2$, whence $$ 5^{\a+1}> 5^{\a+1}-1>\frac{200\cdot 16}{121\cdot 25}\cdot 5^{\a+1}> 1.05 \cdot 5^{\a+1}> 5^{\a+1} $$ and the last statement is clearly false. The only remaining possibility is $p=5$ and $q=7$. If $\a=1$ then $$ \frac{5+1}{5-1}\cdot \frac{1-1/7^{\b+1}}{6^2/7^2}=2 $$ which means $\b=1$ (likewise if $\b=1$, we must have $\a=1$). However in the equation $$ \left(\frac{1-1/5^{\a+1}}{4^2/5^2}\right)\cdot\left(\frac{1-1/7^{\b+1}}{6^2/7^2}\right)=2 $$ both factors on the left are strictly increasing in $\a$ and $\b$ respectively, thus there can be no solutions with $\a>1$ or $\b>1$. \end{proof} \begin{lemma} \label{lem:omega3} If $n\in\cR_2$ and $\o(n)=3$, then $n=1045$. \end{lemma} \begin{proof} Let $p,q,r$ be distinct prime numbers and $\a,\b,\c$ be natural numbers such that $n=p^\a\cdot q^\b\cdot r^\c$ satisfies $\s(n)=2\phi(n)$. We first show $(p,q,r)$ lies in the set of triples \begin{align*} \big\{&(5,11,13),(5,11,17),(5,11,19),(5,11,23),(5,11,29),\\&(5,11,31),(5,13,17),(5,13,19),(5,13,23)\big\}. \end{align*} Note that if $(p,q,r)=(5,11,19)$ then $\a=\b=\c=1$ is certainly a solution. One may assume $p\left(1-\frac{1}{5}\right)\left(1-\frac{1}{11}\right)\left(1-\frac{1}{r}\right) $$ which implies $r< 37$, thence $13\le r\le 31$. We checked each prime triple $(p,q,r)$ with $p\frac{1}{\sqrt{2}}> \left(1-\frac{1}{p}\right)\left(1-\frac{1}{q}\right)\left(1-\frac{1}{r}\right) \ge \left(1-\frac{1}{7}\right)\left(1-\frac{1}{11}\right)\left(1-\frac{1}{13}\right)> 0.719 $$ so we must have $p=5$. (Alternatively simply note that in each of the nine solution classes derived above one always has $p=5$.) We now show how to eliminate each of the eight cases other than $(p,q,r)=(5,11,19)$. To illustrate the method consider $(p,q,r)=(5,11,13)$, and suppose that $n=p^\a\cdot q^\b\cdot r^\c$ solves $\s(n)=2\phi(n)$. Using Equation (\ref{eqn:1}) again, one discovers \begin{eqnarray*} & & -\log\left(1-\frac{1}{5^{\a+1}}\right)-\log\left(1-\frac{1}{11^{\b+1}}\right)- \log\left(1-\frac{1}{13^{\c+1}}\right)\\&=& -\log\left(2\left(1-\frac{1}{5}\right)^2\left(1-\frac{1}{11}\right)^2\left(1-\frac{1}{13}\right)^2\right) =0.103846=:\lambda \text{ say.} \\ \end{eqnarray*} However for $0-\log(1-x) $$ in which case $$ \frac{3}{5^{\min(\a,\b,\c)+1}-1}> \frac{1}{5^{\a+1}-1}+\frac{1}{11^{\b+1}-1}+\frac{1}{13^{\c+1}-1}>\lambda. $$ The latter implies $31>5^{\min(\a,\b,\c)+1}$, and further that $1=\min(\a,\b,\c)$. We next consider in turn each of the cases $\a=1$, $\b=1$ and $\c=1$ leaving the other variables free, thereby eliminating possibilities through a short tree walk. If $\a=1$ then $$ \frac{1}{24}+\frac{1}{11^{\b+1}-1}+\frac{1}{13^{\c+1}-1}>\lambda; $$ consequently $$ \frac{2}{11^{\min(\b,\c)+1}-1}> \lambda-\frac{1}{24} $$ yielding $11^{\min(\b,\c)+1}< 34$, so $\a=1$ is clearly impossible. If however $\b=1$, we obtain $$ 25\le 5^{\min(\a,\c)+1}<\frac{2}{\lam-\frac{1}{120}}+1<22, $$ and if $\c=1$ then $$ 25\le 5^{\min(\a,\b)+1}< \frac{2}{\lam-\frac{1}{168}}+1<22 $$ which is also impossible. Hence there cannot exist solutions with $(p,q,r)=(5,11,13)$. The other remaining seven cases for $(p,q,r)$ are eliminated via similar arguments. \end{proof} The previous three lemmas may be neatly summarised, as follows. \begin{theorem} \label{thm:omega3} If $n\in\cR_2$ has at most $3$ distinct prime factors, then $n=3$, $35$ or $1045$. \end{theorem} More generally, if we make some restrictions on the exponents occurring in the prime decomposition