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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.

  • 1
    The recursion idea wasn't really a number-theoretic insight, and Eric was the one who thought of involving $\Lambda(n)/\log n$ and noticed the factorization pattern, so I don't see why I get so much credit. But now I'm curious if a more analytic derivation can be made in a Riemann-esque way with $z/\log(1-z)$, contour integration, and some kind of divisibility inversion.2011-09-14

2 Answers 2

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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 $\sum_{n\leq M}\left(f*\frac{\Lambda}{\log}\right)(n)=M.$

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Here is the implementation of anon's idea:

enter image description here

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]]]   ] 
  • 0
    @Eric: If you're curious, I've added an explicit description of the coefficients to my question edit. I was fishing a bit with this question, to see if any other techniques existed for solving this problem - which they did, apparently.2011-09-14