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\begin{center}
\vskip 1cm{\LARGE\bf
A Diophantine System Concerning \\
\vskip .11in
Sums of Cubes
}
\vskip 1cm
\large
Zhi Ren\\
Mission San Jose High School\\
41717 Palm Avenue\\
Fremont, CA  94539\\
USA\\
\href{renzhistc69@163.com}{\tt renzhistc69@163.com}
\end{center}

\vskip .2 in


\begin{abstract}
We study the Diophantine system
\[\begin{cases}
x_{1}+\cdots+x_{n}=a,\\
x_{1}^3+\cdots+x_{n}^3=b,
\end{cases}
\]
where $a,b \in \mathbb{Q},ab\neq0,n\geq4$, and prove, using the
theory of elliptic curves, that it has
infinitely many rational parametric solutions depending on $n-3$
free parameters. Moreover, this
Diophantine system has infinitely many positive rational solutions
with no common element for $n=4$, which partially answers a question
in our earlier paper.
\end{abstract}

\section{Introduction}

Ren and Yang \cite{ry} considered the positive integer solutions of
the Diophantine chains
\begin{equation}\label{Eq1}
\begin{cases}
\begin{split}
&\sum_{j=1}^nx_{1j}=\sum_{j=1}^nx_{2j}=\cdots=\sum_{j=1}^nx_{kj}=a,\\
&\sum_{j=1}^nx_{1j}^3=\sum_{j=1}^nx_{2j}^3=\cdots=\sum_{j=1}^nx_{kj}^3=b,\\
&n\geq2,~k\geq2,
\end{split}
\end{cases}
\end{equation}
where $a,b$ are positive integers and determined by $k$ $n$-tuples
$(x_{i1},x_{i2},\ldots,x_{in}),i=1,\ldots,k$.

For $n=2,k=2,$ Eq.~(\ref{Eq1}) has no nontrivial integer solutions
\cite{s}, so we consider $n\geq3$. For $n=3,k=2$, Eq.~(\ref{Eq1})
reduces to
\begin{equation}\label{Eq2}
\begin{cases}
x_{1}+x_{2}+x_{3}=y_{1}+y_{2}+y_{3},\\
x_{1}^3+x_{2}^3+x_{3}^3=y_{1}^3+y_{2}^3+y_{3}^3.
\end{cases}
\end{equation}
Systems like \eqref{Eq2} has been investigated by many
authors, at least since 1915
\cite[p.\ 713]{d}; see
\cite{b,bb,bl,c1,c2,l}. Eq.~(\ref{Eq2}) is interesting because
it reveals the relation between all of the nontrivial zeros of
weight-$1$ $6j$ Racah coefficients and all of its non-negative integer
solutions. More recently, Moreland and Zieve \cite{mz} showed
that ``for triples $(a,b,c)$ of pairwise distinct rational numbers
such that for every permutation $(A,B,C)$ of $(a,b,c)$, the
conditions $(A+B)(A-B)^3\neq(B+C)(B-C)^3$ and $AB^2+BC^2+CA^2\neq
A^3+B^3+C^3$ hold, then the Diophantine system
\[\begin{cases}
x+y+z=a+b+c,\\
x^3+y^3+z^3=a^3+b^3+c^3
\end{cases}\]
has infinitely many rational solutions $(x,y,z)$." This gives a
complete answer to Question 5 in an earlier paper of the author
\cite{ry}.

For $n=3,k\geq3$, Choudhry \cite{c2} proved that Eq.~(\ref{Eq1}) has
a parametric solution in rational numbers, but the solutions are not
all positive. There are arbitrarily long Diophantine chains of the
form Eq.~(\ref{Eq1}) with $n=3$.

For $n\geq 3$, Ren and Yang \cite{ry} obtained a special result of
Eq.~(\ref{Eq1}) with $(x_1,x_2,\ldots,x_{n-3})=(1,2,\ldots,n-3)$,
which leads to Eq.~(\ref{Eq1}) has infinitely many coprime positive
integer solutions for $n\geq3$.

Now we study the case of Eq.~(\ref{Eq1}) for $n\geq 4$ with the
greatest possible generality. For convenience, let us consider the
non-zero rational solutions of the Diophantine system
\begin{equation}\label{Eq3}
\begin{cases}
x_{1}+\cdots+x_{n}=a,\\
x_{1}^3+\cdots+x_{n}^3=b,
\end{cases}
\end{equation}
where $a,b \in \mathbb{Q},ab\neq0,n\geq4$.

