TITLE: Summing the prime power counting function up to equal some value $n$
QUESTION [7 upvotes]: I want to find $c_k$ for
$n = 1 + c_1 \Pi(n) + c_2 \Pi(\frac{n}{2})+ c_3 \Pi(\frac{n}{3})+ c_4 \Pi(\frac{n}{4})+ c_5 \Pi(\frac{n}{5})+...$, 
assuming there are such coefficients, where
$\Pi(n) = \pi(n) + \frac{1}{2}\pi(n^\frac{1}{2})+ \frac{1}{3}\pi(n^\frac{1}{3})+ \frac{1}{4}\pi(n^\frac{1}{4})+...$ and $\pi(n)$ is the prime counting function.
Are there known techniques for solving a problem like this?
EDIT -
I was really asking this to figure out how tough of a question this is.  At least for anon, not very tough, it would seem.
In case any of you are curious, one way to calculate these coefficients is like so:
If $C_k$ are the Gregory Coefficients, the first few terms being $-1, \frac{1}{2}, \frac{1}{12}, \frac{1}{24}, \frac{19}{720}, \frac{3}{160},...$, and we have the strict divisor function such that
$d_0'(j) = 1$ if $n = 1$, $0$ otherwise
$d_1'(j) = 1$ if $n \neq 1$, $0$ otherwise
$d_k'(n) = \sum\limits_{j | n} d_1'(j) d_{k-1}'(n/j )$
then $c_k = \sum\limits_{a=0} -1^a C_a d_a'(k)$
There's a straightforward reason why the Gregory coefficients show up, involving Linnik's identity $\sum\limits_{k=1} \frac{-1^{k+1}}{k} d_k'(n) = \frac{\Lambda(n)}{\log n}$and multiplicative inverses of series coefficients, but I won't go into that.  
Anyway, good job, anon.

REPLY [3 votes]: Here is the implementation of anon's idea:

In case you would like to try the code out:
\[CapitalPi][n : (_Integer | _Rational)] := 
 Module[{k = Ceiling[Log2[n]] + 2, rec}, rec = 1/Range[k]; 
  rec.PrimePi[n^rec]]

FindCoefficientsC[len_] := 
 Module[{mat = 
    PadRight[Table[\[CapitalPi][n/m], {n, 2, 2 len + 1}, {m, 1, n}]]},
  LinearSolve[
    PadRight[mat], Range[1, 2 len]][[;; MatrixRank[mat]]]
  ]

REPLY [3 votes]: This is really a comment, but it is a bit too long.
Notice that you are summing under a hyperbola. Since $$\Pi(x)=\sum_{n\leq x}\frac{\Lambda(n)}{\log n}$$  we see that $$\sum_{k\leq x}c_{k}\Pi\left(\frac{x}{k}\right)=\sum_{k\leq x}c_{k}\sum_{n\leq\frac{x}{k}}\frac{\Lambda(n)}{\log n}=\sum_{nk\leq x}c_{k}\frac{\Lambda(n)}{\log n}.$$ Equivalently we are trying to find $f$, some function on the integers,  satifying $$\sum_{n\leq M}\left(f*\frac{\Lambda}{\log}\right)(n)=M.$$