$\sum_{d|n} \phi (\tau (d))=?$

Question

So I have just taken a course on elementary number theory and started playing around with Eulers totient function $\phi (n)$ and $\tau (n):= \#\{\text{divisors of } n\}$. I read about Gauss theorem which states that for positive integers $n\geq 1$ $$\sum_{d|n} \phi (d)=n$$ and then thought of a "similar" question which is stated in the title, that is $$\boxed {\sum_{d|n} \phi (\tau (d))=?}$$

I tried out some case that would be simpler than finding a general formula for this sum, which is when $n=p^{k}$ where $p$ is prime and $k$ is a positive integer. Reason being that there are simpler indentities to work with regarding $\phi (n)$ and $\tau (n)$ when $n$ is a power of a prime.

I will present what I found to be true on a more heuristic/evidential basis. Meaning that I have not yet proven the results. Here are my results:

Let $p$ be a prime and $k$ a positive integer. Then we have that:

1. $\boxed{\sum_{d|p^{k}} \phi (\tau (d))=\sum\limits_ {i=1}^ {k+1} \phi (i)=\sum\limits_ {i=1}^ {k} \phi (i)+\phi (k+1)}.$
2. $\boxed{\sum_{d|p^{k}} \phi (\tau (d))\stackrel{?}{\equiv} 0 \pmod 2}$, if $k\geq 1$.

Proving the first identity is rather simple. Simply note that the divisors of $p^{k}$ are the following list of integers: $\underbrace{p^{0}, p^{1}, \ldots , p^{k-1}, p^{k}}_{k+1}$. Meaning in particular that $p^{k}$ has exactly $k+1$ divisors. Thus $$\sum_{d|p^{k}} \phi (\tau (d))= \phi (\tau (p^{0}))+\phi (\tau (p^{1}))+\ldots +\phi (\tau (p^{k})) = \phi (1)+\phi (2)+\ldots +\phi(k)+\phi (k+1)=\sum\limits_ {i=1}^ {k+1} \phi (i)$$ which follow from the fact that $\tau (p^{k})=k+1$. Thus $$\sum_{d|p^{k}} \phi (\tau (d))=\sum\limits_ {i=1}^ {k+1} \phi (i)=\sum\limits_ {i=1}^ {k} \phi (i)+\phi (k+1).$$

That $\boxed{\sum_{d|p^{k}} \phi (\tau (d))\equiv 0 \pmod 2}$ I found by experimentation but I dont know exactly how to prove this, given that I have not made any mistakes, that is...

Since $\sum_{d|p^{k}} \phi (\tau (d))=\sum\limits_ {i=1}^ {k+1} \phi (i)$ one finds that:

• If $n=p^{1}$, then $\sum\limits_ {i=1}^ {1+1} \phi (i)=\phi (1)+\phi (2)=1+1=2$.
• If $n=p^{2}$, then $\sum\limits_ {i=1}^ {2+1} \phi (i)=\phi (1)+\phi (2) + \phi (3)=2+2=4$.
• If $n=p^{3}$, then $\sum\limits_ {i=1}^ {3+1} \phi (i)=\phi (1)+\phi (2)+\phi (3)+\phi (4)=4+2=6$
• If $n=p^{4}$, then $\sum\limits_ {i=1}^ {4+1} \phi (i)=\sum\limits_ {i=1}^ {4} \phi (i)+\phi (5)=6+4=10$,

so on and so forth. I made calculations up to $n=p^{6}$ which gives $\sum\limits_ {i=1}^ {6+1} \phi (i)=18$, which suggests that $$\sum_{d|p^{k}} \phi (\tau (d))\stackrel{?}{\equiv} 0 \pmod 2$$ if $k\geq 1$. How would one go about proving this?

Answer 1

The image of the euler totient function is always even, if the input is greater than two. An elegant prove can be found here, which uses $\gcd(k, n) = 1\Rightarrow\gcd(n - k, n) = 1$.

Using this argument, we may write $$\sum\limits_ {i=1}^ {k+1}\phi(i)=\sum\limits_ {i=1}^ {2}\phi(i)+\sum\limits_ {i=3}^ {k+1}\phi(i)=1+1+\text{even number}=\text{even number}.$$

Furthermore, the sum is called "Totient summatory function".

