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 $1$)
