How to define function $f(n)$ for the number of integers in $\mathbb{O}_n$ that are evenly divisible by 11? Consider the set $\mathbb{O}_n$ which contains the odd integers less than or equal to $n$ that are not divisible by 3, 5 or 7 but includes 3, 5 and 7.
$\mathbb{O}_n$ = { 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 121, 127, 131, 137, 139, 143, 149, 151, ... ,$n$ }
The integers 9, 15, 21, … , 3 + 6$i$ where $i$ is an integer, are not in the set since these numbers are divisible by 3.
Likewise, the integers 15, 25, 35, … 5 + 10$i$ and the integers 21, 35, 49, … 7 + 14$i$ where $i$ is an integer, are not in the set.
Let function $f(n)$ equal the number of integers in $\mathbb{O}_n$ that are divisible by 11, excluding 11.
Examples:
$f(127) = 1$ since there is 1 integer in $\mathbb{O}_{127}$  that is evenly divisible by 11 and is less than or equal to 127. That integer is 121.
$f(151) = 2$ since there are 2 integers in $\mathbb{O}_{151}$ that are evenly divisible by 11 and are less than or equal to 151. Those integers are 121 and 143.
$f(191) = 3$ since there are 3 integers in $\mathbb{O}_{191}$ that are evenly divisible by 11 and are less than or equal to 191. Those integers are 121, 143 and 187.
Define function $f(n)$ as a mathematical formula?
Also, prove that as $n \to\infty$, $\frac{f(n)}{|\mathbb{O}_{n}|} = \frac{1}{11}$.
Edit: Sorry for the mistake. The fraction should have been $\frac{f(n)}{|\mathbb{O}_{n}|}$, not $\frac{f(n)}{n}$.
 A: Using Inclusion/Exclusion, I believe this would be:
$$g(n) = \left\lfloor \dfrac{n}{11} \right\rfloor - \left( \left\lfloor \dfrac{n}{22} \right\rfloor + \left\lfloor \dfrac{n}{33} \right\rfloor + \left\lfloor \dfrac{n}{55} \right\rfloor + \left\lfloor \dfrac{n}{77} \right\rfloor\right) + \left(\left\lfloor \dfrac{n}{66} \right\rfloor + \left\lfloor \dfrac{n}{110} \right\rfloor + \left\lfloor \dfrac{n}{154} \right\rfloor + \left\lfloor \dfrac{n}{165} \right\rfloor + \left\lfloor \dfrac{n}{231} \right\rfloor + \left\lfloor \dfrac{n}{385} \right\rfloor\right) - \left(\left\lfloor \dfrac{n}{330} \right\rfloor + \left\lfloor \dfrac{n}{462} \right\rfloor + \left\lfloor \dfrac{n}{770} \right\rfloor + \left\lfloor \dfrac{n}{1155} \right\rfloor\right) + \left\lfloor \dfrac{n}{2310}\right\rfloor$$
However, this counts when 11 is in the set. So, we need to subtract 1 when $n\ge 11$: 
$$f(n) = \begin{cases}g(n), & n<11 \\ g(n)-1, & n\ge 11\end{cases}$$
Testing this out with the numbers you have given:
$$f(127) = 11-(5+3+2+1)+(1+1+0+0+0+0)-(0+0+0+0)+0 - 1 = 1$$
$$f(151) = 13-(6+4+2+1)+(2+1+0+0+0+0)-(0+0+0+0)+0-1 = 2$$
$$f(191) = 17-(8+5+3+2) + (2+1+1+1+0+0) - (0+0+0+0) + 0 - 1 = 3$$
While I believe this is the correct formula for $f(n)$, it does not appear to give the limit you want.
$$\lim_{n \to \infty} \dfrac{f(n)}{n} = \dfrac{1}{11}-\left(\dfrac{1}{22}+\dfrac{1}{33}+\dfrac{1}{55}+\dfrac{1}{77}\right)+\left(\dfrac{1}{66}+\dfrac{1}{110}+\dfrac{1}{154}+\dfrac{1}{165}+\dfrac{1}{231}+\dfrac{1}{385}\right)-\left(\dfrac{1}{330}+\dfrac{1}{462}+\dfrac{1}{770}+\dfrac{1}{1155}\right)+\dfrac{1}{2310} = \dfrac{8}{385} \neq \dfrac{1}{11}$$
A: For each number $k$ that is counted by $f(n)$, the number $\frac k{11}$ is an element of $\Bbb O(n)$ that is $\ge 11$ and $\le \frac n{11}$. We conclude that
$$\tag1 f(n)=\left|\Bbb O_{\lfloor n/11\rfloor}\right|-3.$$
I suppose you already have found an expression for $|\Bbb O_m|$ by the inclusion-exclusion principle?
Even without an exact expression for $|\Bbb O_m|$ note that for $x\le m$, we have $x\in \Bbb O_m$ iff $x\le m$ iff $\gcd(x,2\cdot 3\cdot 5\cdot 7)=1$ with the exception that $1\notin \Bbb O_m$ and $3,5,7\in\Bbb O_m$. We conclude that $\Bbb O_m$ has roughly $\frac{\phi(2\cdot 3\cdot 5\cdot 7)}{2\cdot 3\cdot 5\cdot 7}\cdot m$ elements. Thus from $(1)$ we find
$$ \lim_{n\to\infty}\frac{f(n)}{n}=\frac 1{11}\cdot \frac{\phi(2\cdot 3\cdot 5\cdot 7)}{2\cdot 3\cdot 5\cdot 7}.$$
Perhaps you meant
$$ \lim_{n\to\infty}\frac{f(n)}{|\Bbb O_n|}=\frac1{11}.$$