of a solution $n\in\cR_{a/b}$ (e.g., $\nu_p(n)\leq 1$ for all primes $p$), one can obtain explicit finiteness results on the number of solutions with a bounded value of $\o(n)$. \begin{lemma} \label{lem:7} Let $k,a,b$ be fixed natural numbers, and consider the set $\cR_{a/b}$. Then there are at most a finite number of squarefree $n$ satisfying $\s(n)=a\cdot \phi(n)\big/b$ with $\o(n)=k$. \end{lemma} \begin{proof} We assume $k\ge 2$. In Equation (\ref{eqn:1}) one chooses $m=k$, and for all $i$ take $\a_i=1$ since $n$ is squarefree. Now for $n=p_1\cdots p_k$ one can rewrite this equation as $$ \frac{b}{a}=\prod_{i=1}^k\left(1-\frac{2}{p_i+1}\right) < \prod_{i=1}^{k-1}\left(1-\frac{2}{p_i+1}\right). $$ If $n$ is odd then we apply Lemma \ref{lem:6}, and immediately conclude $$ b\cdot \prod_{i=1}^k \left(\frac{p_i+1}{2}\right) < (b+1)^{2^k} $$ which implies $n+k<2^k\cdot (b+1)^{2^k}/b$. If $n=2m$ is even then $\s(m)/\phi(m)=a/(3b)$ and $m$ is odd, hence $$ m+k-1\le\prod_{i=2}^k (p_i+1)< 2^{k-1}(3b+1)^{2^{k-1}}/(3b) $$ in which case $n+2k-2< 2^k(3b+1)^{2^{k-1}}/(3b)$. \end{proof} Setting $b=1$, and combining the bounds directly above for $n$ even/odd, we obtain \begin{theorem} \label{thm:squarefree} If a squarefree integer $n$ satisfies the divisibility $\phi(n)\big| \s(n)$ with $\o(n)=k$, then $$n< 2^{2^k+k}-k.$$ In particular, there exist at most a finite number of such $n$. \end{theorem} %====================================================================================================== \section{Applications of the product compactification of $\mathbf{\N}$} \label{sec:product-compactification} \def\hN{\widehat{\mathbb{N}}} \def\v{\nu} For each prime $p$, let $\mathbb{N}_p$ denote the one point compactification of $\mathbb{Z}_{\geq 0}$; in particular, each finite point $n\in\mathbb{Z}_{\geq 0}$ is itself an open set, and a basis for the neighborhoods of the point at infinity, $p^{\infty}$ say, is given by the open sets $U_p^{(\epsilon)}=\big\{n\in\mathbb{Z}_{\geq 0}:n\ge 1/\epsilon\big\}\cup\{p^{\infty}\}$ with $\epsilon>0$. If $\mP$ indicates the set of prime numbers, let us write $$ \widehat{\mathbb{N}} := \prod_{p\in\mP} \mathbb{N}_p $$ for the product of these indexed spaces, endowed with the standard product topology. Then $\hN$ is a compact metrizable space so it is sequentially compact, hence every sequence in $\hN$ has a convergent subsequence. Note the useful equivalence that $N_i\longrightarrow N_o$ in $\hN$ if and only if for all primes $p$, $\v_p(N_i)\rightarrow \v_p(N_o)$ in $\mathbb{N}_p$. If $N$ is a positive integer we write $\widehat{N}$ for the corresponding element of $\hN$, namely $(n_p:p\in\P)$ where $n_p=\v_p(N)$. For example, $\widehat{84}=(2,1,0,1,\ldots)$ and $\widehat{1}=(0,0,0,\ldots)$. Let $\mathbb{N}$ have the discrete topology; we shall identify $\N$ with its image in $\cN$ under the embedding $n\rightarrow \widehat{n}$, yielding a dense subset of $\hN$. If $h:\N\rightarrow\R$ is a multiplicative function, and if for $p\in\mP$ each sequence $\big(h(p^n)\big)_{n\in\Z_{\geq 0}}$ is Cauchy in neighborhoods $U_p^{(\epsilon)}$, one may extend $h$ to a function on $\widehat{\N}$ through the formula $$ h\big(\widehat{N}\big):= \lim_{\widehat{n}\rightarrow(\dots,\nu_p(N),\dots)}\left(\prod_{p\in\mP}h\big(p^{n_p}\big)\right) $$ providing the limit exists and the product converges, of course. \begin{remark} We shall call $\hN$ equipped with its topology the {\it product compactification