Tonality diamond conjecture

Suppose we take the n-limit tonality diamond, for some odd integer n.
We may call the smallest step between the notes of this diamond, taken
as octave reduced, the n-diamond comma. On the basis of pathetically
slight numerical data, I am conjecturing that the largest prime
appearing in the factorization of the n-diamond comma is n iff n is a
prime. Since obviously this requires n to be prime, it is equivalent
to conjecture that if n is a prime, it appears in the factorization of the
n-diamond comma. Note that some of the intervals between notes of the
diamond, for prime n, will belong to lower prime limits, so this
conjecture is not vacuous. It is saying that for the *smallest* such
interval, it *does* belong to the full n-limit.

I haven't thought about a proof and it could be easy, but I calculated
the table below, which gives n, the prime limit of the n-diamond
comma, and the n-diamond comma, and noted that we get the n-limit when
n is prime. The ratio between n and the prime limit of the n-diamond
comma is also an interesting thing to look at. Anyway, here is a table of
n-diamond commas up to n=61.

3: 3 9/8
5: 5 25/24
7: 7 50/49
9: 5 81/80
11: 11 121/120
13: 13 169/168
15: 7 225/224
17: 17 289/288
19: 19 361/360
21: 17 442/441
23: 23 529/528
25: 13 625/624
27: 13 729/728
29: 29 841/840
31: 31 961/960
33: 17 1089/1088
35: 17 1225/1224
37: 37 1369/1368
39: 19 1521/1520
41: 41 1682/1681
43: 43 1850/1849
45: 23 2025/2024
47: 47 2209/2208
49: 7 2401/2400
51: 17 2601/2600
53: 53 2809/2808
55: 11 3025/3024
57: 29 3249/3248
59: 59 3481/3480
61: 61 3721/3720

Here (up to the 207 limit) are n-diamond commas where the ratio of the
prime limit to n is 1/3 or less. We have 2401/2400 in the 7-limit,
3025/3024 and 9801/9800 in the 11-limit, 4225/4224 in the 13-limit
etc. The third number is the ratio, and the fourth is the square root
of the denominator; as you can see, the denominators are all squares.
Moreover, 7^4/(7^4-1), 5^6/(5^6-1) and 13^4/(13^4-1) all make their
appearance.

49 2401/2400 1/7 49
51 2601/2600 1/3 51
55 3025/3024 1/5 55
65 4225/4224 1/5 65
69 4761/4760 1/3 69
77 5929/5928 19/77 77
91 8281/8280 23/91 91
99 9801/9800 1/9 99
115 13225/13224 29/115 115
125 15625/15624 31/125 125
129 16641/16640 1/3 129
153 23409/23408 19/153 153
155 24025/24024 1/5 155
161 25921/25920 1/7 161
169 28561/28560 17/169 169
171 29241/29240 43/171 171
175 30625/30624 29/175 175
183 33489/33488 1/3 183
187 34969/34968 47/187 187
189 35721/35720 47/189 189

For most primes, it appears the p-diamond comma is p^2/(p^2-1). However,
in some cases (7, 41, 43, 73, 83, 157, 193, 211 ...) it is of the form
(p^2+1)/p^2. I've not found an example where it is of neither form.
All primes where (p^2+1)/p^2 is minimal must be such that p^2+1
factors into primes of size p or less, but this condition is not
sufficient, 47 being the first counterexample.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:

> I haven't thought about a proof and it could be easy...

I've now thought about it, and the proof is not difficult. Clearly for
any odd n, the n-diamond comma must be at least as small as n^2/(n^2-1),
since this is the ratio separating (n+1)/n and n/(n-1), and these are
both in the n-limit since n+1 and n-1 are even, and hence factor into
primes smaller than n. Hence, the n-diamond comma has to be an interval
between intervals whose numerator or denominator contain n; so that
either n/b < c/n, and the ratio is bc/n^2, or a/n < n/d, and the ratio
is n^2/ad. But these have to be contiguous Farey fractions on the nth
row of the Farey sequence, so that they are (n^2+1)/n^2 or n^2/(n^2-1).
If n is a prime p, this means (p^2+1)/p or p^2/(p^2-1), whose
factorization in either case involves p; hence, p is in the
factorization of the p-diamond comma, and the tonality diamond
conjecture is true. So is the conjecture the n-diamond comma is of the
form n^2/(n^2-1) or (n^2+1)/n^2.

I note the n-diamond comma is superparticular with square(n)
as the numerator, except when n is 7, 21, 41, or 43, when it
is the denominator.

-C.

>Suppose we take the n-limit tonality diamond, for some odd integer n.
>We may call the smallest step between the notes of this diamond, taken
>as octave reduced, the n-diamond comma. On the basis of pathetically
>slight numerical data, I am conjecturing that the largest prime
>appearing in the factorization of the n-diamond comma is n iff n is a
>prime. Since obviously this requires n to be prime, it is equivalent
>to conjecture that if n is a prime, it appears in the factorization of the
>n-diamond comma. Note that some of the intervals between notes of the
>diamond, for prime n, will belong to lower prime limits, so this
>conjecture is not vacuous. It is saying that for the *smallest* such
>interval, it *does* belong to the full n-limit.
>
>I haven't thought about a proof and it could be easy, but I calculated
>the table below, which gives n, the prime limit of the n-diamond
>comma, and the n-diamond comma, and noted that we get the n-limit when
>n is prime. The ratio between n and the prime limit of the n-diamond
>comma is also an interesting thing to look at. Anyway, here is a table of
>n-diamond commas up to n=61.
>
>3: 3 9/8
>5: 5 25/24
>7: 7 50/49
>9: 5 81/80
>11: 11 121/120
>13: 13 169/168
>15: 7 225/224
>17: 17 289/288
>19: 19 361/360
>21: 17 442/441
>23: 23 529/528
>25: 13 625/624
>27: 13 729/728
>29: 29 841/840
>31: 31 961/960
>33: 17 1089/1088
>35: 17 1225/1224
>37: 37 1369/1368
>39: 19 1521/1520
>41: 41 1682/1681
>43: 43 1850/1849
>45: 23 2025/2024
>47: 47 2209/2208
>49: 7 2401/2400
>51: 17 2601/2600
>53: 53 2809/2808
>55: 11 3025/3024
>57: 29 3249/3248
>59: 59 3481/3480
>61: 61 3721/3720

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
> I note the n-diamond comma is superparticular with square(n)
> as the numerator, except when n is 7, 21, 41, or 43, when it
> is the denominator.

Yes, it always must be one or the other.