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\begin{center}
\vskip 1cm{\LARGE\bf 
Partial Sums of Powers of Prime Factors
}
\vskip 1cm
\large
Jean-Marie De Koninck\\
D\'epartement de Math\'ematiques et de Statistique\\
Universit\'e Laval\\
Qu\'ebec G1K 7P4\\
Canada\\
\href{mailto:jmdk@mat.ulaval.ca}{\tt jmdk@mat.ulaval.ca}\\
\ \\
Florian Luca\\
Mathematical Institute, UNAM\\
Ap. Postal 61-3 (Xangari)\\
CP 58 089\\
Morelia, Michoac\'an\\
Mexico\\
\href{mailto:fluca@matmor.unam.mx}{\tt fluca@matmor.unam.mx}
\end{center}

\vskip .2 in
\begin{abstract}
Given integers $k\ge 2$ and $\ell\ge 3$, let $S_{k,\ell}^*$ stand for
the set of those positive integers $n$ which can be written as
$n=p_1^k+p_2^k+\ldots+p_\ell^k$, where $p_1,p_2,\ldots,p_\ell$ are
distinct prime factors of $n$. We study the properties of the sets
$S^*_{k,\ell}$ and we show in particular that, given any odd $\ell\ge
3$, $\displaystyle{\#\bigcup_{k=2}^\infty S_{k,\ell}^*=+\infty}$.
\end{abstract}

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\section{Introduction}

In \cite{kn:approx}, we studied those numbers with at least two distinct prime
factors which can be expressed as the sum of a fixed power $k\ge 2$ of their
prime factors. For instance, given an integer $k\ge 2$, and letting
$$S_k:=\{n: \omega(n)\ge 2 \mbox{ and } n = \sum_{p|n}p^k\},$$
where $\omega(n)$ stands for the number of distinct prime factors of $n$,
one can
check that the following 8 numbers belong to $S_3$:
\begin{eqnarray*}
378 & = & 2 \cdot 3^3 \cdot 7= 2^3+3^3+7^3, \\
2548 & = & 2^2\cdot 7^2\cdot 13= 2^3+7^3+13^3, \\
2\,836\,295 & = & 5\cdot 7 \cdot 11 \cdot 53 \cdot 139= 5^3+7^3+11^3+53^3+139^3,\\
4\,473\,671\,462  &= & 2\cdot 13 \cdot 179 \cdot 593 \cdot 1621= 2^3 + 13^3 + 179^3 + 593^3 + 1621^3 ,\\
23\,040\,925\,705 & = & 5 \cdot 7\cdot 167\cdot 1453 \cdot 2713 =
5^3+ 7^3+ 167^3+ 1453^3 + 2713^3, \\
13\,579\,716\,377\,989 & = & 19 \cdot 157 \cdot 173 \cdot 1103 \cdot 23857= 19^3+ 157^3+ 173^3 + 1103^3 + 23857^3, \\
21\,467\,102\,506\,955 & = & 5 \cdot 7^3 \cdot 313 \cdot 1439 \cdot 27791 = 5^3 + 7^3 + 313^3 + 1439^3 + 27791^3\\
119\,429\,556\,097\,859 & = & 7\cdot 53\cdot 937\cdot 6983\cdot 49199=
7^3+53^3+937^3+6983^3+49199^3.
\end{eqnarray*}
In particular, we showed that 378 and 2548 are the only numbers in $S_3$ with
exactly three distinct prime factors. \vskip 5pt

We did not find any number belonging to $S_k$ for $k=2$ or $k\ge 4$, although
each of these sets may very well be infinite. \vskip 5pt

