Division by $2p+1$ Can $\left\lfloor{\dfrac{x}{2p+1}} \right\rfloor$ be expressed in terms of $\left\lfloor{\dfrac{x}{p}} \right\rfloor$ for prime $p$? 
How to divide by $2p+1$ by only using division by $p$?
EDIT:
The above formulation is wrong. I meant "expressed in terms" in a sense broader that "a function that takes $\left\lfloor{\dfrac{x}{p}} \right\rfloor$  as an argument. 
Different version: let $0\leq a,b < 2p+1$ ($a,b$ known integers) and $x=ab$. How to divide $x$ by $2p+1$ in a way cheaper than just dividing by $2p+1$? Dividing by $p$ is cheaper than dividing by $2p+1$. It doesn't have to be a formula, algorithm is also ok.
 A: $$ \frac{x}{2p+1} = \frac{x}{2p} \frac{1}{1+1/(2p)} = \sum_{j=0}^\infty (-1)^j \frac{x}{(2p)^{j+1}}$$
For $\left\lfloor \dfrac{x}{2p+1} \right\rfloor$, you can stop the series if you come to a point where further terms can't make a difference (which should happen unless $x$ is an integer multiple of $2p+1$).  Thus if $$S_N = \sum_{j=0}^N (-1)^j \dfrac{x}{(2p)^{j+1}}$$
$\left\lfloor \dfrac{x}{2p+1} \right\rfloor = \left\lfloor S_N \right\rfloor$ if 
$N$ is even and $x/(2p)^{N+2} < S_N - \lfloor S_N \rfloor$ or $N$ is odd and 
$x/(2p)^{N+2} < \lceil S_N \rceil - S_N$.
A: I don't think it can, in general. By the time you get $\lfloor{x\over p}\rfloor$, information has already been lost that was required in order to know  $\left\lfloor{x\over 2p+1}\right\rfloor$.
Say for example that you had $p=2$.  You need an expression for $\left\lfloor{x\over 2p+1}\right\rfloor = \left\lfloor{x\over 5}\right\rfloor$ that yields 0 when $x=4$ and 1 when $x=5$.  But no function of $\left\lfloor{x\over 2}\right\rfloor$ can do this, because $\left\lfloor{x\over 2}\right\rfloor = 2$ for both $x=4$ and $x=5$. 
A: Hint $\ $ Reverse engineer this old trick that reduces dividing by $11$ to dividing by $10$.
$\ \ \begin{eqnarray}13717.4\, -\, 1371.74\, &=&\, 12345.66_{\phantom{M^{M^M}}} \\
\Rightarrow\quad \dfrac{137174}{11}\, &=&\, 12345.66 + 123.4566 + 1.234566 + 0.01234566+\,\cdots \\ &=&\, 12470.36... \end{eqnarray}$
