## Noble frequency ratios as prime-count vectors in ℚ(√5)

Dave Keenan
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### Re: Noble frequency ratios as prime-count vectors in ℚ(√5)

"One question for רועי סיני and Dave Keenan - I think in some recent post, we determined that the "canonical" versions of each feudal prime are not necessarily the two that are closest together, right? I had thought we determined that was the criterion that was best for some reason. What criterion are we using instead?"

I tried to reply to you on facebook, twice, but facebook claimed that my replies violate their spam guidelines! Then I tried to just tell you that, on facebook, and apparently that has disappeared too! Then I posted a link to this post. It seems that whatever I post now, vanishes without a trace. I think facebook has lost the plot.

My (supposedly-spammy) answer was:
In the old facebook thread Roee wrote:
"Dave Keenan, In fact, your choice of primes is not always the two closest in size. For example, F29/f29 is 1.733..., while (f29ϕ)/(F29/ϕ) is 1.510...
However, they are always either the best or the second best choice, because they are both between bϕ-b = b/ϕ and bϕ, where b is their equal golden part, which means the quotient between them is less than ϕ², and any time you multiply one of them by ϕ and divide the other by ϕ you either multiply or divide the directed quotient by ϕ². Anyway, I don't think this is a large enough issue to use the other choice instead in this case, because of the consistent normal forms they have."

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battaglia01
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Real Name: Mike Battaglia

### Re: Noble frequency ratios as prime-count vectors in ℚ(√5)

Hi, I'm on here now. Weird that FB has so many problems.

OK, I still don't quite get it though. Originally I'd suggested we go with the unique representatives which are conjugates of one another in the algebraic number field. We used that for a little bit as a normal form.

Then, I *thought* what Dave suggested we use the two which are closest in size instead. I don't remember why, but I remember looking into it for like 3 seconds and agreeing it probably was better.

Now we are saying that Dave's criterion turns out to not always the two which are closest in size. I thought that *was* Dave's criterion. If not, then what criterion are we using instead which is sometimes, but not always the same as the two which are closest in size?
battaglia01
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Real Name: Mike Battaglia

### Re: Noble frequency ratios as prime-count vectors in ℚ(√5)

Also, FWIW, I am still kind of interested in this, but much more interested in the logarithmic approximation stuff, which I felt like was "good enough", but had the benefit of this simple structure practically gave us a one-size-fits-all framework for every possible variation on this we could ever want - noble numbers, silver aristocratic numbers, etc. I am curious what Roeesi thinks of that.
Dave Keenan
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### Re: Noble frequency ratios as prime-count vectors in ℚ(√5)

Welcome to the Sagittal forum, Mike. @battaglia01

See the text beginning "A fundamental prime is" near the middle of this post:
viewtopic.php?p=4606#p4606
רועיסיני
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Real Name: Roee Sinai

### Re: Noble frequency ratios as prime-count vectors in ℚ(√5)

battaglia01 wrote: Fri May 12, 2023 10:49 am OK, I still don't quite get it though. Originally I'd suggested we go with the unique representatives which are conjugates of one another in the algebraic number field. We used that for a little bit as a normal form.

Then, I *thought* what Dave suggested we use the two which are closest in size instead. I don't remember why, but I remember looking into it for like 3 seconds and agreeing it probably was better.
As I said on Facebook, we are using the criterion that they can be represented as -a + bϕ for 0 < a < b. This gives a unique pair of positive primes that multiplies to the ordinary prime where each prime in the pair is the negative conjugate of the other. While they don't have to be the closest pair that multiplies to give the ordinary prime I proved that they have to be either the closest or the second closest in a comment on Dave's post, which was archived, as Dave already wrote, in viewtopic.php?p=4674#p4674 and started by "In fact, your choice".
Every pair of feudal numbers that multiplies to give an ordinary rational number are always either conjugates or negative conjugates of each other so this criterion doesn't narrow the choice down.
battaglia01 wrote: Fri May 12, 2023 10:49 am Also, FWIW, I am still kind of interested in this, but much more interested in the logarithmic approximation stuff, which I felt like was "good enough", but had the benefit of this simple structure practically gave us a one-size-fits-all framework for every possible variation on this we could ever want - noble numbers, silver aristocratic numbers, etc. I am curious what Roeesi thinks of that.
I also think this is good enough, but FWIW I also thought about this algorithm, and wanted to share it.
battaglia01
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Real Name: Mike Battaglia

