I promise it’s not as complex as it sounds.
A couple days ago I was visiting musictheory.net/lessons, trying to reinforce some basic knowledge, and I got to the section on rhythm and meter, namely time signatures.
(The fundamental thing to know about reading time signatures is that the top number represents the number of beats and the bottom number represents the value of a beat.)
Simple meter was simple enough. You have simple duple, which is two notes of a given value; Simple triple, which is three notes of a given value; and simple quadruple, which is four notes of a given value. In each case, you could take each note and break it into two notes, and then take those two notes, and break them into two notes, and take those notes and break them into two, and so on.
It’s when I got to compound and odd meters though that I noticed something similar to a simple thought experiment I did a long time ago.
First, to elaborate a little bit, once you get to a 6 note time signature, for example, let’s say 6/8, you have a situation where you can imagine translating that to a “simpler” time signature, e.g. 3/4. And, in fact, if you were to take three quarter-notes, and split each of them into two, you’d have six eighth notes. Equivalence. Since, you can translate in that way between the two, the “simpler” 3/4 takes “ownership” of that particular grouping of the 6/8 time signature. It’s like a tree, and it presides over all the ways you can split each of those three notes into two, and then four, and then eight, and so on.
But there’s another grouping for the 6/8 time signature. Rather than seeing it as three groups of two notes each, you can also see it as two groups of three notes each. Simple time signatures don’t split their notes into threes, only twos. So above, while the simple duple can have its two singular notes split into two groups of two notes (four notes total), and then two groups of four notes (eight notes total), and then two groups of eight notes (sixteen notes total), and so on, there’s no way to split the simple duple into two groups of three notes. And so, in this way, the 6/8 time signature, taking “ownership” of this grouping, becomes a “compound” time signature, where “compound” implies that you have notes grouped into threes.
Accounting for the other triple-note groupings, you get the 9/8 compound time signature; referred to as the “compound triple” (three groups of three notes); and the 12/8 compound time signature, referred to as the “compound quadruple” (four groups of three notes).
Now, getting to odd time signatures, these are the cases where you have a mix of both simple and compound groupings of beats. So 5/8, for example has one group of two, and one group of three; 7/8 has two groups of two and one group of three; 10/8 has two groups of two and two groups of three; 11/8 has two groups of two and three groups of three; and so on.
Finally, we get to the thought experiment.
The question is, if you were to take a number, and find all of the different ways that you could sum up to it (assuming positive integers), which set of numbers, if you then multiplied them all together instead of adding them, would yield the largest product?
Let’s take the number 5 for example. You can do 0+5, 1+4, 2+3, 1+1+3, 1+2+2, 1+1+1+2, and 1+1+1+1+1. Then, to switch out addition for multiplication, you get 0*5=0, 1*4=4, 2*3=6, 1*1*3=3, 1*2*2=4, 1*1*1*2=2, and 1*1*1*1*1=1. So the maximum product we were able to get was 2*3=6.
If you do this enough, you’ll find a pattern.
Starting at 4, you get 2*2=4. Then for 5, it’s 3*2=6. For 6 it’s 3*3=9. For 7 it’s 3*2*2=12. For 8 it’s 3*3*2=18. For 9 it’s 3*3*3=27. For 10 it’s 3*3*2*2=36. For 11 it’s 3*3*3*2=54. And for 12 it’s 3*3*3*3=81.
The pattern is that you’ll only ever have one or two 2s, and if you have two 2s then for the next number, one of them will be turned into a three, and on the next number, another will be turned into a three, and on the number after that, one of the threes will be split into two 2s.
2*2 -> 3*2 -> 3*3 -> 3*2*2 -> 3*3*2 -> 3*3*3 -> 3*3*2*2 -> …
The length of the set grows whenever you split a three into two 2s.
More importantly though, and the purpose of this post, is that this mostly maps to the time signatures.
It doesn’t work below 4, since with 3 in particular you get the maximum with 2*1=2, which would imply one group of 2 and one group of 1, but with pretty much everything above 4, it works, and you could always hand-wave the 2 and the 3 cases as being “atomic”, or ignoring sets that contain 0 or 1. Interestingly enough, if you were to work backward, with the pattern given above, the 2*2 would fuse into a 3, and then that 3 would become a 2 on the next step back, such that it all works.
Anyway, I just thought that was interesting, and it was an unexpected place to be reminded of that little exercise.