[r-t] Holt's Original for Oxford Bob Triples?

[Robert Bennett]

5040 Oxford Bob Triples
Robert H Bennett, 2021.

   234567
 E 546327-2
   625734-4
(S 436725-1
EB 256743-1
ES 346725-1)
   463257-2
   264357-5
   732645-3
   637245-5
   236745-5
   362457-2
   743625-3
   647325-5
   256347-1
   642735-4
   426357-2
   324657-5
   573624-1
   325467-4
   423567-5
   524367-5
   675324-1
   376524-5
   623457-4
   746235-3
   247635-5
   352647-1
   653247-5
   726534-3
 E 426573-5
   264735-2
   432576-4
   534276-5
   235476-5
   762435-1
   627354-2
   326754-5
   263547-2
   562347-5
   365247-5
   723654-3
   467235-3
   354267-2
   673254-1
   736542-2
   537642-5
   265374-3
   362574-5
   563274-5
   635742-2
   276354-3
   372654-5
   253467-4
   452367-5

repeated, replacing the bracketted calls with a bob at the same lead (-2)
Apart from normal bobs, the other calls used are:
E= 5ths place when the treble is in 3-4 down; S= grandsire single;
EB= E followed by normal bob;   ES = E followed by grandsire single.

This peal is an analogue of Holt's Original peal of Grandsire Triples.
Oxford Bob Triples appears (as far as I can see) to have no exact
equivalent of the bob course (BBB) of Grandsire triples, which is
symmetrical and hence reversible.

Using two types of bobs, there is a reversible two lead course:


1234567
2135476
2314567
3241576
3425167
4352617
4536271
5463721
5647312
6574132 E
6751423
7615432 B
7164523
1765432
1674523  x2

=3.1.5.1.7.1.7.1.7.1.5.1.3.1   x2

This course can be singled into a round block containing the rest of the
leads.
Because both the E bobs and the normal B bobs affect 3 bells, there have to
be at least 6 of each kind of bob in this composition.
Because the round block has an even number of leads (358), a two part
composition is possible.

If the two-lead course is left out, a block of 5012 changes is produced. Is
this the longest possible non peal?


RHB
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