How many pairs $(x,y)$ such that $x+2=y$ have an $x$ or $y$ divisible by 5 and not 3? Let the set $S_{n}$ = {$(x,y):x,y \in \mathbb{O}$} such that $x+2=y$ and $y$ is less than or equal to odd integer $n$ and $\mathbb{O}$ is set of odd integers > 1.
Let us define the function $f(n) = |S_{n}|$ that counts the number of pairs in $S_{n}$.
Examples:  $S_{9} =\{(3,5),(5,7),(7,9)\}$ and $f(9) = 3$.
$S_{13} = \{(3,5), (5,7), (7,9), (9,11),(11,13)\}$ and $f(13) = 5$
$S_{37} = \{(3,5), (5,7), (7,9), (9,11), (11,13), (13,15), (15,17), (17,19), (19,21), (21,23), (23,25), (25,27), (27,29), (29,31),(31,33),(33,35),(35,37)\}$
and $f(37) = 17$
It turns out that $f(n) = ((n-3)/2)$ for odd integers $n$.
Let us define another function $g(n)$ that counts the number of pairs $(x,y)$ such that either $x$ or $y$ is divisible by 5 and $x$ or $y$ is not divisible by 3, and $x \neq 5$ and $y\neq 5$.
Example: $g(37) = 2$ because there are only 2 pairs where $x$ or $y$ is divisible by 5 and not 3, and $x$ or $y$ is not equal to 5.
They are (23,25) and (35,37).
Note, the pairs (3,5) and (5,7) are not included since they contain 5, the pairs (13,15) and (15,17) are not included because 15 is divisible by 3, the pair (25,27) is not included because 27 is divisible by 3, and the pair (33,35) is not included because 33 is divisible by 3.
What is the formula for $g(n)$ in terms of $f(n)$ ?
What is the formula for $g(n)$ in terms of $f(n)$ for limit $n \to\infty$?
Edit:
I believe $g(n)$ approaches $(1/3)(2/5)f(n)$ as $n \to\infty$.
Since $f(n)$ approaches $(n/2)$ as $n$ gets large, $g(n)$ approaches $n/15$ as $n \to\infty$.
I need somebody to confirm this.
The table below compares the actual value for $g(n)$ and $n/15$ for a few values of $n$ and they look comparable.
$g(n)$ approaches $n/15$ as $n \to\infty$">
 A: Integers to be avoided are multiples of $15$ so you have the following sequence for $g(\infty$): $$\{\color{red}{(23,25),(25,27)},\color{blue}{(33,35),(35,37)},\color{green}{(53,55),(55,57)},\cdots\}$$ Suppose that $n$ is odd and is greater than $23$ for non-triviality. The cases $n<23$ can be brute-forced separately, which I will not do so here. By doing some counting, it is possible to obtain the following closed forms modulo $30$.
Note: Please try to derive some of them yourself before asking for further hints.
\begin{array}{c|c}n\pmod{30}&1&3&5&7&9\\\hline g(n)&\frac2{15}(n-1)-2&\frac2{15}(n-3)-2&\frac1{10}(n-5)&\frac2{15}(n-7)&\frac2{15}(n-9)\end{array}
\begin{array}{c|c}n\pmod{30}&11&13&15&17&19\\\hline g(n)&\frac2{15}(n-11)&\frac2{15}(n-13)&\frac2{15}(n-15)&\frac2{15}(n-17)&\frac2{15}(n-19)\end{array}
\begin{array}{c|c}n\pmod{30}&21&23&25&27&29\\\hline g(n)&\frac2{15}(n-21)&\frac2{15}(n-23)&\frac2{15}(n+5)-3&\frac2{15}(n+3)-2&\frac2{15}(n+1)-2\end{array}
Therefore, since $f(n)=\frac{n-3}2$ as you have calculated, for odd $n>23$, $$\lim_{n\to\infty}\frac{g(n)}{f(n)}=\begin{cases}\frac15\,\text{for}\,n\equiv5\pmod{30}\\\frac4{15}\,\text{otherwise}\end{cases}$$
