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A procedure for finding the k*N subgroup which seems to suffice is to take the product m of the odd primes less than or equal to p, and then find the Euler genus Euler(m^i) for integers 1, 2, and so forth. Here Euler(d) for an odd integer d is the set of all divisors of d reduced to an octave, and including 2. For each such genus, select those intervals such that the k*N patent val maps the interval to a number divisible by k, and then find the corresponding normal interval list. When two successive values i and i+1 lead to the same normal list, add to that a basis for the commas of the k*N patent val, and return the normal interval list for that.

For example, to find the 7-limit 2*6 subgroup we first find m = 3*5*7 = 105. The subgroup of Euler(105) consisting of those intervals mapped to an even number by <12 19 28 34| is 2.5.7. The subgroup of Euler(105^2) is 2.9.5.7, not the same. However, the subgroup of Euler(105^3) is again 2.9.5.7. If to this we add 81/80, we still obtain 2.9.5.7. It is clear this is maximal, as 3 is mapped by 12et to an odd number (19), and the rest of the values are prime.