of $\mathbb{N}$}. A nice account detailing some of the properties of this set-up, with the compactification called the `supernatural topology', is given by Pollack in \cite{SN}; it is used, for example, to explore the perfect number problem. The topology was first introduced by Steinitz \cite{steinitz} and used in a number of applications, including \cite{artjuhov} and \cite{borho}. \end{remark} \begin{lemma} \label{lem:continuity1} Let $\cP\subset \mP$ be a finite set of primes. Assume we are given a sequence of natural numbers $(N_i)$ such that $N_i\rightarrow N_o$ in $\cN$, and for all $i\in\N$ one has $\supp (N_i)\subset \cP$. Finally let $h:\mathbb{N}\rightarrow \mathbb{R}$ be a positive multiplicative function, such that for each $p\in\cP$ $$ h(p^\infty):= \lim_{n\rightarrow\infty} h(p^n) $$ exists in $(0,\infty]$, i.e. is strictly positive or infinity. Defining the disjoint sets $$ \mathcal{A}=\{p\in\cP:\v_p(N_o)<\infty\} \quad\text{and}\quad \mathcal{B}=\{p\in\cP: \v_p(N_o)=\infty\} $$ then inside $\mathbb{R}_{>0}\cup\{\infty\}$, $$ \lim_{i\rightarrow\infty} h(N_i)= h(A)\cdot h(B^\infty) $$ where $A:=\prod_{p\in \mathcal{A}} p^{\v_p(N_o)}$ and $B:=\prod_{p\in \mathcal{B}}p$. \end{lemma} \begin{proof} Firstly note that $\supp( N_o)\subset \cP$, so the sets of primes $\mathcal{A}$ and $\mathcal{B}$ are well-defined and finite. We can write $\cP=\mathcal{A}\cup \mathcal{B}$ as a disjoint union, and then decompose $$ N_i=\prod_{p\in\cP} p^{\v_p(N_i)}=\prod_{p\in \mathcal{A}} p^{\v_p(N_i)}\cdot\prod_{p\in \mathcal{B}} p^{\v_p(N_i)}. $$ Because $N_i\rightarrow N_o$ in $\cN$, for each prime $p$ we must have $\v_p(N_i)\rightarrow \v_p(N_o)$ in $\Z_{\geq 0}\cup \{\infty\}$. By assumption $h$ is multiplicative on $\N$, in which case $$ h(N_i)= \prod_{p\in\cP} h\big(p^{\v_p(N_i)}\big)= \prod_{p\in \mathcal{A}}h\big(p^{\v_p(N_i)}\big) \cdot \prod_{p\in \mathcal{B}}h\big(p^{\v_p(N_i)}\big). $$ Since $\mathcal{A}$ is a finite set, then at each $p\in \mathcal{A}$ and for all $i\ge i_1$ say, $\v_p(N_i)=\v_p(N_o)$. It follows that the first term on the right in the above expression is equal to $h(A)>0$. Now for every $p\in \mathcal{B}$, $\v_p(N_i)\rightarrow \infty$ hence $h\big(p^{\v_p(N_i)}\big)\rightarrow h(p^\infty)$. Therefore, as the limit of a product of a finite number of terms is the product of the limits, our hypotheses on $h$ ensure the indeterminate form $0\cdot\infty$ does not occur. As an immediate consequence the second term above must tend to $h(B^\infty)$, and the lemma is proved. \end{proof} \begin{lemma} \label{lem:equivalence} Let $h:\N\rightarrow\R$ be a multiplicative function, and assume $(N_i)\subset\N$ is a sequence of natural numbers such that $\widehat{N_i}\rightarrow \widehat{N_o}$ inside $\cN$, for some positive integer $N_o$. If for all sequences $(M_i)\subset\N$ such that $\widehat{M_i}\rightarrow \widehat{1}$ in $\cN$ we also have $h(M_i)\rightarrow 1$ inside $\R$, then $\lim_{i\rightarrow\infty}h(N_i)= h(N_o)$. \end{lemma} \begin{proof} Given that $\widehat{N_i}\rightarrow \widehat{N_o}$, one defines $\cP:=\supp(N_o)$ which is a finite set as $N_o\in\N$. Let us write $N_i=A_i\cdot M_i$ where $\supp(A_i)\subset \cP$, and $\supp(M_i)\cap \cP=\emptyset$ so that $\gcd(A_i,M_i)=1$. Since $\widehat{M_i}\rightarrow \widehat{1}$ thus $h(\widehat{M_i})\rightarrow 1$ and $\widehat{A_i}\rightarrow \widehat{N_o}$, and it therefore follows $$ h(N_i)=h(A_i)\cdot h(M_i)\longrightarrow h(\widehat{N_o})\cdot 