Triple Totient Quotient

This solution is wrong. I have not reviewed it as yet.

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We will use a well known formula for $\phi(n).$ If $n=p_1^{a_1}\cdot...\cdot p_k^{a_k},$ then $\phi (n)=(p_1-1)p_1^{a_1-1}\cdot ... \cdot (p_k-1)p_k^{a_k-1}=n\cdot \frac{p_1-1}{p_1}\cdot...\cdot \frac{p_k-1}{p_k}$

If $n=2^a,$ $a\geq 1,$ then the $\phi(n)=2^{a-1}.$ So $\frac{n}{\phi(\phi(\phi(n)))}$ can be $8$ (if $a\geq 3$ ), $4,$ $2,$ or $1.$

For $n=p_1^{a_1}\cdot...\cdot p_k^{a_k},$ we denote the order $a_i$ of $p_i$ in $n$ by $ord_{p_i}n.$

Note that for every odd prime $p_i$ the number $p_i-1$ is even. So $ord_2(\phi(n))\geq ord_2 n - 1 + \#(odd\ prime\ factors\ of\ n)$

If $n$ is divisible by four or more distinct odd primes, $ord_2\phi(n)\geq ord_2 n +3.$ So $ord_2 \phi(\phi(\phi(n)))\geq ord_2 \phi(\phi(n))-1 \geq ord_2 \phi(n) -2 \geq ord_2 n +1$ Thus, for $\frac{n}{\phi(\phi(\phi(n)))}$ to be integer, the number of odd primes in $n$ is at most $3$ . Moreover, if $n$ has three odd prime factors, then all the above inequality must be equalities. In particular, $\phi(n)$ must be a power of $2.$ So all odd prime factors of $n$ are of the form $2^i+1$ for some $\ i$ . Since the prime factors are distinct, $ord_2\phi(n)-ord_2(n)\geq 1+2+3,$ so again we get a contradiction.

Suppose $n=2^ap^bq^c,$ where $p<q$ are odd primes and $b$ and $c$ are positive. The same argument as above shows that $ord_2(p-1)+ord_2(q-1)\leq 3.$ Also if $ord_2(p-1)+ord_2(q-1)=3,$ then $\phi(n)$ must be a power of $2.$ So in this sub-case we must have $n=2^a\cdot 3\cdot 5,$ with $a\geq 1,$ and the ratio is $15$ . If $ord_2(p-1)+ord_2(q-1)=2,$ then $p\equiv q\equiv 3 \pmod 4.$ We will prove that in this case $\phi(n)$ can only be of the form $2^i,$ or $2^i\cdot 3.$ Indeed, otherwise $ord_2\phi(\phi(\phi(n)))\geq ord_2\phi(n),$ because an other odd factor of $\phi(n)$ will either create an odd prime factor for $\phi(\phi(n))$ or extra power of $2$ in $\phi(\phi(n))$ . Clearly, the only possible case here is $n=2^a\cdot 3^b \cdot 7,$ which gives the ratio of $21$ .

Now suppose that $n=2^a\cdot p^b,$ where $p$ is an odd prime. First of all, if $b\geq 2,$ then the prime $p$ must be $3$ . In this case $b$ can be anything, and we get ratio $27$ if $b \geq 3$ and $18$ or $9,$ depending on $a,$ if $b=2.$

Suppose now that $b=1$ . If $\phi(n)$ is a power of $2,$ then $p-1$ can only be $2,4,8,$ thus $p=3$ or $p=5.$ These give ratios $12,$ $6,$ $3;$ $10$ and $5$ (depending on $a$ ). If $\phi(n)$ is not a power of $2$ then, clearly, it can only be of the form $2^a\cdot 3,$ $2^a\cdot 3^2,$ or $2^a\cdot 5.$ This means that $p$ can only be $7,$ $11$ and $13$ . This gives ratios $14,$ $7,$ $11,$ and $13$ .

Putting it all together, the possible integer ratios are $8,4,2,1,15,21,27,18,9,12,6,3,10,5,14,7,11,13.$ They add up to $(1+2+...+15)+18+21+27=171$