Recall that the statement of this included a requirement that the

block be ordered by the val v dual to the block's basis for the basis

to be valid. There is a simple sufficient condition for this.

If {b_3, b_5, ..., b_p} is the block's basis, we may form the Graham

matrix whose rows are given by the vector corresponding to 2 (in the

usual prime notation, that would be [1,0, ..., 0]) and the basis row

vectors. If we invert this matrix the columns will be vals, the first

of which is the val v/n, where v(2) = n (which we may recall is

positive, and gives the number of notes an octave of the block. If

the other vals are {u_3, u_5, ..., u_p} we have an expression for any

note r in this notation as

r = 2^v(r)/n * b_3^u_3(r) * ... * b_p^u_p(r).

The note r will be in the block if 0 <= v(r) < n and if

-1/2 <= u_q(r) < 1/2

for each odd prime q<=p.

A sufficient condition for the notes of the block to be ordered by v

is that the product b_3^u_3(r) ... b_p^u_p(r) is not greater than

2^(1/2n) and given the bounds on the exponents u_q(r) a sufficient

condition in turn for this is that the product (c_3 c_5 ... c_p)^n,

(where c_q = exp(|ln b_q|) is the multiplicative absolute value of

b_q) is less than 2.