On Saturday, 20 February 2021, Ian McCulloch <<a href="mailto:ianmcc@physics.uq.edu.au" target="_blank">ianmcc@physics.uq.edu.au</a>> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><br>
On Sat, 20 Feb 2021, Martin Bright wrote:<br>
<br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
<br>
<br>
On Sat, 20 Feb 2021 at 17:18, Philip Saddleton <<a href="mailto:cantab@saddleton.org.uk" target="_blank">cantab@saddleton.org.uk</a>> wrote:<br>
On Fri, 2021-02-19 at 15:33 +0100, Martin Bright wrote:<br>
> Proof: It's enough to prove it in the case h = (2 4 6 .. n n-1 ... 5<br>
> 3) in cycle notation.<br>
<br>
Is it? It depends whether 'Plain Bob Lead Heads' means all of them.<br>
It's not true for methods with a 1-lead course, but what if n-1 is not<br>
prime?<br>
<br>
<br>
I was assuming “Plain Bob lead heads” meant a course of n-1 leads. You’re right that this is the point at which it doesn’t work in other cases.<br>
</blockquote>
<br>
But what actaully happens in Royal with 3-lead courses? A method with plain-bob lead ends, 3-lead plain course, and not 12 or 10 lead end change would be quite interesting - a novice band can learn 3 leads and ring a plain course, and it has some theoretical interest too. Is this possible?<br>
<br>
Cheers,<br>
Ian<br>
</blockquote><div><br></div><div><br></div><div>My suspicion is that if you take h to be either of the two none rounds plain bob lead heads you actually have to have for a 3 lead course of Royal (namely 1795038264, 1860492735) the rest of Martin's argument can still be made to go through but I've not checked this...</div><div><br></div><div>It is possible to get a two lead course of Triples where the lead head notation is not 1 or 127 (the equivalent thing for odd stages): eg</div><div>7.1.7.123.7.1.347,145<br></div><div>So care does need to be taken with courses shorter than n-1 leads for sure.</div><div><br></div><div>Cheers,</div><div>Alan</div>