 4 16 8 3 9 12 2 10 5 13 6 14 7 15 1 11 4 16 8 3 9 6 2 11 5 10 14 7 13 1 15 12 4 16 4 10 15 3 8 14 7 11 2 6 12 9 13 5 1 16 2 3 7 14 6 4 13 5 10 8 11 9 12 15 1 16 8 4 14 3 15 10 7 11 2 6 12 9 13 5 1 16 2 9 3 10 5 15 6 11 1 13 7 14 8 12 4 16 8 2 13 3 14 7 9 12 6 10 5 15 1 11 4 16 8 12 9 3 2 10 5 13 6 14 7 15 1 11 4 16 8 2 13 1 10 5 11 6 9 3 14 7 15 12 4 16 8 12 3 13 2 9 7 14 6 10 5 15 1 11 4 16 8 12 9 15 10 5 3 13 7 1 11 2 14 6 4 16 4 10 15 9 2 14 7 11 8 12 6 3 13 5 1 16 8 4 14 6 12 10 7 11 2 15 3 9 13 5 1 16 2 9 3 10 8 12 5 13 6 14 7 15 1 11 4 16 4 10 6 12 8 14 7 11 2 15 3 9 13 5 1 16 8 12 3 13 4 14 7 11 2 10 5 15 6 9 1 16 4 10 15 3 6 2 14 7 11 8 12 9 13 5 1 16 2 9 3 10 5 15 12 6 13 7 14 8 1 11 4 16 4 2 11 3 9 15 12 6 13 7 14 8 10 5 1 16 8 12 6 14 4 10 7 11 2 15 3 9 13 5 1 16 8 12 9 15 6 11 7 14 2 4 13 3 10 5 1
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.
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@plastereddragon.com> To: "Richard Guy" <rkg@cpsc.ucalgary.ca>; "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 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@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.
I won't repeat the whole message. This is just to say that the 37, 1, ... was chosen by Paul Vaderlind 'cos 37 and (37 + 1)/2 are both prime, restricting the ends of the chain. Such primes will give smaller numbers of chains when n is small, but the effect probably diminishes for large n. R.
participants (4)
-
Chuck Seggelin -
Edwin Clark -
Eugene Salamin -
Richard Guy