An extension of Dirichlet's Divisor Problem

**Kiwi_Dave** · November 4th 2008, 08:04 PM

Let $d_3(n)$ denote the number of ordered triples of integers whose product is n. Prove that:

$\sum^N_{n=1}d_3(n)=N \log^2 N+O(N \log N)$

HINT: Count the lattice points with positive coordinates under or on the surface defined by xyz=N.

Something I know:
$\sum^N_{n=0}d(n)=N \log N+cN+O(\sqrt N)$ for some constant c.

I can see two possible ways of aproaching this.
Method 1. If I imagine a unit cube placed onto each point (x,y,z) that lies between the curve xyz=N and the three planes x=1, y=1 and z=1 then the volume (and hence number of) these cubes will exceed the volume under the curve xyz=N. By calculus (very rusty, could be wrong) I find that the volume is $0.5 N \log^2 N$ . Suggesting that once I consider the negative integer solutions I would get an answer of more than $2 N \log^2 N$ plus a big O term. Wrong by a factor of 2 and I don't know how to calculate the big O terms!

Method 2.
$\sum^N_{n=1}d_3(n)=\sum^N_{z=1}([N/z] \log [N/z]+c[N/z]+O(\sqrt {[N/z]}))$ where [N/z] represents the integer part of N/z. But I have had no luch working through the detail of this idea either!

**PaulRS** · November 5th 2008, 05:26 AM

Let $ d_3 ^ +$ be the number of triplets $ \left( {x,y,z} \right) \in \left( {\mathbb{Z}^ + } \right)^3 $ such that $ x \cdot y \cdot z = n $ , note that: $ d_3 = 4 \cdot d_3 ^ + $ because indeed: $ 3 \cdot d_3 ^ + = d_3 - d_3 ^ + $ ( intepret the RHS as the number of triplets $ \left( {x,y,z} \right) \in \mathbb{Z}^3 $ such that $ x \cdot y \cdot z = n $ and at least one of those factors is negative- therefore we must have 2 negatives since $ n \in \mathbb{Z}^ + $ )

Note that: $ \sum\limits_{k = 1}^n {f\left( k \right) \cdot \sum\limits_{j = 1}^{\left\lfloor {\tfrac{n} {k}} \right\rfloor } {g\left( j \right) \cdot \sum\limits_{s = 1}^{\left\lfloor {\tfrac{n} {{k \cdot j}}} \right\rfloor } {h\left( s \right) \cdot x^{k \cdot j \cdot s} } } } = \sum\limits_{k = 1}^n {\left( {\sum\limits_{a \cdot b \cdot c = k;a,b,c > 0} {f\left( a \right) \cdot g\left( b \right) \cdot h\left( c \right)} } \right) \cdot x^k } $

Thus: $ \sum\limits_{k = 1}^n {\sum\limits_{j = 1}^{\left\lfloor {\tfrac{n} {k}} \right\rfloor } {\sum\limits_{s = 1}^{\left\lfloor {\tfrac{n} {{k \cdot j}}} \right\rfloor } 1 } } = \sum\limits_{k = 1}^n {d_3 ^ + \left( k \right)} $ that is: $ \sum\limits_{k = 1}^n {\sum\limits_{j = 1}^{\left\lfloor {\tfrac{n} {k}} \right\rfloor } {\left\lfloor {\tfrac{n} {{k \cdot j}}} \right\rfloor } } = \sum\limits_{k = 1}^n {d_3 ^ + \left( k \right)} $

By the definition of the floor function: $ \sum\limits_{k = 1}^n {d_3 ^ + \left( k \right)} \leqslant \sum\limits_{k = 1}^n {\sum\limits_{j = 1}^{\left\lfloor {\tfrac{n} {k}} \right\rfloor } {\tfrac{n} {{k \cdot j}}} } < \sum\limits_{k = 1}^n {d_3 ^ + \left( k \right)} + \sum\limits_{k = 1}^n {\sum\limits_{j = 1}^{\left\lfloor {\tfrac{n} {k}} \right\rfloor } 1 } $

That is: $ \sum\limits_{k = 1}^n {d_3 ^ + \left( k \right)} \leqslant \sum\limits_{k = 1}^n {\tfrac{n} {k} \cdot H_{\left\lfloor {\tfrac{n} {k}} \right\rfloor } } < \sum\limits_{k = 1}^n {d_3 ^ + \left( k \right)} + \sum\limits_{k = 1}^n {\left\lfloor {\tfrac{n} {k}} \right\rfloor } $

