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I am stuck with one formula at page 35 of The Ultimate Challange the 3x+1 problem.

$T^{(i)}(n)\equiv x_i(n) \pmod 2$

$x_i$ form the so called parity vector of the Collatz sequence.

Example of parity vector taken from E. Rosendaal page: For n= 17 we find v0 = 1, v1 = 0, v2 = 1, v3 = 0, v4 = 0, v5 = 1, etc.

Now

$T^{(i)}(n)=\lambda_k(n)\cdot n+\rho_k(n)$

where

$\lambda_k(n)=\frac {3^{x_0(n)}+...+3^{{k-1}(n)}}{2^k}$

and

$\rho_k(n)=\sum_{i=0}^{k-1} x_i(n) \frac{3^{x_{i+1}(n)+...+x_{k-1}(n)}}{2^{k-i}}$

T is the iteration function, n is the starting value of the sequence, i is the i-th iteration of the T function.

How to prove this formula?

I am completely stuck.

Enzo Creti
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  • Perhaps it is worth to moreover mention that $T(n)$ actually has a definition. I could get google books to show me page 35 of the mentioned book, where $T(n)$ has been defined $T(n)={3n+1 \over 2 }$ if $n$ is odd, $T(n)=n/2$ if $n$ is even. This is nontrivial to mention, since the elementary Collatz-iteration $T(n)$ has -at least- three different definitions. A second, very common (and I think the original), one is $T(n)={3n+1 }$ if $n$ is odd, $T(n)=n/2$ if $n$ is even. A third well known one is also known as "Syracuse-iteration": $T(n)={3n+1 \over 2^{\nu_2(3n+1)} }$ – Gottfried Helms Dec 10 '20 at 17:08
  • What are you trying to prove? A parity vector is what it is, just a parity vector. It gives the least significant bit-state in base 2. So its unclear what you're asking. –  Dec 12 '20 at 08:38
  • @NaturalNumberGuy - when I started looking at the Collatz-problem I'd once proved, that over the iterations, beginning from some $n$, all intermediate as well as the final iterates can be written in the form $a_k \cdot n + r_k$ that means that for instance $n$ does not evolve to a polynomial in $n$ or the like. Surely this is not a big proof, just bookkeeping when iteration is calculated, but likely a thing like this is meant. Unfortunately the notations of the definitions of the $\lambda$ and the $\rho$ are uncomfortably obfuscated... a thing which always had let me avoid such articles... – Gottfried Helms Dec 12 '20 at 12:25
  • Perhaps a look in my essay http://go.helms-net.de/math/collatz/aboutloop/collloopintro_main.htm in section "some general remarks" at the point of "a canonical representation" the explanation is sufficient to be reformatted to that formulae in your question (and thus might give a route to your own proof) – Gottfried Helms Dec 12 '20 at 12:40

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