In this paper, we examine the sets
$$S_k^*:=\{n: \omega(n)\ge 2 \mbox{ and } n={\sum_{p|n}}^*p^k\} \qquad (k=2,3,\ldots) ,$$
where the star next to the sum indicates that it runs over some subset of primes
dividing $n$. For instance, $870\in S_2^*$, because
$$870=2\cdot 3 \cdot 5 \cdot 29=2^2 + 5^2 +29^2.$$
Clearly, for each $k\ge 2$, we have  $S_k^*\supseteq S_k$. Moreover, given
integers $k\ge 2$ and $\ell\ge 3$, let $S_{k,\ell}^*$ stand for the set of those
positive integers $n$ which can be written as $n=p_1^k+p_2^k+\ldots+p_\ell^k$,
where $p_1,p_2,\ldots,p_\ell$ are distinct prime factors of $n$,
so that for each
integer $k\ge 2$,
$$S_k^* = \bigcup_{\ell =3}^\infty S^*_{k,\ell}.$$
We study the properties of the sets $S^*_{k,\ell}$ and we show in particular
that, given any odd $\ell\ge 3$, the set $\displaystyle{\bigcup_{k=2}^\infty
S_{k,\ell}^*}$ is infinite. We treat separately the cases $\ell=3$ and $\ell\ge
5$, the latter case being our main result.


In what follows, the letter $p$, with or without subscripts, always denotes a
prime number. 

\section{Preliminary results}

We shall first consider the set $S_2^*$. Note that if $n\in S_2^*$, then $P(n)$,
the largest prime divisor of $n$, must be part of the partial sum of primes which
allows $n$ to belong to $S_2^*$. Indeed, assume the contrary, namely that, for
some primes $p_1<p_2<\ldots<p_r$,
$$n=p_1^{\alpha_1} \ldots  p_r^{\alpha_r}
=p_{i_1}^2 + \ldots + p_{i_\ell}^2 \quad \in S_2^*,$$ where $i_1<i_2< \ldots <
i_\ell\le  r-1$, with $r \ge 3$. Then
$$p_1\ldots p_{r-2}p_{r-1} p_r \le n <\ell p_{i_\ell}^2 \le r p_{r-1}^2 ,$$
so that $p_1 \ldots p_{r-2}p_r < r p_{r-1}< r p_r$, which implies that $p_1\ldots
p_{r-2} < r$, which is impossible for $r\ge 3$.

While by a parity argument one can easily see that each element of $S_k$ (for any
$k\ge 2$) must have an odd number of prime factors, one can observe that elements
of $S_k^*$ can on the contrary be written as a sum of an even number of prime
powers, as can be seen with $298995972\in S^*_2$ (see below). 

We can show that if Schinzel's Hypothesis is true (see Schinzel \cite{kn:schin}),
then the set $S_3^*$ is infinite. We shall even prove more. 


\begin{thm}
If Schinzel's Hypothesis is true, then $\#S_{3,3}^*=+\infty$.
\end{thm}

\begin{proof}
Assume that $k$ is an even integer such that $r=k^2-9k+21$ and $p=k^2-7k+13$
  are both primes, then $n=2rp(r+k)\in S_{3,3}^*$. Indeed, in this case, one can see that
\begin{equation} \label{eq:1}
n=2rp(r+k) = 2^3 + r^3 + p^3,
\end{equation}
since both sides of (\ref{eq:1}) are equal to
$2k^6-48k^5+492k^4-2752k^3+8844k^2-15456k+11466$. Now Schinzel's Hypothesis
guarantees that there exist infinitely many even $k$'s such that the
corresponding numbers $r$ and $p$ are both primes.
\end{proof}


Note that the first such values of $k$ are $k=2$, 6, 10, 82 and 94. These yield
the following four elements of $S_{3,3}^*$ (observing that $k=2$ and $k=6$
provide the same number, namely $n=378$):
\begin{eqnarray*}
378 & = &  2 \cdot 3^3 \cdot 7= 2^3+3^3+7^3, \\
109306 & = & 2\cdot 31 \cdot 41 \cdot 43 = 2^3+31^3+43^3, \\
450843455098 & = & 2\cdot 6007 \cdot 6089 \cdot 6163 = 2^3 + 6007^3+6163^3,\\
1063669417210 & = & 2\cdot 8011 \cdot 8105 \cdot 8191 = 2^3+8011^3 + 8191^3.
\end{eqnarray*}
Not all elements of $S_3^*$ are generated in this way. For instance, the
following numbers also belong to $S_3^*$:
\begin{eqnarray*}
23391460 & = & 2^2\cdot 5 \cdot 23 \cdot 211 \cdot 241 = 2^3+ 211^3+241^3,\\
173871316 & = & 2^2 \cdot 223 \cdot 421 \cdot 463 = 2^3+ 421^3+463^3, \\
126548475909264420 & = & 2^2 \cdot 3 \cdot 5 \cdot 11 \cdot 83 \cdot 101 \cdot 45569 \cdot 501931\\
& = & 2^3 + 5^3 + 83^3 + 45569^3 + 501931^3,
\end{eqnarray*}
as well as all those elements of $S_3$ mentioned in Section 1. 