### Re: Noble frequency ratios as prime-count vectors in ℚ(√5)

OK, that makes sense. Then I'd ask, what is the benefit of this normal form rather than the two that were closest in size? One benefit of using the two closest in size is it makes something like a "complexity" metric relatively easy (just use log(size of prime) weighting and take the L1 norm)
רועיסיני
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Real Name: Roee Sinai

### Re: Noble frequency ratios as prime-count vectors in ℚ(√5)

battaglia01 wrote: Sat May 13, 2023 3:35 am OK, that makes sense. Then I'd ask, what is the benefit of this normal form rather than the two that were closest in size? One benefit of using the two closest in size is it makes something like a "complexity" metric relatively easy (just use log(size of prime) weighting and take the L1 norm)
The main advantage of this normal form, IMO, is that you can "see at a glance" whether or not a prime is normalized – just check that the wooden part is negative, that the golden part is positive and that the second is greater in magnitude than the first. In return, this form gives you that these primes are always positive, that the ordinary prime is the product of the two feudal primes and that they are relatively close to each other – the quotient between them is less than ϕ², which means they are either the closest pair that multiplies to the ordinary prime or the second closest one, and they are the closest pair for the first 3 split primes. They also have the property that if you conjugate one of them, it is always minus the other one, which is likely to make computations more elegant.
On the other hand, if you insist on choosing the closest pair of positive feudal primes that multiply to the ordinary prime, it seems like you demand more but get less – you use 3 different demands on different properties of the primes, and get neither a visually elegant form nor a consistent relation between one prime and the conjugate of the other, as 11 factors into (-1+3ϕ) * (-2+3ϕ), which look similar and are negative conjugates of each other, while 29 factors into (6-ϕ) * (5+ϕ), where the wooden parts are now positive, the golden part is positive for one prime and negative for the other and both are smaller in magnitude than the wooden parts, and the primes are now usual conjugates of each other.

That being said, if you prefer the pair which are the closest in size, then the algorithms are not that hard to fix. In the first one, in the end of the normalization procedure you consider the prime and its negative conjugate and if the larger one is smaller than the smaller one multiplied by ϕ you return the prime you wanted to return and otherwise in the case you needed the smaller one you return it multiplied by ϕ and in the case you needed the larger one you return it divided by ϕ. In addition, every time I wrote "negative norm" or "minus its norm" and "negative conjugate" or "minus its conjugate" this should be replaced by "the absolute value of the norm" and "the conjugate if the norm is positive, otherwise its minus". In the second algorithm, instead of scanning all the numbers of the form a+bϕ for every b, you only scan the numbers a+bϕ where a is between b/ϕ³ and b-b/ϕ³, and in addition scan the numbers b+aϕ and b+a-aϕ where a < b/ϕ³.
battaglia01
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Real Name: Mike Battaglia

### Re: Noble frequency ratios as prime-count vectors in ℚ(√5)

OK. FWIW, my view is that different bases are simply useful for different things. For instance, in regular JI, both the {2/1, 3/1, 5/1, 7/1, ...} basis and the {2/1, 3/2, 5/4, 7/4, ...} basis are useful.

Certainly, the conjugate basis is useful for various mathematical reasons, and these other two also seem to be useful. I do think, however, that choosing the basis w/ the two pairs closest in size and putting the L1 norm on that basis, weighted by the size of the primes in question, is a good idea.
רועיסיני
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Real Name: Roee Sinai

### Re: Noble frequency ratios as prime-count vectors in ℚ(√5)

There is indeed no problem with different bases, as long as there is an easy way to convert between them (and there indeed is between Dave's basis and the closest pairs basis). However, I'm not sure that the choice of the closest pair necessarily makes more sense than Dave's basis when using them to calculate the complexity of noble numbers, since Dave's basis also has the property that if a noble number is the quotient of two feudal primes, these primes have associates in the basis whose quotient also gives the same noble number.
Also, what do you mean by "the conjugate basis"? Every two feudal numbers with the same norm that multiply to an ordinary rational number, and in particular every two feudal primes that multiply to an ordinary prime, are either conjugates or minus conjugates of each other.