1=h(N_o). $$ \end{proof} \begin{theorem} \label{thm:support} Let $\cP$ be a finite set of primes. Then there exist at most a finite number of positive integers $N\in\N$ with $\supp(N)\subset \cP$ such that $\s(N)\big/\phi(N)=2$. \end{theorem} \begin{proof} Let $\cN$ be endowed with the product topology, and suppose there exists an infinite sequence of distinct integers $(N_i)_{i\ge 1}$ in $\cR_2$ for which $\supp(N_i)\subset \cP$ for each $i\in\N$. Then there is a subsequence, also denoted $(N_i)$, and a limit $N_o\in\cN$ such that $N_i\rightarrow N_o$. Let $h(n):=\s(n)/\phi(n)$ for $n\in \mathbb{N}$ and $h(p^\infty):=p^2/(p-1)^2$, so that $h(p^n)\rightarrow h(p^\infty)$ for all $p\in\P$. If $N_o$ was divisible by a prime not in $\cP$, then we would have an $N_i$ divisible by that prime, which is false. Hence $N_o$ has finite support. By Lemma \ref{lem:continuity1} we can write $$ N_o=A\cdot B^\infty $$ where $A$ and $B$ each supported by $\cP$ are odd, have $\gcd(A,B)=1$, are such that $B$ is squarefree, and $h(N_i)\rightarrow h(A)\cdot h(B^\infty)$. However this means $2=h(A)\cdot h(B^\infty)$. Since $A\|N_o$ we must have $A\|N_i$ for all sufficiently large $i$, and because the $N_i$ are distinct, we must have $A=N_i$ for at most one $i$. Thus for $i\ge i_1$ say we have $A$ a proper unitary divisor of $N_i$. For such an $i$, put $N_i=A\cdot B_i$ so $B_i>1$. Let us write $B_i=\prod_{p|B_i}p^{e_{p,i}}$ where $e_{p,i}=\v_p(B_i)$ tends to $\v_p(B^{\infty})= +\infty$ with $i$. Indeed, by passing to a suitable subsequence, without loss of generality one can assume each exponent $e_{p,i}$ is monotonically increasing. In particular, $$ 2 = h(N_i) = h(A)\cdot h(B_i) = \frac{\s(A)}{\phi(A)}\cdot \prod_{p\mid B_i}\frac{p^2}{(p-1)^2}\cdot\!\left(1-\frac{1}{p^{e_{p,i}+1}}\right). $$ The right-hand side is monotonic increasing with $i$, whilst the left-hand side is constant; this yields an immediate contradiction, therefore no such infinite sequence $(N_i)$ can have existed in the first place. \end{proof} \begin{theorem} \label{thm:Omega} Let $k\ge 1$ be a given natural number. There exist at most a finite number of positive integers $n$ with $\Omega(n)\le k$ satisfying $\s(n)\big/\phi(n)=2$. \end{theorem} \begin{proof} Let $h:\N\rightarrow \R$ be defined by $h(n)=\s(n)/\phi(n)$. Suppose there exist an infinite number of distinct positive integers $N_i$ satisfying both the equality $2=h(N_i)$ and the constraint $\Omega(N_i)\le k$ for all $i$. By passing to a suitable subsequence, also denoted $(N_i)$, one can assume $N_i\rightarrow N_o$ in $\cN$. \begin{remark} (i) We must have $N_o$ supported by at most $k$ primes, otherwise this would apply to at least one $N_i$ yielding $\Omega(N_i)\ge \omega(N_i)>k$ which is false. (ii) Furthermore, all the components of $N_o$ in $\cN$ must be bounded above by $k$ since at each prime $p$, one has $k\ge \v_p(N_i)\rightarrow \v_p(N_o)$. (iii) It follows that this limit point $N_o$ must correspond to a natural number. \end{remark} \medskip For $i$ sufficiently large, we can therefore write $N_i=N_o\cdot B_i$ where $B_i$ is an integer supported by at most $k$ primes, and $\gcd(N_o,B_i)=1$. As there are at most a finite number of primes dividing $N_o$, we can assume for all $p\in\supp(N_o)$ that $\v_p(N_o)=\v_p(N_i)$ for all indices $i\ge i_o$ say. Then the support of $B_i$ consists of all primes dividing $N_i$ which do not divide $N_o$. We claim that there is a sequence of natural