[seqfan] Re: Numbers in a circle
Ron Hardin
rhhardin at att.net
Tue Dec 7 01:37:18 CET 2010
I assume an all-10 solution isn't wanted either for L=10
Excluding L but allowing 1 in the solutions, I get solutions for L=2..
1 2 3 4 13 6 33 32 141 21 853 ... unless I've misunderstood
for L=10 these are some (as column vectors)
..6....5....1....5....5....5....2....5....2....3....4....5....5....9....3....5
..6....2....5....5....5....5....4....6....2....5....8....2....1....9....5....5
..6....5....8....6....3....1....2....5....2....3....4....5....5....9....5....8
..6....5....5....5....5....5....4....8....2....5....2....5....8....9....6....5
..6....1....5....5....2....8....2....5....2....5....4....6....5....9....5....5
..6....5....1....5....5....5....4....5....2....3....2....5....5....9....3....5
..6....2....5....5....5....5....2....6....2....5....6....2....1....9....5....5
..6....5....8....6....3....1....4....5....2....3....8....5....5....9....5....8
..6....5....5....5....5....5....2....8....2....5....6....5....8....9....6....5
..6....1....5....5....2....8....4....5....2....5....2....6....5....9....5....5
..6....2....5....5....5....8....5....8....6....5....5....5....6....5....5....5
..5....8....8....1....8....4....5....5....4....5....5....5....2....5....8....3
..2....2....5....5....5....2....5....7....6....5....6....8....6....1....5....5
..5....8....5....3....5....4....5....5....4....2....5....5....2....5....5....5
..5....2....5....5....8....2....5....5....6....5....3....3....6....3....6....6
..6....8....5....5....5....6....5....8....4....5....5....5....2....5....5....5
..5....2....8....1....8....8....5....5....6....5....5....5....6....5....8....3
..2....8....5....5....5....6....5....7....4....5....6....8....2....1....5....5
..5....2....5....3....5....2....5....5....6....2....5....5....6....5....5....5
..5....8....5....5....8....4....5....5....4....5....3....3....2....3....6....6
..5....5....3....5....8....5....5....4....2....3....5....5....2....8....5....5
..5....5....5....7....5....8....6....2....6....5....5....5....5....5....7....7
..1....5....5....5....2....5....5....6....2....5....5....2....5....5....5....5
..5....6....5....5....5....5....2....8....6....1....5....5....6....8....5....5
..7....5....5....8....5....1....5....6....2....5....8....5....5....5....6....3
..5....5....3....5....8....5....5....2....6....3....5....5....2....8....5....5
..5....5....5....7....5....8....6....4....2....5....5....5....5....5....7....7
..1....5....5....5....2....5....5....8....6....5....5....2....5....5....5....5
..5....6....5....5....5....5....2....4....2....1....5....5....6....8....5....5
..7....5....5....8....5....1....5....2....6....5....8....5....5....5....6....3
..3....6....5....5....6....4....5....8....5....5....8....8....5....5....5....5
..5....5....5....5....2....8....7....5....5....5....5....8....1....1....6....5
..8....8....5....5....4....4....5....5....6....5....8....8....5....5....5....8
..5....5....5....7....8....8....5....1....5....3....5....8....5....2....3....5
..5....5....7....5....4....4....5....5....7....5....5....8....5....5....5....8
..3....6....5....5....2....8....5....8....5....5....8....8....5....5....5....5
..5....5....5....5....4....4....7....5....5....5....5....8....1....1....6....5
..8....8....5....5....2....8....5....5....6....5....8....8....5....5....5....8
..5....5....5....7....6....4....5....1....5....3....5....8....5....2....3....5
..5....5....7....5....8....8....5....5....7....5....5....8....5....5....5....8
..3....5....5....5....5....6....5....8....5....5....7....8....5....7....5....3
..5....6....5....5....8....8....2....6....5....5....5....6....5....5....3....3
..5....5....7....8....5....6....5....2....1....8....5....8....5....5....5....3