\begin{thm}
$\displaystyle{\# \bigcup_{k=2}^\infty S^*_{k,3}=+\infty}$.
\end{thm}

\begin{proof}
This follows immediately from the fact that for each element $n\in S_{k,3}^*$,
one can find a corresponding element $n' \in S^*_{k(2r+1),3}$ for $r=1,2,\ldots$.
Indeed, if $n\in S_{k,3}^*$, then it means that
$$n= p_1^k + p_2^k + p_3^k$$
for some distinct primes divisors $p_1,p_2,p_3$ of $n$. In particular, it means
that $p_a|(p_b^k+p_c^k)$ for each permutation $(a,b,c)$ of the integers 1, 2 and
3. We claim that, given any positive integer $r$, the number
$$n' := p_1^{k(2r+1)} + p_2^{k(2r+1)} + p_3^{k(2r+1)}$$
belongs to $S^*_{k(2r+1),3}$. Indeed, we only need to show that $p_a|
(p_b^{k(2r+1)} + p_c^{k(2r+1)})$ for each permutation $(a,b,c)$ of the integers
1, 2 and 3. But this follows from the fact that $ (p_b^k + p_c^k) $  divides
$(p_b^{k(2r+1)} + p_c^{k(2r+1)})$; but since $p_a$ divides $ (p_b^k + p_c^k) $,
we have that $p_a$ divides $(p_b^{k(2r+1)} + p_c^{k(2r+1)})$ and therefore that
$n' \in S^*_{k(2r+1),3}$. Since $378\in S^*_{3,3}$, the proof is complete.
\end{proof}

\noindent {\bf Remark.} It clearly follows from Theorems 1 and 2 that
$\displaystyle{\#S^*_{3(2r+1),3} = + \infty}$ for any $r\ge 1$.



\section{Proof of the main result}


\begin{thm} \label{th:th5}
Given any odd integer $\ell \ge 5$,
$$\# \bigcup_{k=2}^\infty S^*_{k,\ell}=+\infty.$$
\end{thm}

\noindent  This is an immediate consequence of the following two lemmas.


\begin{lemma}
\label{lem:1} Let $t=2s\ge 2$ be an even integer and $p_1,\dots,p_t$ be primes
such that \begin{itemize}
\item[(i)] $p_i\equiv 3\pmod {4}$ for all $i=1,\dots,t$.
\item[(ii)] $\gcd(p_i,p_j-1)=1$ for all $i,j$ in $\{1,\dots,t\}$.
\item[(iii)] $\gcd(p_i-1,p_j-1)=2$ for all $i\ne j$ in $\{1,\dots,t\}$.
\end{itemize}
Assume furthermore that $a_1,\dots,a_t$ are integers and $n_1,\dots,n_t$ are odd
positive integers such that \begin{itemize}
\item[(iv)] $\gcd(2n_i+1,p_i-1)=1$ for all $i=1,\dots,t$.
\item[(v)] $p_i\mid \sum_{j=1}^t p_j^{n_i}+a_i^{n_i}$ for all $i=1,\dots,t$.
\item[(vi)] $s=t/2$ of the $t$ numbers ${\displaystyle{\left(\frac{a_i}{p_i}\right)}}$ for $i=1,\dots,t$ are equal to $1$ and the other $s$ are equal to $-1$.
\end{itemize}
Then there exist infinitely many primes $p$ such that $S^*_{\frac{p-1}2,t+1}$
contains at least one element.
\end{lemma}