numbers $(x_i)_{i\ge 1}$ tending to infinity, such that $\supp(B_i)\subset [x_i, \infty)$. If not, then there exists a prime $q$ for which $\v_q(B_i)>0$ infinitely often, which would force $\v_q(N_o)$ to be positive even though $q\not\in\supp(N_o)$! Therefore our claim concerning $\supp(B_i)$ must be true, in which case $$ 1\le h(B_i)= \prod_{p\mid B_i}\left(\frac{1-\frac{1}{p^{\v_p(B_i)+1}}}{\left(1-\frac{1}{p}\right)^2}\right) < \prod_{p\mid B_i} \left(1-\frac{1}{p}\right)^{-2} \le \left(1-\frac{1}{x_i}\right)^{-2k} \longrightarrow 1. $$ By Lemma \ref{lem:equivalence} it follows $2=h(N_i)=h(N_o)\cdot h(B_i)\rightarrow h(N_o)$, and therefore $h(N_o)=2$. But the latter is impossible as the $N_i$ being distinct means $N_o$ is a proper divisor of $N_i$ for at least one $i$, thereby implying that $h(N_o)<2$ which gives us a contradiction. \end{proof} \section{Density of the union of the $\cR_a$'s} \label{sec:density} Paul Erd\H{o}s was very interested in the so-called {\bf primitive} sequences of natural numbers, i.e., sequences $A=(a_n)_{n\in\N}$ with $a_1 0$ let us define $A(x)= \sum_{a_n\le x} 1$. In 1935, Erd\H{o}s showed the lower natural density of $A$ was necessarily zero, namely that $$ \liminf_{x\rightarrow\infty} \frac{A(x)}{x}=0. $$ It follows from Lemma \ref{lem:monotone} that each sequence of solutions to $\s(n)=a\cdot \phi(n)$ yields a primitive sequence, and therefore the set $\cR_a$ itself has trivial lower density. \begin{remark} We have a further item of evidence that the density of $\cR_a$ is zero, via a theorem of Erd\H{o}s and Davenport from 1937. If for some sequence $(a_n)$ one has $$ \limsup_{x\rightarrow\infty}\frac{1}{\log x}\cdot\sum_{a_n\le x} \frac{1}{a_n}>0 $$ then there exists a subsequence $(a_{n_i})\subset(a_n)$ such that $a_{n_i}\big| a_{n_{i+1}}$ for all indices $i\in\N$. Hence if the $a_n$'s comprise all the elements in $\cR_a$ ordered by `$<$' say, then one deduces $$ \limsup_{x\rightarrow\infty}\frac{1}{\log x}\cdot\sum_{a_n\le x} \frac{1}{a_n}=0. $$ Thus $\cR_a$ has lower asymptotic density zero, and also upper logarithmic density zero. Unfortunately, in general, this alone is not sufficient to imply density zero for the set (see the much celebrated example of Besicovitch \cite{besi}). \end{remark} However the issue of whether solution sets have density zero can be easily resolved upon noting that by Lemma \ref{lem:erdos-winter}, $f(n):= \log \left(\s(n)/\phi(n)\right)$ has a continuous increasing distribution function, hence the same is true for $\s(n)/\phi(n)$. Therefore, for any given positive integers $a$ and $b$, the set $\cR_{a/b}$ has density zero. In fact much more than this is true in the case $b=1$, i.e., when $\phi(n)\mid \s(n)$. The authors are grateful to the referee for the following proof, based on ideas from \cite[Theorem 3.1]{banks} (which we labelled Lemma \ref{lem:phi-values}). \begin{theorem} \label{thm:density} If $N(x):= \#\{n\le x: \phi(n)\mid \s(n)\}$ then as $x\rightarrow\infty$, one has the bound $$ N(x)\ll x \exp\left(-\frac{1}{2}\sqrt{\log x}\right). $$ \end{theorem} \begin{proof} Let $n\le x$ satisfy $\phi(n)\mid \s(n)$, and write $p$ for the largest prime factor of $\phi(n)$. Assume that $p>y$ where $y>0$ is a large parameter to be chosen later, and write $A$ for the subset of such $n$ with $p\le y$. Now one has $p \mid \phi(n) \mid \sigma(n)$, therefore there exists a prime power $q^e \| n$ with $p \| \s(q^e)$. If $e=1$ then $p \mid q+1$; otherwise