Eric Roosendaal, who worked to try to disprove the Conjecture, wrote an entire page of terms he coined to describe the properties of numbers in Collatz sequences. I was either mostly or completely confused reading through his definitions and proofs. When I looked at the tables and attempted to work out the formulas he presented, I struggled to find a connection between the Collatz trajectories and what he talked about.

What does delay and residue mean and what do those values describe about the behavior of the Collatz Conjecture? Why does 993 have the largest residue value?

It may be he is using fundamental concepts I have never heard of or thought about before, especially if I never came across them in school yet. I do not have any formal degrees in mathematics.


1 Answer 1


Delay(N): Delay $D(N)$ is the number of all iterations which are needed to arrive at $1$.

Res(N): For residue, consider the following, where for my own convenience I translate all into my style of writing, especially using small letters for elements of the iterations and reserve capital letters for numbers in the exponents. So instead of $N$ let's call the first element $a=a_1$ . Then redefine $N$ for the number of odd iterations and $S$ for the sum of all even iterations. (This says also, that $D(a)=S+N$ in my notation, see below).

Ok. Let's moreover redefine one Collatz iteration on odd positive numbers $a$ as $$a_{k+1}= {3a_k+1\over 2^{A_k}} $$ where the exponent $A_k$ is taken such that the result is one odd integer again (see also "syracuse"-notation of the Collatz-problem in wikipedia).

Look at the product of all odd $a_k$ $$ a_2 \cdot a_3 \cdot ... \cdot a_N \cdot 1 = {3a_1+1\over 2^{A_1}} \cdot {3a_2+1\over 2^{A_2}} \cdot ... \cdot {3a_{N-1}+1\over 2^{A_{N-1}}} \cdot {3a_{N}+1\over 2^{A_{N}}} $$ It seems convenient to multiply the initial number $a=a_1$ to that equation too to get the more symmetric one $$ a_1 \cdot a_2 \cdot a_3 \cdot ... \cdot a_N \cdot 1 = a_1 \cdot {3a_1+1\over 2^{A_1}} \cdot {3a_2+1\over 2^{A_2}} \cdot ... \cdot {3a_{N-1}+1\over 2^{A_{N-1}}} \cdot {3a_{N}+1\over 2^{A_{N}}} $$

Now let $S=\sum_{k=1}^N A_k$ the sum of all exponents $A_k$ which is also the number of all "divide-by-2" steps and multiply: $$ 2^S \cdot a_1 \cdot a_2 \cdot a_3 \cdot ... \cdot a_N \cdot 1 = a_1 \cdot (3a_1+1) \cdot (3a_2+1) \cdot ... \cdot (3a_{N-1}+1) \cdot (3a_{N}+1) $$ Next divide that by the $a_k$-product on the lhs and cancel on rhs: $$ 2^S \cdot 1 = a_1 \cdot (3+{1\over a_1}) \cdot (3+{1\over a_2}) \cdot ... \cdot (3+{1\over a_{N-1}}) \cdot (3+{1\over a_{N}}) $$ Divide the equation by $3^N$ where $N$ is the number of iterations on odd numbers, and finally by the initial number $a_1$: $$ {2^S\over 3^N} \cdot \frac1{a_1}= (1+{1\over 3a_1}) \cdot (1+{1\over 3a_2}) \cdot ... \cdot (1+{1\over 3 a_{N-1}}) \cdot (1+{1\over 3 a_{N}}) = \text{Res}(a_1) $$ This is the formula $$2^{E(N)} = 3^{O(N)} N \cdot \text{Res}(N) $$ or, better rearranged $${2^{E(N)}\over 3^{O(N)}}\cdot \frac1N =\text{Res}(N) $$ in Rosendaal's page. The rationale behind that formula seems to be the consideration, that any number $a_1$ could be expressed by some $2^X/3^Y$ where $X$ and $Y$ are related by the $log_2(3)$ and the number $a_1$ shall be related to this basic relations respective some "jitter", which occurs because we iterate in integer numbers and have not real values like in $X$ and $Y$ .

With that interpreation one might say: $a=993$ has the largest relative "jitter"-coefficient (but I didn't yet check that numerically)

In my analyses I arrived at the product-formula at the rhs in the study of conditions for the existence of cycles: we need then, that the lhs (with the $\frac1{a_1}$ removed) equals the rhs for some $a_1$. (The removal occurs, because we assume by the full transformation that instead of the number $1$ the number $a_1$ occurs in the final step, so we have instead of $\cdot \frac 1{a_1}$ the value $ \cdot \frac{a_1}{a_1}$ and this cancels to the removable factor $1$)

  • $\begingroup$ Thank you for your explanation! This makes a lot more sense. $\endgroup$ Oct 23, 2017 at 22:17

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