King, Peter R peter.king at imperial.ac.uk
Thu Dec 2 15:38:22 UTC 2004

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> Your right that there are only 6 regular lead heads for
> Royal, but this does not mean the answer is not still
> (n-2)!.  Try writing down the place bell order for one of
> the short-course regular Royal lead heads.  It will be
> something like
>
>   2 -> 8 -> 7 -> 2,
>   3 -> 6 -> 9 -> 3,
>   4 -> 0 -> 5 -> 4.
>
> That is, it is (obviously) not a single list sequence.  Any
> 9-cycle can be written as a single sequence of place bells
> starting say with the tenor.  Tenths place bell can become
> any of the 9 other place bells, which can become any of 8
> remaining place bells, until only one remains.  That place
> bell becomes 10ths place bell.  So the number of 9-cycle
> lead heads is still (n-2)! even though there are few regular
> lead heads than might be expected.

I agree entirely that on n bells there are n-1 (regular) lead heads, but
if n-1 is not prime some of these may lead to cycles less than n-1
(which are obviously factors of n-1) and so lead to courses of less than
n-1 leads. Whilst there is nothing wrong with this I chose explcitly to
disallow these (entirely my choice) and so not to consider
differentials. Then I presume I reduce the total by the prime factors of
n-1 (or some such number). I suppose I should have asked how many
non-differntial lead heads are there. I am probably being a bit opaque
in my terminology (or perhaps my understanding).

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