as $2q^e > \sigma(q^e) \ge p$ and $n\neq q^e$, the integer $n$ would have a proper prime power divisor $q^e$ say, with $q^e > p/2 > y/2$. Now using partial summation, the set of $n\le x$ with such a prime power divisor, $B$ say, is of size $O(x/\sqrt{y})$; exploiting the fact $p \mid \phi(n)$ and a similar argument, we see that either there is a prime $r \mid n$ with $p \mid r-1$, or instead $n$ belongs to an exceptional set, $C$ say, of size $O(x/y)$. Suppose $n\not\in A\cup B\cup C$. Then $n$ has prime factors $r\equiv 1 (\bmod p)$ and $q \equiv -1 (\bmod p)$, and the number of such $n \le x$ is bounded above by $x/(rq)$. Summing over both $r$ and $q$ in the given progressions shows that for a given $p$, the number of such $n$ is $O(x (\log x)^2/p^2)$; secondly, summing over primes $p > y$ implies the number of $n$ up to $x$ is $O(x (\log x)^2/y)$. Putting everything together, this argument establishes that the total number of $n \le x$ for which $\phi(n)\mid \s(n)$ satisfies the upper bound $$ N(x) \ll \#\big\{n\le x: \text{$\phi(n)$ has only prime factors $\le y$}\big\} + \frac{x}{\sqrt{y}} +\frac{ x}{y} + \frac{x (\log x)^2}{y}. $$ For $x$ and $u$ sufficiently large, by Lemma \ref{lem:phi-values} one has $\Phi(x,y)\le xe^{-u/2}$, and minimizing $$ xe^{-\frac{u}{2}}+ \frac{x}{\sqrt{y}} $$ we obtain at the minimum $$\frac{\log y}{2}+ \log\log x-2\log\log y= \frac{\log x}{2\log y}. $$ Neglecting the ``loglog" terms yields $\log y= \sqrt{\log x}$, $u=\sqrt{\log x}$ and $y=\exp(\sqrt{\log x})$. Checking these estimates, one then deduces $$ \Phi(x,y)\le x\exp(-(1+o(1))u\log\log u)\le x\exp(-\frac{u}{2})\le \frac{x}{\sqrt{y}} $$ so for $x$ sufficiently large, $N(x)\ll x \exp\big(\!-\frac{1}{2}\sqrt{\log x}\big)$. The result now follows. \end{proof} For a given $m\in\N$, the number of solutions $n$ to `$m=\s(n)=2\phi(n)$' seems very small. One might contrast this with the equation $m=\phi(n)$, where the number of solutions $c_m$ has been shown to satisfy $c_m>m^\delta$ for infinitely many $m$, and a range of values of $\delta$ \cite{woolridge, pomerance1}. \section{Unsolved problems} \label{sec:problems} \begin{enumerate} \item If $\s(n)=2\phi(n)$ then it seems likely that $d(n)\mid \s(n)$, and we have shown numerically that this is true for all $n$ up to $2.1\times 10^9$. The divisibility $d(n)\mid \s(n)$ follows easily when $n$ is squarefree, and most solutions to $\s(n)=2\phi(n)$ appear to have this shape. \item It seems that the number of solutions to $\s(n)=2\phi(n)$ is infinite. An easier problem would be to find an infinite set of integers $a$ such that $\s(n)=a\cdot\phi(n)$ for at least one $n$. \item No square satisfies $\s(n)=a\cdot\phi(n)$ for any $a\ge 2$. 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Soc.} {\bf 76} (1979), 229--234. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2010 {\it Mathematics Subject Classification}: Primary 11A25; Secondary 11A41, 54H99, 11N60, 11N25, 11N64. \noindent \emph{Keywords: } sum of divisors, Euler's totient function, product compactification. \bigskip \hrule \bigskip \noindent (Concerned with sequences \seqnum{A062699}, \seqnum{A068390}, and \seqnum{A104901}.) \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received June 13 2013; revised versions received August 4 2013; September 19 2013; September 26 2013. Published in {\it Journal of Integer Sequences}, October 13 2013. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document} \end{document}