Using the theory of elliptic curves, we prove the following theorems:

\begin{theorem}\label{Thm1} For $n\geq 4$, the Diophantine
system (\ref{Eq3}) has infinitely many rational parametric solutions
depending on $n-3$ free parameters.
\end{theorem}

\begin{theorem}\label{Thm2} For $n=4$, the Diophantine system (\ref{Eq3}) has infinitely many positive
rational solutions.
\end{theorem}

From these two theorems, we have

\begin{corollary}\label{Cor3} For $n\geq 4$ and every positive integer $k$,
there are infinitely many primitive sets of $k$ $n$-tuples of
polynomials in $\mathbb{Z}[t_1,t_2,\ldots,t_{n-3}]$ with the same
sum and the same sum of cubes.
\end{corollary}

\begin{corollary}\label{Cor4}
For $n=4$ and every positive integer $k$, there are infinitely many
primitive sets of $k$ $4$-tuples of positive integers with the same
sum and the same sum of cubes.
\end{corollary}

\section{The proofs of the theorems}\label{TPT}

In this section, we give the proofs of our theorems, which are
related to the rational points of some elliptic curves. The proof of
Theorem~\ref{Thm1} is inspired by the method of \cite{u}.

\begin{proof} In view of the homogeneity of Eq.~(\ref{Eq3}), we
let $a,b \in \mathbb{Z},ab\neq0$. First, we prove it for $n=4$ and
then deduce the solution of Eq.~(\ref{Eq3}) for all $n\geq 5$. In
the following Diophantine system
\begin{equation}\label{Eq4}
x_{1}+x_2+x_3+x_{4}=a,x_{1}^3+x_2^3+x_3^3+x_{4}^3=b,
\end{equation}
eliminating $x_4$ from the first equation and letting $x_3=tx_2$, we get
\begin{equation}\label{Eq5}
\begin{split}
&3(tx_2+x_2-a)x_1^2+3(tx_2+x_2-a)^2x_1+3t(t+1)x_2^3\\
&-3a(t+1)^2x_2^2+3a^2(t+1)x_2+b-a^3=0.
\end{split}
\end{equation}
To prove Theorem \ref{Thm1} for $n=4$, it is enough to show that the
set of $x_2\in \mathbb{Q}(t)$, such that Eq.~(\ref{Eq5}) has a
solution (with respect to $x_1$), is infinite. Then we need to show
that there are infinitely many $x_2\in \mathbb{Q}(t)$ such that the
discriminant of Eq.~(\ref{Eq5}) is a square, which leads to the
problem of finding
infinitely many rational parametric solutions on the following curve
\[\begin{split}
C:~y^2=&9(t^2-1)^2x_2^4+36at(t+1)x_2^3\\
&-18a^2(t+1)^2x_2^2+12(a^3-b)(t+1)x_2-3a(a^3-4b).
\end{split}\]

The discriminant of $C$ is
\[\begin{split} \Delta(t)=&-5038848(t+1)^4\big((-b+a^3)t^2+(-2b-a^3)t-b+a^3\big)^2\\
&\big((9b^2+a^6-10a^3b)t^4+(-36b^2+14a^3b-2a^6)t^3+(54b^2-24a^3b+3a^6)t^2\\
&+(-36b^2+14a^3b-2a^6)t+9b^2+a^6-10a^3b\big),\end{split}\] and is
non-zero as an element of $\mathbb{Q}(t)$. Then $C$ is smooth.

By \cite[Prop.\ 7.2.1, p.\ 476]{co}, we can transform the curve $C$
into a family of elliptic curves
\[\begin{split}
E:~&Y^2=X^3-18a^2(1+t)^2X^2\\
&+108a(1+t)^2((a^3-4b)t^2+(2a^3+4b)t+a^3-4b)X\\
&-648(1+t)^2((a^6-8ba^3-2b^2)t^4+(-8ba^3+4b^2+4a^6)t^2+a^6-8ba^3-2b^2),
\end{split}\]
by the inverse birational map $\phi:~(x_2,y)\longrightarrow(X,Y)$.
Because the coordinates of this map are quite complicated, we omit
these equations.