Answer 2

After a bit more thought, I wanted to follow up on my comment above. Your question about the even parity has been answered, but I wanted to keep digging into the more general question you proposed. First, a quick lemma:

$$n = \prod_{i=1}^{\omega(n)} p_i^{a_i} \implies \tau(n) = \prod_{i=1}^{\omega(n)} (a_i +1)$$

Basic proof at Wikipedia. The function $\omega(n)$ is the number of distinct prime divisors of $n$, that is, the cardinality of the set of values of $p_i$.

Note that $\tau(n)$ does not depend at all on the values taken by $p_i$. But even further, the set of divisors takes on the same form independent of $p_i$. Each divisor is the product of the tuples of the Cartesian product

$$\{1, p_1, p_1^2, \cdots p_1^{a_1}\} \times \{1, p_2, p_2^2, \cdots p_2^{a_2}\} \times \cdots \times \{1, p_i, p_i^2, \cdots p_i^{a_i}\}$$

Let's take a simple, but not overly simple, example. Consider $n = p^4q^2r$, using $p, q$ and $r$ (all prime) to save time typing subscripts. Then our Cartesian product is:

$$\{1, p, p^2, p^3, p^4\} \times \{1, q, q^2\} \times\{1, r\}$$

And there are $(4+1)(2+1)(1+1)=30$ divisors--remember, they're the products of the tuples--which we can put in a 3-D array (with the third dimension being two separate tables) (deep breath):

$$ % outer vertical array of arrays \begin{array}{c} % inner horizontal array of arrays \begin{array}{cc} % inner array of minimum values \begin{array}{c|ccccc} 1 & 1 & p & p^2 & p^3 & p^4\\ \hline 1 & 1 & p & p^2 & p^3 & p^4\\ q & q & pq & p^2q & p^3q & p^4q\\ q^2 & q^2 & pq^2 & p^2q^2 & p^3q^2 & p^4q^2 \end{array} & % inner array of maximum values \begin{array}{c|ccccc} r & 1 & p & p^2 & p^3 & p^4\\ \hline 1 & r & pr & p^2r & p^3r & p^4r\\ q & qr & pqr & p^2qr & p^3qr & p^4qr\\ q^2 & q^2r & pq^2r & p^2q^2r & p^3q^2r & p^4q^2r \end{array} \end{array} \end{array} $$

But once again, the formats don't depend on the identities of $p, q$, or $r$. It's true whether we're using $(2, 3, 5) \rightarrow n = 720$ or $(103, 107, 109) \rightarrow n = 140\ 456\ 858\ 986\ 021$. And since we now want $\tau(d)$ for each, well, that again depends only on the exponents in each divisor. Putting the exponents on the "axes," the array for $\tau(d)$ looks like this:

$$ % outer vertical array of arrays \begin{array}{c} % inner horizontal array of arrays \begin{array}{cc} % inner array of minimum values \begin{array}{c|ccccc} 0 & 0 & 1 & 2 & 3 & 4\\ \hline 0 & 1 & 2 & 3 & 4 & 5\\ 1 & 2 & 4 & 6 & 8 & 10\\ 2 & 3 & 6 & 9 & 12 & 15 \end{array} & % inner array of maximum values \begin{array}{c|ccccc} 1 & 0 & 1 & 2 & 3 & 4\\ \hline 0 & 2 & 4 & 6 & 8 & 10\\ 1 & 4 & 8 & 12 & 16 & 20\\ 2 & 6 & 12 & 18 & 24 & 30 \end{array} \end{array} \end{array} $$

Which is actually a lot more regular than I was expecting initially. In fact, the left table is literally just a multiplication table with an offset of $1$, thanks to the $+1$ in that first lemma. Perhaps we should have labeled with the exponent $+ 1$. And adding another layer gives a 3-D multiplication table, essentially. Which means that for any given $n$, the individual $\tau(d)$ values are just an $a_1$ by $a_2$ by $\cdots a_i$ array in $i$ dimensions. Then the values of $\tau(d)$ are a set with this recursive definition:

$$S_1 := \{1, 2, 3, \cdots a_1 + 1\} \ ; \ \ S_2 := S_1 \cup 2S_1 \cup \cdots (a_2+1)S_1 \ ; \ \ S_i := S_{i-1} \cup 2S_{i-1} \cup \cdots (a_i + 1)S_{i-1}$$ where $kS$ represents scalar multiplication of each element in the set by $k$. Can we boil this down farther? Perhaps:

$$S_i := \bigcup_{k_1=1}^{a_1+1} \bigcup_{k_2=1}^{a_2+1} \cdots \bigcup_{k_i=1}^{a_i+1} \{ k_1k_2 \cdots k_i \}$$