In addition, if you want to measure the importance of precision for an interval in an Euclidean-like norm, there are other ways to do that, some of which are basis-agnostic. for example, you can map every interval to a vector showing the (ordinary) factorisation of the absolute value of its (arithmetic) norm, a vector that counts every time each class of associate primes, except for the class of √5, is used in the number's factorisation, minus the number of times primes from the conjugate class are used, and a real number which is ½logᵩ(|x/x̅|), that you can think of as a sort of "phi count", and then order these coordinates in some way, give them fitting weights (where large parts in the prime class vector and the log φ number will make the number less important than the same parts in the arithmetic norm factors vector) and calculate the (weighted L2) norm of that. You can represent that triple of vectors and number by listing the norm factorisation, a semicolon and then the phi count and the prime class vector. For example:
• 3 has norm 9 so it will get [0, 2〉 for the norm factorisation (with infinite zeros in the end, as every other vector in these examples), it doesn't have any purely feudal primes in its factorisation so it'll get 0 for the prime class vector and it equals its conjugate so the last number will be 0 as well, so it will be [0, 2;〉.
• φ has norm -1 which has absolute value 1 so its norm factorisation is 0, its factorisation has no primes so the prime class vector is also 0, and its φ count is ½logᵩ(|φ/φ̅|) = ½logᵩ(φ²) = 1, so it will be [; 1〉.
• -1+3φ has norm -11 which has absolute value 11 so it will get [0, 0, 0, 0, 1〉 for the norm factorisation, it is a purely feudal prime, so if we say that its class gets the positive sign it'll get [1〉 for the prime class vector, and for the phi count we have ½logᵩ(|(-1+3φ)/(2-3φ)|) ≈ 0.31, so it will be [0, 0, 0, 0, 1; 0.31, 1〉.
• -2+3φ also has norm -11 so its norm factorisation is [0, 0, 0, 0, 1〉 as well, it is of the conjugate association class to -1+3φ so its prime class vector is [-1〉, and its φ count is ½logᵩ(|(-2+3φ)/(1-3φ)|) ≈ -0.31, so it will be [0, 0, 0, 0, 1; -0.31, -1〉.
• 11 has norm 121, so its norm factorisation is [0, 0, 0, 0, 2〉, it is factored to two purely feudal primes of conjugate association classes so its prime class vector is 0, and is conjugate to itself so its phi count is also 0, so it will be [0, 0, 0, 0, 2;〉. The last two parts are similarly 0 for all of the ordinary integers. This also makes sense with the fact that 11 = (-1+3φ)*(-2+3φ) as the vectors sum up to give the same result.
• √5 has norm -5, so its norm factorisation is [0, 0, 1〉, it is a prime that is not counted so its prime class vector is 0 and is conjugate of its minus so its φ count is ½logᵩ(|-1|) = ½logᵩ(1) = 0, so it will be [0, 0, 1;〉, which makes sense with the fact that its square is 5, an ordinary integer.
• -1+5φ has norm -29 so its norm factorisation vector is [0, 0, 0, 0, 0, 0, 0, 0, 0, 1〉, its prime count vector, if we decide that its associate class gets the positive sign, is [0, 0, 1〉, and its phi count is ½logᵩ(|(-1+5φ)/(4-5φ)|) ≈ 0.57, so it will be [0, 0, 0, 0, 0, 0, 0, 0, 0, 1; 0.57, 0, 0, 1〉.
• 6-φ has norm 29 so its norm factorisation vector is again [0, 0, 0, 0, 0, 0, 0, 0, 0, 1〉, it is an associate of -1+5φ, so its prime class vector is also [0, 0, 1〉, and its phi count is ½logᵩ(|(6-φ)/(5+φ)|) ≈ -0.43, so it will be [0, 0, 0, 0, 0, 0, 0, 0, 0, 1; -0.43, 0, 0, 1〉. Crucially, this method shows no inherent preference to one of these primes over the other, but treats then as equivalent.
The disadvantages of this method is that it uses fractional values and the unit vectors, except for φ's and √5's, don't correspond to any musical interval, which means it probably won't be good for existing temperament finders. I'll probably be able to write a new one that will fit this method, if you want that and if I'll know how the existing ones work.