..8....7....5....5....2....2....5....4....5....5....1....6....5....8....8....3
..5....5....5....2....5....4....8....8....2....7....5....8....3....5....5....3
..3....5....5....5....5....8....5....4....5....5....7....6....5....7....5....3
..5....6....5....5....8....4....2....2....5....5....5....8....5....5....3....3
..5....5....7....8....5....2....5....4....1....8....5....6....5....5....5....3
..8....7....5....5....2....4....5....2....5....5....1....8....5....8....8....3
..5....5....5....2....5....2....8....6....2....7....5....6....3....5....5....3
..5....5....5....5....7....5....2....1....2....7....5....6....5....1....8....2
..3....3....5....3....5....8....5....5....5....5....5....5....5....5....5....4
..5....5....3....5....5....5....5....5....5....5....1....7....6....2....5....2
..5....3....5....5....6....7....8....5....3....3....5....5....5....5....3....6
..5....5....8....8....5....5....5....5....5....5....5....5....8....5....5....8
..5....5....5....5....7....5....2....1....2....7....5....6....5....1....8....6
..3....3....5....3....5....8....5....5....5....5....5....5....5....5....5....2
..5....5....3....5....5....5....5....5....5....5....1....7....6....2....5....4
..5....3....5....5....6....7....8....5....3....3....5....5....5....5....3....8
..5....5....8....8....5....5....5....5....5....5....5....5....8....5....5....4
rhhardin at mindspring.com
rhhardin at att.net (either)
----- Original Message ----
> From: N. J. A. Sloane <njas at research.att.com>
> To: seqfan at seqfan.eu
> Cc: njas at research.att.com
> Sent: Mon, December 6, 2010 5:18:52 PM
> Subject: [seqfan] Numbers in a circle
>
>
> Numbers in a circle
>
> Enough about licenses! Here is a problem that has been keeping
> me awake at night: Fix a positive integer L, say 10.
> Is there a directed circle of L numbers,
> all in the range 1 thru L, not necessarily all distinct,
> with the property that a(n+1) tells you how far back around the circle
> you have to go before you see another copy of a(n)?
>
> Here's an example which doesn't quite work,
> but which will explain what I'm looking for:
> (The "arrows" are just to show the direction.)
>
> ->
> 1
> 2 6
> 2 2 |
> ^ 4 4 v
> | 2 1
> 4
> <-
>
> You see 10 numbers around a circle, arranged clockwise. Here L is 10.
>
> Start with the 1 at the top.
> That this is 1 tells you that the previous term, 2,
> can also be found 1 step back.
> The "2" at 10 o'clock tells you that the "2" at 9:30 o'clock is matched by
> another 2 2 steps back, and indeed it is (the latter 2 being at 7 o'clock).
> The 4 at 9 o'clock says that the term before it, 2, has a mate 4 steps back,
> and it does.
>
> If the terms are labeled a(0) through a(L-1), going
> clockwise around the circle, then for all n, a(n+1)
> is the number of steps back round the circle before
> you see another a(n).
>
> If a(n) only appears once, then a(n+1)=L, and conversely.
>
> The condition fails for the 1 at 5 o'clock, and for the 2 at 2:30 o'clock
> (which should be 10 since the 6 is unique)
>
> The only solution I know, of any length L, is a circle with L 1's.
> The problem is to show that there are no other solutions!
> Can anyone help?
>
> [I've tried a few things: for example, let A be the matrix
> in which A_{i,j} is the number of times there is an i immediately
> followed by a j. The sum of the i-th row is the number
> of i's in the loop. The sum j A_{i,j} is L, for any i.
> The matrix A has a lot of nice properies. Doesn't seem
> to lead anywhere though.]
>
> This question arises when one tries to analyze sequence A181391
> and its generalizations.
>
> Neil
>
>
>
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