And then we can write your initial function as: $$f(n) = f\left(\prod_{i=1}^{\omega(n)} p_i^{a_i}\right) = \sum_{d \mid n} \varphi (\tau(d)) = \color{green}{\sum_{k_1=1}^{a_1+1} \sum_{k_2=1}^{a_2+1} \cdots \sum_{k_i=1}^{a_i+1} \varphi(k_1k_2 \cdots k_i)}$$

Sadly, I'm unsure how to simplify farther from here. We can't take advantage of $\varphi$ being multiplicative, because there will be plenty of cases where pairs of $k$ are not coprime. There's probably some mileage in $\varphi(2m) = \varphi(m)$ for odd $m$, and $\varphi(2m) = 2\varphi(m)$ for even $m$, but I'm not sure how to get something from that. Alas. To finish, though, here are tables of values of $f(n)$ for $a_1 \leq 10, a_2 \leq 5, a_3 \leq 2$:

\begin{array}{c|cccccccccc} a_3 = 0 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline 0 & 1 & 2 & 4 & 6 & 10 & 12 & 18 & 22 & 28 & 32 & 42 \\ 1 & 2 & 5 & 9 & 15 & 23 & 29 & 41 & 53 & 65 & 77 & 97 \\ 2 & 4 & 9 & 19 & 29 & 45 & 57 & 81 & 101 & 131 & 151 & 191 \\ 3 & 6 & 15 & 29 & 47 & 71 & 91 & 127 & 163 & 205 & 241 & 301 \\ 4 & 10 & 23 & 45 & 71 & 115 & 143 & 203 & 255 & 321 & 377 & 477 \\ 5 & 12 & 29 & 57 & 91 & 143 & 183 & 255 & 323 & 407 & 479 & 599 \end{array}

\begin{array}{c|ccccccccccc} a_3 = 1 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline 0 & 2 & 5 & 9 & 15 & 23 & 29 & 41 & 53 & 65 & 77 & 97 \\ 1 & 5 & 14 & 24 & 42 & 62 & 80 & 110 & 146 & 176 & 212 & 262 \\ 2 & 9 & 24 & 46 & 76 & 112 & 148 & 202 & 262 & 328 & 388 & 478 \\ 3 & 15 & 42 & 76 & 130 & 190 & 250 & 340 & 448 & 550 & 658 & 808 \\ 4 & 23 & 62 & 112 & 190 & 290 & 374 & 512 & 668 & 818 & 986 & 1216 \\ 5 & 29 & 80 & 148 & 250 & 374 & 494 & 668 & 872 & 1076 & 1292 & 1582 \end{array}

\begin{array}{c|ccccccccccc} a_3 = 2 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline 0 & 4 & 9 & 19 & 29 & 45 & 57 & 81 & 101 & 131 & 151 & 191\\ 1 & 9 & 24 & 46 & 76 & 112 & 148 & 202 & 262 & 328 & 388 & 478 \\ 2 & 19 & 46 & 98 & 152 & 228 & 300 & 414 & 522 & 678 & 786 & 976 \\ 3 & 29 & 76 & 152 & 246 & 362 & 482 & 656 & 844 & 1072 & 1260 & 1550 \\ 4 & 45 & 112 & 228 & 362 & 558 & 726 & 996 & 1264 & 1612 & 1900 & 2350 \\ 5 & 57 & 148 & 300 & 482 & 726 & 966 & 1308 & 1672 & 2128 & 2512 & 3082 \end{array}

OK, two final notes, which you might glean from the tables. First, $f(n)$ is even if $\omega(n)$ is odd, and odd if $\omega(n)$ is even. Also, the tables are diagonally symmetrical (along the diagonal of any pair of dimensions).

I hope this was, at least, interesting to you!

Answer 3

Since $ \phi(n)$ is even for any $n>2$ and $\phi(1) + \phi(2) = 2$ implies that $\sum_{i = 1} ^{k+1} \phi(i)$ is even.

In general if $ x_i> 2,\text{for } i = 1,2,\cdots ,n$ are natural numbers then $$ \sum_{i = 1}^{n} \phi(x_i) = 0 \mod 2$$