An easy calculation shows that the point
\[\begin{split}P=&\bigg(18a^2(t^4+1)/(t-1)^2,36\big((a^3-b)t^6+(2b+a^3)t^5\\
&+(b-a^3)t^4+(4a^3-4b)t^3+(b-a^3)t^2+(2b+a^3)t-b+a^3\big)/(t-1)^3\bigg)\end{split}\]
lies on $E$. To prove that the group $E(\mathbb{Q}(t))$ is infinite,
it is enough to find a point on $E$ with infinite order. By the
group law of the elliptic curves, we can get $[2]P$. Let $[2]P_2$ be
the point of specialization at $t=2$ of $[2]P$. The $X$-coordinate
of $[2]P_2$ is
\[\frac{ 18a^2(-567b^2+2322ba^3+80937a^6)}{(-9b+111a^3)^2}.\]
Let $E_2$ be the specialization of $E$ at $t=2$, i.e.,
\[E_2:~Y^2=X^3-162a^2X^2+972a(9a^3-12b)X-192456a^6+979776ba^3+104976b^2.\]
There are two cases we need to discuss.

1. For $b=37a^3/3$, the curve $E_2$ becomes
\[Y^2=X^3-162a^2X^2-135108a^4X+27859464a^6.\]
Now $[2]P_2$ is the point at infinity on $E_2$, and we need find a point of
infinite order. Let $Y'=Y/a^3,X'=X/a^2$. We have an elliptic curve
\[E'_2:~Y'^2=X'^3-162X'^2-135108X'+27859464.\] It is easy to show that $Q=(234,-432)$ is a point
of infinite order on $E'_2$. Then there are infinitely many rational
points on $E'_2$ and $E$.

2. For $b\neq 37a^3/3$, when the numerator of the $X-$coordinate of
$[2]P_2$ is divided by the denominator with respect to $b$, the
remainder equals
\[r=69984a^5(-3b+43a^3).\]

1. For $a\neq 0$ and $b\neq 43a^3/3$, we see that $r$ is not zero. By the
Nagell-Lutz theorem (\cite[p.\ 56]{st}), $[2]P_2$ is a point of
infinite order on $E_2$.
Thus $P$ is a point of infinite order on $E$.

2. For $a\neq 0$ and $b=43a^3/3$, the curve $E_2$ becomes
\[Y^2=X^3-162a^2X^2-158436a^4X+35417736a^6.\] Let
$Y'=Y/a^3,X'=X/a^2$. We have an elliptic curve
\[E'_2:~Y'^2=X'^3-162X'^2-158436X'+35417736.\] It is easy to show
that $R=(306,-648)$ is a point of infinite order on $E'_2$. Then
there are infinitely many rational points on $E'_2$ and $E$.

In summary, for $a,b \in \mathbb{Z},ab\neq0$, there are infinitely
many rational points on $E$. By the birational map $\phi$, we can
get infinitely many rational solutions of Eqs.~(\ref{Eq5}) and
(\ref{Eq4}). This completes the proof of Theorem~\ref{Thm1} for
$n=4$.

Next, we will deal with Eq.~(\ref{Eq3}) for $n\geq 5$. Let
$x_5',x_6',\ldots,x_n'$ be rational parameters and set
\[a'=\sum_{i=5}^{n}x_i',~b'=\sum_{i=5}^{n}x_i'^3.\]
From the proof of the previous part, we know that Eq.~(\ref{Eq4})
has infinitely many rational solutions
\[(x'_{1j},x'_{2j},x'_{3j},x'_{4j}),j\geq1,\] depending on one
parameter $t$ for $A=a-a'$ and $B=b-b'$. This leads to the
conclusion that for each $j\geq 1$, the $n$-tuple of the following
form
\[x_1=x'_{1j},x_2=x'_{2j},x_3=x'_{3j},x_4=x'_{4j},x_i=x'_i,i\geq 5\]
satisfies Eq.~(\ref{Eq3}). \end{proof}

\begin{example} For $n=4$, from the point $[2]P$, we get
\[\begin{split}
x_1=&-\frac{q(t)}{3a^2t(t+1)(t^2-t+1)(t-1)^2p(t)},\\
x_2=&\frac{ah(t)}{(t+1)(t-1)^2p(t)},\\
x_3=&~tx_2,\\
x_4=&a-x_1-x_2-x_3=\frac{s(t)}{3a^2t(t+1)(t^2-t+1)(t-1)^2p(t)},
\end{split}\]
where $q(t)$ and $s(t)$ have degree 13 as a polynomial of
$\mathbb{Q}(t)$, $h(t)$ has degree 8, and $p(t)$ has degree 6.
\end{example}

