Formulas for Sequences Removing Multiples of 2, 3, and 5 First off, I am a programmer so please excuse if some of the terms I use are not the correct mathematical terms. I was working on devising a function to improve one of my prime number generation algorithms. With this in mind, I first set out to find the formulas for a sequence removing multiples of 2 and 3:
\begin{array}{c|c}
x&y\\ \hline
0&5\\ \hline
1&7\\ \hline
2&11\\ \hline
3&13\\ \hline
4&17\\ \hline
5&19\\ \hline
6&23\\ \hline
7&25\\ \hline
8&29\\ \hline
\vdots&\vdots
\end{array}
The equations that I came up for for this sequence are as follows:
$$y = 3x + 5 - x \bmod 2$$
$$x = \left\lfloor\frac{y - 5 + [y \bmod 3 \neq 0]}{3}\right\rfloor$$
After this, I tried to do the same for a sequence removing multiples of 2, 3, and 5:
\begin{array}{c|c}
x&y\\ \hline
0&7\\ \hline
1&11\\ \hline
2&13\\ \hline
3&17\\ \hline
4&19\\ \hline
5&23\\ \hline
6&29\\ \hline
7&31\\ \hline
8&37\\ \hline
9&41\\ \hline
10&43\\ \hline
11&47\\ \hline
12&49\\ \hline
13&53\\ \hline
\vdots&\vdots
\end{array}
While I think I found an equation to get $y$ from a value of $x$, I cannot find a way to get the value of $x$ from a given value $y$. 
$$y = 4x + 7 - 2\left\lfloor\frac{1}{8}x\right\rfloor - 2\left[\{2, 3, 6\} \ \text{contains}\ (x \bmod 8)\right] - 4\left[\{4, 5, 7\} \ \text{contains}\ (x \bmod 8)\right]$$
$$x =\ ?$$
I am wondering if an equation that produces the corresponding value of $x$ for a given value of $y$ for the aforementioned sequence exists, and if indeed it does, what the equation is.

The first one is:
$$y=5+6\bigg\lfloor \frac{x}{2} \bigg\rfloor + 2(x\mod2)$$
Although arguably $x(0)=1$, so:
$$y=1+6\bigg\lfloor \frac{x+1}{2} \bigg\rfloor - 2(x\mod2)$$
The second, with $x(0)=1$, is:
$$y=15+30\bigg\lfloor \frac{x}{8} \bigg\rfloor +4(x\mod8-3.5)-\frac{(x\mod8-3.5)}{|x\mod8-3.5|}\cdot2(\bigg\lfloor\frac{(|x\mod8-3.5|+0.5)}{2}\bigg\rfloor\mod2)$$