Now consider that: $ \left\{ \begin{gathered} \sum\limits_{k = 1}^n {\left\lfloor {\tfrac{n} {k}} \right\rfloor } = \sum\limits_{k = 1}^n {d\left( k \right)} = n \cdot \log \left( n \right) + O\left( n \right) \hfill \\ H_{\left\lfloor {\tfrac{n} {k}} \right\rfloor } = \log \left( {\left\lfloor {\tfrac{n} {k}} \right\rfloor } \right) + O\left( 1 \right) \hfill \\ \end{gathered} \right. $

Then we get: $ \sum\limits_{k = 1}^n {\tfrac{n} {k} \cdot \log \left( {\left\lfloor {\tfrac{n} {k}} \right\rfloor } \right)} - \sum\limits_{k = 1}^n {d_3 ^ + \left( k \right)} = O\left( n \cdot \log \left( n \right)\right) $

Let's see what happens with that logarithm on the LHS:
$ \sum\limits_{k = 1}^n {\tfrac{n} {k} \cdot \log \left( {\tfrac{n} {k}} \right)} - \sum\limits_{k = 1}^n {\tfrac{n} {k} \cdot \log \left( {\left\lfloor {\tfrac{n} {k}} \right\rfloor } \right)} = \sum\limits_{k = 1}^n {\tfrac{n} {k} \cdot \log \left( {\tfrac{n} {{k \cdot \left\lfloor {\tfrac{n} {k}} \right\rfloor }}} \right)} $

Using the definition of the floor function again: $ 0 = \log \left( 1 \right) \leqslant \log \left( {\tfrac{n} {{k \cdot \left\lfloor {\tfrac{n} {k}} \right\rfloor }}} \right) \leqslant \log \left( {1 + \tfrac{1} {{\left\lfloor {\tfrac{n} {k}} \right\rfloor }}} \right) \leqslant \tfrac{1} {{\left\lfloor {\tfrac{n} {k}} \right\rfloor }} $

Thus: $ 0 \leqslant \sum\limits_{k = 1}^n {\tfrac{n} {k} \cdot \log \left( {\tfrac{n} {k}} \right)} - \sum\limits_{k = 1}^n {\tfrac{n} {k} \cdot \log \left( {\left\lfloor {\tfrac{n} {k}} \right\rfloor } \right)} \leqslant \sum\limits_{k = 1}^n {\tfrac{n} {k} \cdot \tfrac{1} {{\left\lfloor {\tfrac{n} {k}} \right\rfloor }}} = O\left( n \right) $

That is: $ \sum\limits_{k = 1}^n {\tfrac{n} {k} \cdot \log \left( {\tfrac{n} {k}} \right)} + O\left( n \right) = \sum\limits_{k = 1}^n {\tfrac{n} {k} \cdot \log \left( {\left\lfloor {\tfrac{n} {k}} \right\rfloor } \right)} $

Thus: $ \sum\limits_{k = 1}^n {\tfrac{n} {k} \cdot \log \left( {\tfrac{n} {k}} \right)} - \sum\limits_{k = 1}^n {d_3 ^ + \left( k \right)} = O\left( n \cdot \log \left( n \right) \right) $

Now we have: $ \sum\limits_{k = 1}^n {\tfrac{n} {k} \cdot \log \left( {\tfrac{n} {k}} \right)} = n \cdot \log \left( n \right) \cdot \sum\limits_{k = 1}^n {\tfrac{1} {k}} - n \cdot \sum\limits_{k = 1}^n {\tfrac{{\log \left( k \right)}} {k}} = \tfrac{n} {2} \cdot \log ^2 \left( n \right) + O\left[ {n \cdot \log \left( n \right)} \right] $ (use integrals to approximate the second sum)

Thus: $ \sum\limits_{k = 1}^n {d_3 ^ + \left( k \right)} = \tfrac{n} {2} \cdot \log ^2 \left( n \right) + O\left[ {n \cdot \log \left( n \right)} \right] $

So this would yield: $ \sum\limits_{k = 1}^n {d_3 \left( k \right)} = 2\cdot n\cdot \log ^2 \left( n \right) + O\left[ {n \cdot \log \left( n \right)} \right] $

PS: When you put $\sum^N_{n=1}d(n)=N \log N+cN+O(\sqrt N) $ , be careful, that $d(n)$ represents the number of positive divisors of n

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