Song Study: E Squared

Greetings to this fine community.
I present to you my latest work.

In this song study, superparticular ratios (2/1, 3/2, 4/3, …) are raised to powers from 0 to 2n+1 in rows separated by a pause. The first row deals with the first superparticular ratio 2/1, the second row with 3/2, and so forth.
A superparticular ratio is the interval from the nth harmonic to the (n+1)th harmonic,
written as (n+1)/n.

The final note of the nth row is ((n+1)/n)^(2n+1). These final notes converge quickly to e^2 which is about 3462 cents.

The nth superparticular ratio may be produced on a fretless string instrument by stopping the string at the same location where you play the (n+1)th harmonic closest to nut (but beware of deviations from this ideal).

The first row is 3 octaves,
C3 C4 C5 C6
or
(2/1)^0 (2/1)^1 (2/1)^2 (2/1)^3
whose final note exceeds e^2 by about 138 cents.

The second and third rows are the Pythagorean 12-note scale plus 13th note comma.
C G D A E B
C F Bb Eb Ab Db Gb Cb
The final note Cb is about 23.46 cents lower than B, the famous Pythagorean comma.
The final note Cb is also about 23.85 cents higher than e^2.
Each row’s final note gets even closer to e^2.

The fourth row is based on the 5-limit just major third (5/4), and the 5th row on the 5-limit just minor third (6/5).
The rest of the rows involve just intervals from just intonations, usually from higher limits.
The eighth row is based on 3-limit Pythagorean major second, 9/8.
The final row is based on the 5-limit just semitone, 16/15.
The final note of the final row is within 2 cents of e^2.

Second note   Last note (of a row)
      cents               cents
2/1  1200.000 (2/1)**3    3600.000
3/2   701.955 (3/2)**5    3509.775
4/3   498.045 (4/3)**7    3486.315
5/4   386.313 (5/4)**9    3476.823
6/5   315.641 (6/5)**11   3472.054
7/6   266.870 (7/6)**13   3469.321
8/7   231.174 (8/7)**15   3467.611
9/8   203.910 (9/8)**17   3466.470
10/9  182.403 (10/9)**19  3465.670
11/10 165.004 (11/10)**21 3465.088
12/11 150.637 (12/11)**23 3464.652
13/12 138.572 (13/12)**25 3464.316
14/13 128.298 (14/13)**27 3464.052
15/14 119.442 (15/14)**29 3463.841
16/15 111.731 (16/15)**31 3463.669

e**2  3462.468 cents
e     1731.234 cents

Where is e?
It lies approximately half way between the 2 middle notes of each row.
The first of the 2 middle notes is ((n+1)/n)^n.
Written in a different way, this is (1 + 1/n)^n of Jacob Bernoulli’s famous limit definition of e which is also the original definition.
When n goes to infinity in the limit, this converges to e from below.
This convergence is very slow and you must listen to the infinite version of this song study for a long time before it gets perceptibly close.
Also, the next note, the second of the 2 middle notes, ((n+1)/n)^(n+1), also converges to e
as n goes to infinity but from above.
The harmonic average of these two notes, or alternatively, the square root, ((n+1)/n)^(n+1/2),
provides a good approximation of e.

Playing the source version below in Supercollider plays all rows to infinity.

// ROLLIN RYAN (Ryan Alexander Alam)
// Contact:
// Try rollin.ryan@hotmail.com or
// Try LODZ CONNEXION TIBAFE thisisbutafreeemail@tutanota.com or
// Knock big house door or post mail to
// 184 Brzezinska, 92-703, Lodz, Poland
// Phone number: +48600537542, t.me/rollin_ryan
// 2024/11/10

// Copy and paste this file into SuperCollider (made in version 3.13.0)
// Click on top menu bar: Language -> Evaluate File

(
var bpm = 120;
var ff = 110*(2**(3/12)); // fundamental freq at C3
TempoClock.tempo = (1/8)*(bpm/60); //bps
s.boot;
s.waitForBoot {
    var stream = Routine.new({
	(1..).do({|i|
		(2*i+2).do({|j|
		postln([i+1,i,j]);
			if (j!==(2*i+1),
				{[(((i+1)/i)**j),1/8].yield;},
				{[(((i+1)/i)**j),1].yield;});
		});});});
SynthDef(\saw,{|freq=440,gate=1,out=0,pan=0,amp=1|
	var sound = Saw.ar(freq);
	var env = EnvGen.kr(Env.linen(0.001,0.0,4.8,1.0,-20),gate,doneAction: 2);
	Out.ar(out,Pan2.ar(sound*env,pan,amp));
}).add;
s.sync;
Routine({
	loop({
			var sn = stream.next;
			Synth(\saw,[\freq,sn[0]*ff,\out,0]);
			sn[1].wait;
	})
}).play;
}
)

Questions and comments are welcome.