[r-t] Plain Bob leadheads

[Ian McCulloch]

On Fri, 19 Feb 2021, Mark Davies wrote:

> Please can someone produce, or point me at, a concise proof of this lemma:
>
> PB1: In a palindromic treble-dominated method with Plain Bob leadheads, 
> on an even number of bells, the lead end place notations must be 12 or 1n.

Not very consise, but a bit easier than brute-force:

The lead-end of a palindromic method must be a product of disjoint 
2-cycles (ie, the work of every bell is either itself symmetric, or in a 
mirror pair with another bell).

Working backwards from each plain bob lead head, we can calculate the 
possible lead ends.  Eg from 135264 on 6 bells, the possible lead ends 
are, with their lead end place notation,

135246  (12)
153246  (14)
153624  (16)
135246  (1234)
135624  (1236)
132564  (1256)
153264  (1456)

The only ones of these that are products of disjoint 2-cycles are (12) and 
(16).  Hence the others are not possible with a palindromic method.

There is surely a more elegant proof associated with the properties of the 
permutation (5,3,2,4,6) but I don't know enough algebra.

Cheers,
Ian