\begin{proof}
Let $a$ be such that
\begin{equation}
\label{eq:cong} a\equiv 2n_i+1\pmod {(p_i-1)/2},\qquad a\equiv 3\pmod 4,\qquad
a\equiv a_i\pmod {p_i}
\end{equation}
for all $i=1,\dots,t$.  The fact that the above integer $a$ exists is a
consequence of the Chinese Remainder Theorem and conditions (i)-(iii) above.
Since $n_i$ is odd, $(p_i-1)/2$ is also odd and $a\equiv 3\pmod 4$, we conclude
that the congruence  $a\equiv 2n_i+1\pmod {(p_i-1)/2}$ implies $a\equiv
2n_i+1\pmod {2(p_i-1)}$.

Now let $\displaystyle{M=4\prod_{i=1}^t \frac{p_i(p_i-1)}2}$. Note that the
number $a$ is coprime to $M$ by conditions (i)-(iv). Thus, by Dirichlet's Theorem
on primes in arithmetic progressions, it follows that there exist infinitely many
prime numbers $p$ such that $p\equiv a \pmod M$. It is clear that these primes
satisfy the same congruences \eqref{eq:cong} as $a$ does. Let $p$ be such a prime
and set
$$
n=\sum_{i=1}^t p_i^{(p-1)/2}+p^{(p-1)/2}.
$$
Note that since $p\equiv 2n_i+1\pmod {2(p_i-1)}$, we get that $(p-1)/2\equiv
n_i\pmod {p_i-1}$. Therefore by Fermat's Little Theorem and condition (v) we get
$$
n\equiv \sum_{j=1}^t p_j^{n_i}+p^{n_i}\equiv \sum_{j=1}^t p_j^{n_i}+
a_i^{n_i}\equiv 0 \pmod {p_i}
$$
for all $i=1,\dots,t$. Finally, conditions (i), (v) and the Quadratic Reciprocity
Law show that from the $t=2s$ numbers
$$
\left(\frac{p_i}{p}\right)=-\left(\frac{p}{p_i}\right)=
-\left(\frac{a_i}{p_i}\right),
$$
exactly half of them are $1$ and the other half are $-1$. Thus, half of the
numbers $p_i^{(p-1)/2}$ are congruent to $1$ modulo $p$ and the other half are
congruent to $-1$ modulo $p$ which shows that $n$ is a multiple of $p$. Hence,
$n$ is a multiple of $p_i$ for $i=1,\dots,t$ and of $p$ as well, which implies
that $n\in S^*_{\frac{p-1}2,t+1}$.
\end{proof}

\begin{lemma}
\label{lem:2} If $s\ge 2$ then there exist primes $p_i$ and integers $a_i,~n_i$
for $i=1,\dots, t$ satisfying the conditions of Lemma \ref{lem:1}.
\end{lemma}