From the above example, it seems too difficult to prove that these
rational parametric solutions are positive, so we need a new idea to
prove Theorem~\ref{Thm2}.

\begin{proof} In the proof of Theorem \ref{Thm1}, for $n=4$ we
get the curve
\[\begin{split}
C:~y^2=&9(t^2-1)^2x_2^4+36at(t+1)x_2^3\\
&-18a^2(t+1)^2x_2^2+12(a^3-b)(t+1)x_2-3a(a^3-4b).
\end{split}\]

The discriminant of $C$ is
\[\begin{split} \Delta(t)=&-5038848(t+1)^4\big((-b+a^3)t^2+(-2b-a^3)t-b+a^3\big)^2\\
&\big((9b^2+a^6-10a^3b)t^4+(-36b^2+14a^3b-2a^6)t^3+(54b^2-24a^3b+3a^6)t^2\\
&+(-36b^2+14a^3b-2a^6)t+9b^2+a^6-10a^3b\big).\end{split}\]
Let us
consider $\Delta(t)=0$, so that $C$ has multiple roots. Put
\[(-b+a^3)t^2+(-2b-a^3)t-b+a^3=0,\] and solving for $t$, we
get\[t=\frac{2b+a^3\pm \sqrt{12ba^3-3a^6}}{-b+a^3}.\] In order to
make $t$ be a rational number, take
\[12ba^3-3a^6=c^2,\]
where $c$ is a rational parameter. Then we have
\[b=\frac{3a^6+c^2}{12a^3},~t=\frac{3a^3+c}{3a^3-c},~or~\frac{3a^3-c}{3a^3+c}.\]
According to the symmetry of $t$, consider
\[t=\frac{3a^3+c}{3a^3-c}.\]

Let\[Y_1=Y+\frac{6atX}{t-1}+36(a^3-b)(t-1)(t+1)^2,\]
we get
\[\begin{split}
E':~&Y_1^2=X^3-18a^2(t+1)^2X^2-108a(t+1)^2((a^3-4b)t^2+(4b+2a^3)t+a^3-4b)X\\
&-648(t+1)^2((a^6-8ba^3-2b^2)t^4+(-8ba^3+4b^2+4a^6)t^2+a^6-8ba^3-2b^2).
\end{split}\]
Substituting \[b=\frac{3a^6+c^2}{12a^3},~t=\frac{3a^3+c}{3a^3-c}\]into
$E'$, we get
\[Y_1^2=\frac{((3a^3-c)^2X+72a^2c^2)((3a^3-c)^2X-36a^2(c^2+9a^6))^2}{(3a^3-c)^6}.\]
To get infinitely many solutions of $(Y_1,X)$, put
\[(3a^3-c)^2X+72a^2c^2=d^2,\]
which leads to
\[X=\frac{d^2-72a^2c^2}{(3a^3-c)^2}.\]
Then
\[Y=-\frac{d(d+12ca)(27a^7+9ac^2-dc)}{c(3a^3-c)^3}.\]

