# The Inverse of the Euler Totient Function

How can we calculate the cardinality of the inverse of Totient function of any positive integer n ? I tried going through this paper, but I couldn't understand the procedure.

Thanks

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Your link seems to be broken. – Eric Naslund Oct 13 '12 at 4:06

See also my recent paper "Computing the (number or sum of) inverses of Euler's totient and other multiplicative functions", which presents a generic dynamic programming algorithm for finding the inverses of a multiplicative function for a given integer value.

Please let me know if something is unclear.

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I liked your preprint. I would like it more if you gave a gentler introduction to your main equation (the one with the big O symbols) and defined or set up more components before displaying the equation. One way that might work is to claim and write down an equation for Min, then say Min is a context and that the Min example is a special case of, and then present the general form. My mind appreciates a graspable example just before a syntactically similar generalization. – The Masked Avenger Jan 24 '14 at 17:34
@The Masked Avenger: Thank you for insightful comments. This preprint was composed in rush (for a conference submission) and I plan to improve it significantly in further revisions. – Max Alekseyev Jan 24 '14 at 17:45
This impresses me as a (potentially very instructive) example of applied category theory. Even if it turns out to be an instance of something that (say) Michael Barr wrote up some time ago, I could see this as motivation to learn more category theory. – The Masked Avenger Jan 24 '14 at 18:17
I've just posted an updated version to arxiv. Please let me know if it addresses your concerns. – Max Alekseyev Apr 22 '14 at 0:30

Of course you should also take a look at This OEIS entry, and to the references within.

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f you are interested in a computational approach, there is software that can compute $\varphi^{-1}(n)$.

PARI/GP Scripts for Miscellaneous Math Problems by Max Alekseyev check invphi.gp.

The original invphi.gp appears for quite old pari/gp and doesn't run on current pari, I ported it here.

Here is a sample session:

? \r invphi2.gp
? n=2*13*17;v=invphi(n);#v
%1 = 2
? v
%2 = [443, 886]
? eulerphi(v[1])==n
%3 = 1

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I've tested invphi.gp on PARI/GP 2.6.2 and it works fine. Could you please tell me what exactly the problem you are experiencing with my code? Thanks! – Max Alekseyev Jan 21 '14 at 18:27
@MaxAlekseyev I suppose I backported it to older pari/gp for backwards compatibility (pari/gp versions are not fully compatible and the code didn't work for me at the time). Your code indeed works on new pari/gp. – joro Jan 21 '14 at 18:48

Here is a naive attempt, which can be refined to give an upper bound on the cardinality. I will only look at the case that the inverse n is odd, the even case being mildly more complicated.

So given $p$, I want to find how many odd $n$ satisfy $\phi(n)=p$. Let $p=r2^w$ with $r$ odd. To make things interesting assume $w>0$. Pick $b \geq w$ and assume $n$ has at most $b$ factors. Place the $w$ 2's in $b$ buckets. If a prime factor $q$ of $r$ is not going into a bucket, then we must put in one bucket enough to make $q-1$. Otherwise, distribute the prime factors of $r$ into the buckets. Each bucket will contain those primes which multiply to form $q-1$, where $q$ is a factor of $n$. If $r$ has $c$ not necessarily distinct prime factors, there are then at most $c^{b+1}$ ways to distribute the factors of $r$, and not all of them will work.

A recursive version is to assume the least prime factor of $n$ is $q$, and then try to find solutions to $\phi(m)=p/[(q-1)q^s]$ for appropriate values of $s$. This may be quicker to implement but harder to analyze.

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I assume you are asking about $N(m)$, the number of distinct integers $n$ which satisfy $\phi(n)=m$ where $\phi$ is the Euler Totient function.

There are many results regarding upper and lower bounds for the size of $N(m)$, as well as the mean and variance. In particular, Carmichael conjectured that $N(m)$ is never equal to $1$.

Pomerance gave the upper bound $$N(m)\leq m\exp{-(1+o(1))\log m \frac{\log \log \log m}{\log \log m}}$$ and also showed that there are infinitely many $m$ for which $$N(m)\geq m^{\frac{5}{9}}.$$

Bateman showed that $$\sum_{m\leq x} N(m)=\frac{\zeta(2)\zeta(3)}{\zeta(6)}x+O\left(xe^{-c\sqrt{\log x\log \log x}}\right),$$ and we also have that $$\sum_{m\leq x} N(m)-\frac{\zeta(2)\zeta(3)}{\zeta(6)}x=\Omega\left(x^\frac{5}{9}\right)$$

For more details, see the following paper of Pomerance: Popular Values of Euler's Function.

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thanks Eric, but for calculations what's o(1) ? – pranay Oct 13 '12 at 8:04