\begin{proof}
Observe that $t-1\ge 3$. Choose primes $p_1,\dots,p_{t-1}$ such that $p_i\equiv
11\pmod {12}$ for all $i=1,\dots,t-1$, $\gcd(p_i,p_j-1)=1$ for all $i,j$ in
$\{1,\dots,t-1\}$, $\gcd(p_i-1,p_j-1)=2$ for all $i\ne j$ in $\{1,\dots,t-1\}$
and finally $p_1+\dots+p_{t-1}$ is coprime to $p_1\dots p_{t-1}$. Note that
$N=p_1+\dots+ p_{t-1}$ is an odd number. Such primes can be easily constructed
starting with say $p_1=11$ and recursively defining $p_2,\dots,p_{t-1}$ as primes
in suitable arithmetic progressions. Take $n_i=1$ for $i=1,\dots,t$. Let
$q_1,\dots,q_\ell$ be all the primes dividing $N$. Pick some integers
$a_1,\dots,a_{t-1}$ such that $s$ of the numbers
${\displaystyle{\left(\frac{-a_i}{p_i}\right)}}$ are $-1$ and the other $s-1$ are
$1$. Now choose a prime $p_t$ such that $p_t\equiv 11\pmod {12}$, $p_t$ is
coprime to $p_i-1$ for $i=1,\dots,t-1$, $p_t\equiv -a_i-N\pmod {p_i}$ for
$i=1,\dots, t-1$, and ${\displaystyle{\left(\frac{q_i}{p_t}\right)=1}}$ for all
$i=1,\dots,\ell$. For these last congruences to be fulfilled, we note that it is
enough to choose $p_t\equiv 1\pmod {q_u}$ if $q_u\equiv 1\pmod 4$ and $p_t\equiv
-1\pmod {q_u}$ if $q_u\equiv 3\pmod 4$, where $u=1,\dots,\ell$. Notice that this
choice is consistent with the fact that $p_t\equiv 11\pmod {12}$ if it happens
that one of the $q_u$ is $3$. So far, the primes $p_1,\ldots, p_t$ satisfy
conditions (i)--(iii) of Lemma \ref{lem:1}. Finally, put $a_t=-N$. Note that
${\displaystyle{\left(\frac{-a_t}{p_t}\right)=\prod_{u=1}^{\ell}
\left(\frac{q_u}{p_t}\right)^{\alpha_u}=1}}$. Here, $\alpha_u$  is the exact
power of $q_u$ in $-a_t$. Hence, exactly $s$ of the numbers
${\displaystyle{\left(\frac{-a_i}{p_i}\right)}}$ are $1$ and the others are $-1$
and since all primes $p_i$ are congruent to $3$ modulo $4$ the same remains true
if one replaces $-a_i$ by $a_i$. Thus, condition (vi) of Lemma \ref{lem:1} holds.
Now one checks immediately that (v) holds with $n_i=1$ for all $i=1,\dots,t$,
because for all $i=1,\dots,t-1$ we have
$$
\sum_{j=1}^t p_j^{n_j}+a_i^{n_i}\equiv N+p_t+a_i\pmod {p_i}\equiv 0 \pmod {p_i},
$$
while
$$
\sum_{j=1}^t p_j^{n_j}+a_t=N+p_t-N\equiv 0\pmod {p_t},
$$
and (iv) is obvious because $2n_i+1=3$ and $p_i-1\equiv 10\pmod {12}$ is not a
multiple of $3$ for $i=1,\dots,t$.
\end{proof}

\medskip

\noindent {\bf Remark.} The above argument does not work for $s=1$. Indeed, in
this case $t-1=1$, therefore $p_1+\dots+p_{t-1}=p_1$ and this is {\bf not}
coprime to $p_1$.




\section{Computational results and further remarks}

To conduct a search for elements of $S_k^*$, one can proceed as follows. If $n\in
S_{k,\ell}^*$, then there exists a positive number $Q$ and primes
$p_1,p_2,\ldots,p_\ell$ such that
$$n=Qp_1 \ldots p_{\ell-1} p_\ell = p_1^k + \ldots + p_{\ell-1}^k + p_{\ell}^k,   $$
so that a necessary condition for $n$ to be in $S_{k,\ell}^*$ is that
$p_\ell|(p_1^k+\ldots + p_{\ell-1}^k)$. (Note that some of the $p_i$'s may also
divide $Q$.)

For instance, in order to find $n\in S_{k,3}^*$, we examine the prime factors $p$
of $r^k+q^k$ as $2\le r<q$ run through the primes up to a given $x$, and we then
check if $\displaystyle{Q:= \frac{r^k+q^k+p^k}{rqp}}$ is an integer. If this is
so, then the integer $n=Qrqp$ belongs to  $S_{k,3}^*$.

Proceeding in this manner (with $\ell=3,4$), we found the following elements of
$S_2^*$:
\begin{eqnarray*}
870 & = & 2\cdot 3 \cdot 5 \cdot 29=2^2 + 5^2 +29^2,\\
188355 & = & 3\cdot 5 \cdot 29 \cdot 433=5^2+29^2+433^2,\\
298995972 & = & 2^2\cdot  3\cdot 11 \cdot 131 \cdot 17291=3^2+11^2+131^2+17291^2,\\
1152597606 & = & 2\cdot  3\cdot 5741 \cdot 33461=2^2+5741^2+33461^2, \\
1879906755 & = & 3\cdot  5\cdot 2897 \cdot 43261=5^2+2897^2+43261^2, \\
5209105541772& = &2^2\cdot 3\cdot 11\cdot 17291\cdot 2282281=
3^2+11^2+17291^2+2282281^2.
\end{eqnarray*}

