I have been working on the Collatz problem for a while now, and have made this efficient cobweb plot function for it, where it automatically does all the dividing by two and always returns an odd number to be input back into it: $$ C\left(x\right)=\frac{x\cos^{2}\left(\frac{\pi x}{2}\right)}{2^{\left(\sum_{m=1}^{\infty}\left(m\sin^{2}\left(\frac{\pi}{2^{m+1}}x\right)\prod_{l=0}^{m-1}\left(\cos^{2}\left(\frac{\pi}{2^{m-l}}x\right)\right)\right)\right)}}+\sum_{k=1}^{\infty}\left(\frac{3x+1}{2^{k}}\cos^{2}\left(\frac{\pi x}{2^{k+1}}-\left(-1\right)^{k}\frac{\left(2^{k}-\left(-1\right)^{k}\right)\pi}{3\cdot2^{k+1}}\right)\prod_{n=1}^{k}\left(\sin^{2}\left(\frac{\pi x}{2^{n}}+\left(-1\right)^{n}\frac{\left(2^{n-1}-\left(-1\right)^{n-1}\right)\pi}{3\cdot2^{n}}\right)\right)\right) $$

The spot where the Jacobsthal numbers show is in the second sine and cosine functions, where $\frac{\left(2^{k}-\left(-1\right)^{k}\right)\pi}{3\cdot2^{k+1}}$ can be rearranged into $\frac{2^{k}- (-1)^{k}}{3} \cdot \frac{\pi}{2^{k+1}}$ and $\frac{2^{n-1}- (-1)^{n-1}}{3} \cdot \frac{\pi}{2^{k+1}}$ can be found in $\frac{\left(2^{n-1}-\left(-1\right)^{n-1}\right)\pi}{3\cdot2^{n}}$. Those two functions when $\frac{\pi}{2^{k+1}}$ is taken out can calculate the Jacobsthal numbers, with the latter being one step behind form the former.

Now my question. I found out that the Jacobsthal numbers occurred in this by sheer brute force, doing enough to get a few numbers and sticking them in the OEIS, showing me the working sequence of A001045. Because I did this the brute force way, I don't understand why they work, just that they do work, and I would really appreciate if someone could see and explain why they show up.

Desmos graph: https://www.desmos.com/calculator/1fkdrnwm3e

  • 1
    $\begingroup$ Just on a mobile at the mo but I noticed this previously. I'm pretty sure they share the same recurrence relation. In fact any equivalence class of odd integers the same number of steps (counting odd numbers only) can be expressed as a linear combination of a Lucas sequence and its partner. Jacobsthal numbers are a special case of such a sequence. $\endgroup$ Jul 20, 2021 at 13:58
  • $\begingroup$ Sorry, that should have read "...the same number of steps fron 1..." $\endgroup$ Jul 22, 2021 at 11:45
  • 1
    $\begingroup$ I’m voting to close this question because this question is asking for tuition. $\endgroup$
    – postmortes
    Oct 10, 2021 at 6:27


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.