Tracing back, we get
\[\begin{split}
x_1=&\frac{(-3a^3+c)d^2+(54a^7+18ac^2)d+108a^2(3a^3+c)(3a^6+c^2)}{72a^3(dc+27a^7+9c^2a)},\\
x_2=&\frac{d(d+12ca)(3a^3-c)}{72a^3(dc+27a^7+9c^2a)},\\
x_3=&\frac{d(d+12ca)(3a^3+c)}{72a^3(dc+27a^7+9c^2a)},\\
x_4=&\frac{(-3a^3-c)d^2+(-54a^7-18ac^2)d+108a^2(3a^3-c)(3a^6+c^2)}{72a^3(dc+27a^7+9c^2a)}.
\end{split}\]
To prove $x_i>0,i=1,2,3,4$, assume that $a>0,c>0,d>0$. Then we have
\[72a^3(dc+27a^7+9c^2a)>0,x_3>0,\] so we just need to consider the
numerators of $x_1,x_2,x_4$. Moreover, set $3a^3-c>0$, we have
$x_2>0$, and the discriminants of
\[(-3a^3+c)d^2+(54a^7+18ac^2)d+108a^2(3a^3+c)(3a^6+c^2)\]and \[(-3a^3-c)d^2+(-54a^7-18ac^2)d+108a^2(3a^3-c)(3a^6+c^2)\]
are $108(-c^2+45a^6)(3a^6+c^2)a^2>0$. We see that the intervals of $d$
such that $x_1>0,x_4>0$ are given by
\[\bigg(\frac{3(9a^6+3c^2-\sqrt{\delta})a}{3a^3-c},\frac{3(9a^6+3c^2+\sqrt{\delta})a}{3a^3-c}\bigg)\]
and\[\bigg(\frac{3(-9a^6-3c^2-\sqrt{\delta})a}{3a^3+c},\frac{3(-9a^6-3c^2+\sqrt{\delta})a}{3a^3+c}\bigg),\]
respectively, where $\delta=405a^{12}+126a^6c^2-3c^4$. It is easy to
show that
\[\frac{3(-9a^6-3c^2+\sqrt{\delta})a}{3a^3+c}>0,\frac{3(9a^6+3c^2-\sqrt{\delta})a}{3a^3-c}<0,\] and
\[\frac{3(9a^6+3c^2+\sqrt{\delta})a}{3a^3-c}>\frac{3(-9a^6-3c^2+\sqrt{\delta})a}{3a^3+c}.\] Hence if
\[d\in \bigg(0,\frac{3(-9a^6-3c^2+\sqrt{\delta})a}{3a^3+c}\bigg),\] we
have $x_1,x_4>0$. This completes the proof of Theorem~\ref{Thm2}.
\end{proof}

\begin{example}\label{Eg6} If we take $a=c=1$, then
$t=2$, $b=1/3$, and
\[
x_1=\frac{-d^2+36d+864}{36(d+36)}, x_2=\frac{d(d+12)}{36(d+36)},
x_3=\frac{d(d+12)}{18(d+36)}, x_4=\frac{-d^2-18d+216}{18(d+36)},\]
where $d\in (0,-9+3\sqrt{33}\approx 8.233687940)$ and $d$ is a
rational number. Taking $d=1,2,3,4,5,6$,\ $7,8$, we get eight
$4$-tuples of positive rational solutions with the same sum $1$ and
the same sums of cubes $1/3$, which are as follows:
\[\begin{split}(x_1,x_2,x_3,x_4)=&\bigg(\frac{899}{1332},\frac{13}{1332},\frac{13}{666},\frac{197}{666}\bigg),\bigg(\frac{233}{342},\frac{7}{342},\frac{7}{171},\frac{44}{171}\bigg),\\
&\bigg(\frac{107}{156},\frac{5}{156},\frac{5}{78},\frac{17}{78}\bigg),\bigg(\frac{31}{45},\frac{2}{45},\frac{4}{45},\frac{8}{45}\bigg),\bigg(\frac{1019}{1476},\frac{85}{1476},\frac{85}{738},\frac{101}{738}\bigg),\\
&\bigg(\frac{29}{42},\frac{1}{14},\frac{1}{7},\frac{2}{21}\bigg),\bigg(\frac{1067}{1548},\frac{133}{1548},\frac{133}{774},\frac{41}{774}\bigg),\bigg(\frac{68}{99},\frac{10}{99},\frac{20}{99},\frac{1}{99}\bigg).\end{split}\]
\end{example}

\section{The proofs of the corollaries}\label{TPC}

In this section, we give the proofs of the corollaries and two
examples.