Although we could not find any elements of $S_4$, we did find some elements of
$S_4^*$, but they are quite large. Here are six of them:
\begin{eqnarray*}
107827277891825604 & = & 2^2\cdot 3 \cdot 7 \cdot 31 \cdot 67 \cdot 18121 \cdot 34105993  =  3^4 + 31^4 + 67^4 + 18121^4,  \\
48698490414981559698 &= & 2\cdot 3^4 \cdot 7 \cdot 13 \cdot 17 \cdot 157\cdot
83537\cdot 14816023  =   2^4 + 17^4 + 83537^4 ,
\end{eqnarray*}
\begin{eqnarray*}
& & 3137163227263018301981160710533087044 \\
& & \qquad = 2^2\cdot 3^2 \cdot 7\cdot 11\cdot 191\cdot 283\cdot 7541\cdot 1330865843\cdot 2086223663996743 \\
&  & \qquad =   3^4 + 7^4 + 191^4 + 1330865843^4 , \\
& & 129500871006614668230506335477000185618 \\
& & \qquad = 2\cdot 3^2 \cdot 7 \cdot 13^2 \cdot 31\cdot 241\cdot 15331\cdot
  21613 \cdot 524149\cdot 1389403 \cdot 3373402577\\
& & \qquad = 2^4 + 241^4 + 3373402577^4 ,\\
& & 225611412654969160905328479254197935523312771590\\
& & \qquad = 2\cdot 3^2 \cdot 5 \cdot 7 \cdot 13^2 \cdot 37\cdot 41\cdot 109 \cdot 113 \cdot 127\cdot 151\cdot 541\cdot 911\cdot 5443\\
& & \qquad \qquad \qquad \qquad \cdot 3198662197\cdot 689192061269\\
& & \qquad = 5^4+7^4+113^4+127^4+911^4+ 689192061269^4 ,\\
& & 17492998726637106830622386354099071096746866616980 \\
& & \qquad = 2^2\cdot 5 \cdot 7 \cdot 23 \cdot 31\cdot 97\cdot 103\cdot 373\cdot 1193\cdot 8689 \cdot 2045107145539 \cdot 2218209705651794191\\
&  & \qquad =   2^4 + 103^4 + 373^4 + 1193^4 + 2045107145539^4 .
\end{eqnarray*}
Note that these numbers provide elements of $S_{4,3}^*$, $S_{4,4}^*$, $S_{4,5}^*$
and $S_{4,6}^*$. 

The smallest elements of $S_2^*$, $S_3^*$, \ldots, $S_{10}^*$ are the following:
\begin{eqnarray*}
870  & = &  2\cdot 3 \cdot 5 \cdot 29=2^2+5^2+29^2 \\
378 & = & 2\cdot 3^3 \cdot 7 = 2^3 + 3^3 + 7^3 \\
107\,827\,277\,891\,825\,604 & = &  2^2\cdot 3 \cdot 7 \cdot 31 \cdot 67 \cdot
18121 \cdot 34105993
      = 3^4 + 31^4 + 67^4 + 18121^4  \\
178\,101 & = & 3^2\cdot 7 \cdot 11 \cdot 257 = 3^5+7^5+11^5 \\
594\,839\,010 & = & 2\cdot 3 \cdot 5 \cdot 17 \cdot 29 \cdot 37\cdot 1087=2^6+5^6+29^6 \\
275\,223\,438\,741 & = & 3 \cdot 23 \cdot 43 \cdot 92761523=3^7 + 23^7 + 43^7\\
26\,584\,448\,904\,822\,018 & = & 2\cdot 3\cdot 7 \cdot 17 \cdot 19\cdot 113
\cdot 912733109
  =  2^8 + 17^8 + 113^8 \\
40\,373\,802  & = &  2\cdot 3^4 \cdot 7 \cdot 35603 = 2^9+3^9+7^9 \\
420\,707\,243\,066\,850  & = & 2 \cdot 3^2 \cdot 5^2 \cdot 29 \cdot 32238102917 = 2^{10} + 5^{10} + 29^{10}. \\
\end{eqnarray*}


