The 3n+1 Problem & Beatty Sequences

[Thomas Klemm]

From Wikipedia 6.3 As a parity sequence I followed the footnote [19] to find a rather old paper:
Terras, Riho (1976), "A stopping time problem on the positive integers".

The theorem 1.1 (Remainder representation) provides a formula to calculate the last value in the modified Collatz sequence:

\[ T^kn=\lambda_k(n)n+\rho_k(n) \]

where \(\lambda_i(n)=\frac{3^{S_i(n)}}{2^i}\) and \(\rho_k=\frac{\lambda_k}{2}(\frac{X_0}{\lambda_1}+\frac{X_1}{\lambda_2}+\ldots+​\frac{X_{k-1}}{\lambda_k})\).
Don't let you confuse by the formula. The important thing is that \(\lambda_k(n)\) is a factor < 1 and \(\rho_k(n)\) is considered small. With a little handwaving I'd say that the factor is the result of the multiplication by 3 and division by 2 and the remainder keeps track of the occasional addition of 1.

In case of the example of your slides in the video \(i=10\) and \(S_i(n)=6\) thus \(\lambda_k(n)=\frac{3^6}{2^{10}}\approx 0.7119140625\).
I just calculated \(\lambda_k(n)n\) and dare to say that the approximation isn't bad:

Code:

507 1101101100 362 ≈ 360.940
347 1101110100 248 ≈ 247.034
923 1101111000 658 ≈ 657.097
583 1110101100 416 ≈ 415.046
423 1110110100 302 ≈ 301.140
999 1110111000 712 ≈ 711.202
975 1111001100 695 ≈ 694.116
815 1111010100 581 ≈ 580.210
367 1111011000 262 ≈ 261.272
735 1111100100 524 ≈ 523.257
287 1111101000 205 ≈ 204.319
575 1111110000 410 ≈ 409.351

This factor \(\lambda_k(n)\) must be < 1 of course. Thus we have \(\frac{3^p}{2^q}<1\Rightarrow 3^p<2^q\Rightarrow p\log(3)<q\log(2)\Rightarrow p\log(3)-q\log(2)<0\Rightarrow \frac{\log(3)}{\log(2)}p-q<0\). But that's exactly the value that you calculate by the method you described.

Cool that you figured that out by yourself. Don't be disappointed that it's already known. At least to me it was new and I enjoyed your video a lot.

Cheers
Thomas