\begin{proof} Take any $k$ rational parametric
solutions $(x_{i1},\ldots,x_{i,n}),i\leq k$ of Eq.~(\ref{Eq3}),
where $x_{i5}=t_2,\ldots,x_{in}=t_{n-3},i\leq k$ are parameters. Let
$c=\lcm_{i,j}(x_{ij},j=1,\ldots,n,i\leq k)$, and write
\[x_{ij}=\frac{y_{ij}}{c},y_{ij}\in
\mathbb{Z}[t_1,t_2,\ldots,t_{n-3}],\]with $(\gcd_{i,j}(y_{ij},c))=1$
and $c \in \mathbb{Z}[t_1,t_2,\ldots,t_{n-3}]$, where $t_1=t$. Then
\[\sum_{j=1}^{n}y_{ij}=ac,\sum_{j=1}^{n}y_{ij}^3=bc^3.\]
Hence
\[\gcd_{i,j}~(y_{ij})=1.\]
For two sets of solutions $\{(x_{i1},\ldots,x_{in}),i\leq k\}$ and
$\{(x'_{i1},\ldots,x'_{in}),i\leq k\}$, if the sets of $n$-tuples
$\{(y_{i1},\ldots,y_{in}),i\leq k\}$ and
$\{(y'_{i1},\ldots,y'_{in}),i\leq k\}$ coincide, then $d=d'$ and the
$n$-tuples coincide. Since there are infinitely many choices of $k$
elements, for every $k$ there are infinitely many primitive
sets of $k$ $n$-tuples of polynomials with the same sum and the same
sum of cubes. This finishes the proof of Corollary~\ref{Cor3}.
\end{proof}

\begin{example} For $n=4$, we have the rational
parametric solutions
\[\begin{split}
x_1=&-\frac{q(t)}{3a^2t(t+1)(t^2-t+1)(t-1)^2p(t)},\\
x_2=&\frac{ah(t)}{(t+1)(t-1)^2p(t)},\\
x_3=&~tx_2,\\
x_4=&a-x_1-x_2-x_3=\frac{s(t)}{3a^2t(t+1)(t^2-t+1)(t-1)^2p(t)}.
\end{split}\]
Multiply the least common  multiple of the denominator of
$x_i,i=1,\ldots,4$. When $a,b \in \mathbb{Z}$, we get that
\[x_1=-q(t),x_2=3a^3t(t^2-t+1)h(t), x_3=3a^3t^2(t^2-t+1)h(t),x_4=s(t)\]
are the $4$-tuples of polynomials in $\mathbb{Z}[t]$ satisfying Eq.~(\ref{Eq3}).
\end{example}

\begin{proof} The proof of Corollary \ref{Cor4} is similar to the proof
of Corollary \ref{Cor3}, so we omit it.
\end{proof}

\begin{example} From the eight $4$-tuples of positive
rational solutions of Example~\ref{Eg6}, we get the following eight
$4$-tuples of positive integers
\[\begin{split}(y_1,y_2,y_3,y_4)=&(150719584015,2179482305,4358964610,66055079090),\\
&(152140218230,4570736170,9141472340,57460683280),\\
&(153169889565,7157471475,14314942950,48670806030),\\
&(153837920236,9925027112,19850054224,39700108448),\\
&(154170771755,12860172325,25720344650,30561821290),\\
&(154192385490,15950936430,31901872860,21267915240),\\
&(153924475705,19186462295,38372924590,11829247430),\\
&(153386782640,22556879800,45113759600,2255687980)\end{split}\]with
the same sum 223313110020 and the same sum of cubes
3712114854198399246457100577\ 336000. \end{example}

\section{A remaining question}\label{ARQ}

Ren and Yang \cite[Ques.\ 4]{ry} raised the following question:

\begin{question}\label{ques9}
 Are there infinitely many
$n$-tuples of positive integers, having no common element, with the same
sum and the same sum of their cubes for $n\geq4$?
\end{question}

It's easy to calculate that any $4$-tuples $(x_1,x_2,x_3,x_4)$,
given by the same method from Example \ref{Eg6}, have no common
element for $d\in \mathbb{Q}\bigcap(0,-9+3\sqrt{33})$. This gives a
positive answer to Question~\ref{ques9} for $n=4$. When $n\geq 5$,
it seems out of our reach. However, we conjecture that the answer to
Question~\ref{ques9} is yes.

\section{Acknowledgment}

The author would like to thank the referee for his valuable comments
and suggestions.


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\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}: Primary
11D25; Secondary 11D72, 11G05.

\noindent \emph{Keywords: } Diophantine system, $n$-tuple, elliptic
curve.

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received August 4 2013;
revised version received  September 4 2013.
Published in {\it Journal of Integer Sequences}, September 8 2013.
Minor revision, November 1 2013.

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