Re: Noble frequency ratios as prime-count vectors in ℚ(√5)
Posted: Sun Sep 25, 2022 4:18 am
Well that's a tough act to follow...
And Dave specifically told me not to do this. So, I'll do it just a bit
I did some comma-hunting, using some rudimentary bounds (like on the counts of unique primes, the max individual prime count, and product complexity), and then just scanning big output lists with my eyes for remarkable-looking results. From the Facebook post above, it looks like Mike Battaglia is working on some way more legit complexity measurements specialized to this situation, which could be used to much more effectively zero in on points of interest here, so I look forward to seeing what he comes up with that.
And Dave specifically told me not to do this. So, I'll do it just a bit
I did some comma-hunting, using some rudimentary bounds (like on the counts of unique primes, the max individual prime count, and product complexity), and then just scanning big output lists with my eyes for remarkable-looking results. From the Facebook post above, it looks like Mike Battaglia is working on some way more legit complexity measurements specialized to this situation, which could be used to much more effectively zero in on points of interest here, so I look forward to seeing what he comes up with that.
- five ~833.1 ¢ F1 (0+1ɸ) [; 1⟩ are 14.0 ¢ off from three ~1393.2 ¢ f5 √5 (-1+2ɸ) [; 0 1⟩, so that comma is [; -5 3⟩. In terms of nobles, we can say that four ~560.1 ¢ f5/F1 (-1+2ϕ)/(0+1ϕ) [; -1 1⟩ are 14.0 ¢ off from one ~2226.2 ¢ f5/f1 (-1+2ϕ)/(-1+1ϕ) [; 1 1⟩.
- four ~422.5 ¢ f11/f5 (-2+3ϕ)/(-1+2ϕ) [; 0 -1 1⟩ are close to three ~560.1 ¢ f5/F1 (-1+2ϕ)/(0+1ϕ) [; -1 1⟩. So that comma is ~9.7 ¢ (-5+10ɸ)/(3+5ɸ) [; 3 -7 4⟩. I'm not sure how to reduce it further.
- five ~339.3 ¢ f19/f11 (-3+4ϕ)/(-2+3ϕ) [; 0 0 -1 0 1⟩ are close to three ~560.1 ¢ f5/F1 (-1+2ϕ)/(0+1ϕ) [; -1 1⟩. So that comma is ~16.5 ¢ [; 3 -3 -5 0 5⟩ and I'm too lazy to get that into (a+bɸ)/(x+yɸ) form.
- one ~833.1 ¢ F1 (0+1ɸ) [; 1⟩ is 7.4 ¢ off from one ~840.5 ¢ 13/8 (13+0ϕ)/(2+0ϕ)³ [-3 0 0 0 0 0 1 ;⟩. So that comma is [-3 0 0 0 0 0 1; -1⟩.
- one ~833.1 ¢ F1 (0+1ɸ) [; 1⟩ is 15.8 ¢ off from one ~848.9 ¢ F79/F31 (-5+8ϕ)/(-3+5ϕ) [; 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 1⟩. So that comma is [; -1 0 0 0 0 0 0 0 -1 0 0 0 0 0 1⟩.
- lots of commas like 3ɸ/5, 5ɸ/8, 8ɸ/13, or 3ɸ²/8, 5ɸ²/13, etc...
- 64/3f11². So it equates two f11/4 ~248.7 ¢ diminished thirds with 4/3, and these are very very close, off by only about half a cent: [6 -1 0 0 -2 ;⟩
- 20ϕ²f5/189 equates three f5/3 ~324.2 ¢ minor thirds with 7/4, off by only about 4¢ [2 -3 3 -1 ;⟩
- 48/f5³ ... so tempering this out equates three 4/f5 neutral 2nds ~173.8 ¢ each with 4/3, which is a porcupine-like situation. That's about 23 ¢ and is [4 1 -3 ;⟩.
- 243/F11³, a 3.5 ¢ comma between five compound fifths and three F11 at ~3168.7 ¢ each, so that's [0 5 0 0 0 -3 ;⟩.