--
Suppose a scale is defined as any unique combination of notes between a given note at the bottom and that same note, one octave up, at the top. Every scale has a minimum of two notes separated by one octave (low "do", high "do"), and a maximum of 12 (the chromatic scale). (It may be possible to generalize to any number of divisions, but I don't know how.) Represented in binary form with 0 representing "there is no note in this slot" and 1 representing "there is a note here" we therefore have
100000000001 (Do, Do)
and
111111111111 (chromatic)
At this point you should be seeing what I didn't for far too long, which is that we obviously have a 10-bit binary array and there are 2^10 unique combinations. But since I want a sum series, I'll go on...
It becomes useful now to qualify the possible scales in terms of the number of filled states, where n=0 is (Do, Do) and n=10 is (chromatic). In such a notation, filled-note-quantity N(0+n) is identical to N(10-n) insofar as its multiplicity is concerned, because whether we have the 1 as the 'default' and vary the 0s or vice versa makes no difference for counting the total number of possible scales. The interesting values are then
0
1
2
3
4
5
and the total number of possible scales is [2*(N0+N1+N2+N3+N4) + N5]
0 and 1 are easy to count: N(0)=1, N(1)=10.
The rest are a bit more complicated.
N(2) has
111000000001
110100000001
110010000001
etc
and
101100000001
etc
Obviously we could just use 2^u and be done. But lets go on.
There are 9 possibilities for "first note filled", 8 possibilities for "first note empty, second note filled", etc, down to 1 possibility for "first eight notes filled". The formula for sum of all numbers: ∑(F,L)= ( L² - F² + F + L)/2 becomes useful - in this case, we have N(2)=∑(1,9)=45.
Now consider the n=3 case.
For "first note filled" there are ∑(1,8) possibilities. For "first note empty, second note filled" there are ∑(1,7) and so on to "first seven notes empty" with ∑(1,1). Our solution is thus:
N(3) = ∑_{m=1}^{8}∑(1,m) = 36+28+21+15+10+6+3+1= 120
N(4) continues according to similar logic. For "first note filled" there are m=1 to m=7 ∑∑(1,m) possibilities, and so on. "First note open, second note filled" has m=1 to m=6 ∑∑(1,m). Thus we have
N(4) = ∑_{m=1}^{7}∑_{1}^{m}∑(1,m) = 28+2*21+3*15+4*10+5*6+6*3+7*1 = 210
Now we are getting somewhere. Of course, for our 12-note-maximum, we are almost done.
N(5) = ∑_{p=1}^{p=6}∑_{m=1}^{m=p}∑_{1}^{m}∑(1,m) = [21+2*15+3*10+4*6+5*3+6*1]+[15+2*10+3*6+4*3+5*1]+[10+2*6+3*3+4*1]+[6+2*3+3*1]+[3+2*1]+[1] = 252
[2*(N0+N1+N2+N3+N4) + N5] = 2*(1+10+45+120+210)+252 = 1024
--
OBSERVATIONS
That was where I realized I was an idiot and could have just raised 2^10 and been done. Nevertheless, there are some interesting things in the above.
First, none of the components of the final sum are powers of two, except for the trivial case of N(1)=N(10)=2^0.
Second, it appears to follow a predictable pattern and generate some interesting identities: we know that N(6) = N (4), and continuing on from N(5) we get
N(6)=∑_{q=1}^{q=5}∑_{p=1}^{p=q}∑_{m=1}^{m=p}∑_{1}^{m}∑(1,m) = N(4) = ∑_{m=1}^{7}∑_{1}^{m}∑(1,m) \
I haven't tested this one because it would take forever to write out all the sums and it is getting late.
So on for N(6) and N(3) and so on.
Third, the number of interest in each N(n) is (10+1-n). Perhaps the above can be generalized to other powers of 2^u with the number of interest being (u+1-n) and identical formulas:
N(2)=∑(1,u-1)
N(3) = ∑_{m=1}^{u-2}∑(1,m)
And so on. I have not tested this.
Fourth, we can keep going forever using the method that shows up in N(4), N(5), etc, just by replacing the highest value of the farthest-left sum with another variable, and then taking the sum of that that sum while the new variable goes from 1 to u+1-n.
QUESTIONS
Is there a general expression for 2^u as a series sum (basically expressing my last paragraph with math, assuming it's not in error)? What about other bases? Non-integer numbers?
Have I made mistakes?
On a scale of one to 2^10, how frivolous of a waste of time is this?
