Squib
Joined 25 April 2025
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23-limit 24 & 34: 24 & 34 & 41(g), 24 & 34 & 53, 24 & 34 & 94, 24 & 34 & 217 | 23-limit 24 & 34: 24 & 34 & 41(g), 24 & 34 & 53, 24 & 34 & 94, 24 & 34 & 217 | ||
==Intervals with monzos containing only ones== | |||
===Non-subgroup monzos=== | |||
Superparticular intervals: | Superparticular intervals: | ||
*[[2/1]] | *[[2/1]] | ||
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41: 17490603/17395070 | 41: 17490603/17395070 | ||
===Subgroup monzos=== | |||
A superparticular interval of this type exists if and only if the square root of 4n+1 is an integer, where n is the product of all primes in the subgroup. The result is the sum of the numerator and denominator of the superparticular interval. | |||
(This method also works for intervals containing any number of the same prime. For example, with factors 2, 2, 2, 2, 3, and 5, n is 240 and (4n+1)^0.5 is 31, which is an integer. So these factors can form a superparticular interval whose numerator and denominator add to 31: [[16/15]].) | |||
(For subgroups with rational or non-prime elements, split them into prime factors and multiply all together to get n, then determine if the final result is in the subgroup. For example, for the 11/2.13.15.19 subgroup, n is 81510 and (4n+1)^0.5 is 571, so the resulting superparticular interval is 286/285, but this is not in the subgroup because 11 and 2 are on the same side of the fraction. So no superparticular interval exists in the subgroup.) | |||
(note about commas like 245/243) | |||
(this should probably get its own page lol) | |||
====All superparticular intervals with no duplicate primes, by prime limit==== | |||
Found by applying this method to every possible subgroup in the prime limit. | |||
2-limit: | |||
*[[2/1]] | |||
3-limit: | |||
*[[3/2]] | |||
5-limit: | |||
*[[6/5]] | |||
7-limit: | |||
*todo | |||