Below is a table of the smallest element  $n\in S^*_{k,\ell}$ for
$\ell=3,4,5,6,7$ (with a convenient $k$): 

\noindent
\begin{tabular}{|l|l|}\hline
$\ell$ & $n$   \\ \hline
3 & $378  =  2 \cdot 3^3 \cdot 7 =2^3+3^3+7^3$ \\
4 &  $298995972  =  2^2\cdot  3\cdot 11 \cdot 131 \cdot 17291 =3^2+11^2+131^2+17291^2$\\
5 & $2\,836\,295  =  5\cdot 7 \cdot 11 \cdot 53 \cdot 139 =5^3+7^3+11^3+53^3+139^3$ \\
6 & (a 48 digit number)  $=5^4+7^4+113^4+127^4+911^4+ 689192061269^4$ \\
7 & (a 145 digit number)  $=2^{14} + 3^{14} + 5^{14} + 11^{14} + 29^{14} +
149^{14} + 19809551197^{14}$   \\ \hline
\end{tabular}


Theorem 3 provides a way to construct infinitely many elements of $S^*_{k,\ell}$
given any fixed positive odd integer $\ell$. However, in practice, the elements
obtained are very large. Indeed, take the case $k=5$. With the notation of Lemma
1, we have $t=4$; one can then choose $\{p_1,p_2,p_3,p_4\} = \{11,47,59,227\}$.
As suggested in Lemma 2, let $n_i=1$ for $i=1,2,3,4$. An appropriate set of
integers $a_i$'s is given by $\{a_1,a_2,a_3,a_4\} = \{8,32,10,110\}$, which gives
$\displaystyle{\left\{ \left( \frac{a_i}{p_i} \right): i=1,2,3,4\right\}
=\{-1,1,-1,1\}}$. Looking for a solution $a$ of the set of congruences
$$\left\{ \begin{array}{l}
                 a\equiv 3 \pmod{ \frac{p_i-1}2 } \quad (i=1,2,3,4) \\
                       a \equiv 3 \pmod 4 \\
                       a\equiv a_i \pmod{p_i} \quad (i=1,2,3,4) \\
                   \end{array} \right.,$$
we obtain $a=4\,619\,585\,064\,883$. With $\displaystyle{M=4 \prod_{i=1}^4
\frac{p_i(p_i-1)}2 = 10\,437\,648\,923\,020}$, we notice that indeed
gcd$(a,M)=1$. As the smallest prime number $p\equiv a \pmod M$, we find
$p=10M+a=108\,996\,074\,295\,083$. This means that the smallest integer $n\in
S^*_{k,5}$ constructed with our algorithm is given by
$$n=\sum_{i=1}^4 p_i^{\frac{p-1}2} + p^{\frac{p-1}2},$$
which is quite a large integer since
$$n \approx p^{\frac{p-1}2} \approx (10^{14})^{\frac 12 \cdot 10^{14}}
\approx 10^{7 \cdot 10^{14}}.$$ 


\begin{thebibliography}{9}
\bibitem{kn:approx} J. M. De Koninck and F. Luca, Integers representable as the sum of powers of their prime
factors, {\it Functiones et Approximatio} {\bf 33} (2005), 57--72.

\bibitem{kn:schin} A. Schinzel and W. Sierpi\'nski,
Sur certaines hypoth\`eses concernant les nombres premiers, {\it Acta Arith.}
{\bf 4} (1958), 185--208. Corrigendum: {\bf 5} (1959), 259.
\end{thebibliography}


\bigskip
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\noindent 2000 {\it Mathematics Subject Classification}:
Primary 11A41; Secondary 11A25.

\noindent \emph{Keywords: } prime factorization.

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\vspace*{+.1in}
\noindent
Received January 5 2006;
revised version received December 30 2006.  
Published in {\it Journal of Integer Sequences}, December 30 2006.

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\noindent
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\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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