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Henry Bushby Transcription pages

VM533B94v4-p071.jpg

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71
Take S4 6b/18c z = 2
Here it is clear by inspection that
there are 2 exclusions, 4 & 8.
B/z - 1 = 6/2 - 1 = 3 - 1 = 2 [checkmark]
M2n = M x 8 = 8, [16 exceeds 2B]
(M2n = 8, 16, 24, 32, 40 &c.
(M'B = 6, 12, 18, 24, 30, &c.
The rule won't work thus.
M2n - 2B = 8 - 12 [??]
16 - 12 = 4 [checkmark]
24 - 12 (exceeds
equals 2B].
The rule stands without the 3rd
[??]exception, though that seems
to apply when z = 1. But even
then it won't cover the 4 in S6 8b/40c
Test the rule as given on p 67
with S4 9b/27c, z = 1
M2n = M x 8 = 8, 16, [24 exceeds 2B]
M2n - 2B = 6, 24, [22 -------
M2n -2 = 6, 14, [22 ------
B/2 - 1= 8. Exclusions are 6, 8, 14, 16, 22

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