1 Introduction and preliminaries
Let p and q be two integers such that \(d=p^{2}-4q\neq 0\) (to exclude a degenerate case). We set two integer sequences \(U_{n}\) and \(V_{n}\) by
for \(n\geq 2\) with initial values \(U_{0}=0\), \(U_{1}=1\) and \(V_{0}=2\), \(V_{1}=p\). The characteristic equation of (1) is \(x^{2}-px+q=0\), and hence its roots are
Their Binet formulas are
for \(n\geq 0\).
$$ U_{n}=U_{n}(p,q)=pU_{n-1}-qU_{n-2} \quad \mbox{and} \quad V_{n}=V_{n}(p,q)=pV_{n-1}-qV_{n-2} $$
(1)
$$ \alpha =\frac{p+\sqrt{d}}{2} \quad \mbox{and} \quad \beta =\frac{p-\sqrt{d}}{2}. $$
(2)
$$ U_{n}=\frac{\alpha ^{n}-\beta ^{n}}{\alpha -\beta} \quad \mbox{and} \quad V_{n}=\alpha ^{n}+\beta ^{n} $$
(3)
Note that, in (1), \(U_{n}(1,-1)=F_{n}\), Fibonacci numbers (sequence A000045 in OEIS), \(V_{n}(1,-1)=L_{n}\), Lucas numbers (sequence A000032 in OEIS), \(U_{n}(2,-1) = P_{n}\), Pell numbers (sequence A000129 in OEIS), and \(V_{n}(2,-1)=Q_{n}\), Pell-Lucas numbers (sequence A002203 in OEIS) (for further details, see [1‐6]).
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Balancing numbers have been defined in [7] and [8]. A positive integer n is called a balancing number if the Diophantine equation
holds for some positive integer r, which is called a cobalancing number. From (4) we have \(\frac{(n-1)n}{2}=rn+\frac{r(r+1)}{2}\), and so
$$ 1+2+\cdots +(n-1)=(n+1)+(n+2)+\cdots +(n+r) $$
(4)
$$ r=\frac{-(2n+1)+\sqrt{8n^{2}+1}}{2} \quad \mbox{and} \quad n=\frac{2r+1+\sqrt{8r^{2}+8r+1}}{2}. $$
(5)
Let \(B_{n}\) denote the nth balancing number, and let \(b_{n}\) denote the nth cobalancing number. Then by (5) \(B_{n}\) is a balancing number if and only if \(8B_{n}^{2}+1\) is a perfect square, and \(b_{n}\) is a cobalancing number if and only if \(8b_{n}^{2}+8b_{n}+1\) is a perfect square. So \(C_{n}=\sqrt{8B_{n}^{2}+1}\) and \(c_{n}=\sqrt{8b_{n}^{2}+8b_{n}+1}\) are integers, called the nth Lucas-balancing and nth Lucas-cobalancing number, respectively. The Binet formulas for balancing numbers are \(B_{n}=\frac{\alpha ^{2n}-\beta ^{2n}}{4\sqrt{2}}\), \(b_{n}=\frac{\alpha ^{2n-1}-\beta ^{2n-1}}{4\sqrt{2}}-\frac{1}{2}\), \(C_{n}=\frac{\alpha ^{2n}+\beta ^{2n}}{2}\), and \(c_{n}=\frac{\alpha ^{2n-1}+\beta ^{2n-1}}{2}\), respectively, where \(\alpha =1+\sqrt{2}\) and \(\beta =1-\sqrt{2}\) (for further details, see [9‐12]).
Later balancing numbers were generalized to the t-balancing numbers (see [13]) for an integer \(t\geq 2\). A positive integer n is called a t-balancing number if
for some positive integer r, which is called a t-cobalancing number. From (6) we observe that
$$ 1+2+\cdots +n=(n+1+t)+(n+2+t)+\cdots +(n+r+t) $$
(6)
$$ \begin{gathered}r = \frac{-(2n+2t+1)+\sqrt{8n^{2}+8n(1+t)+(2t+1)^{2}}}{2}, \\ n = \frac{(2r-1)+\sqrt{8r^{2}+8tr+1}}{2}. \end{gathered} $$
(7)
Let \(B_{n}^{t}\) denote the nth t-balancing number, and let \(b_{n}^{t}\) denote the nth t-cobalancing number. Then from (7) we see that \(B_{n}^{t}\) is a t-balancing number if and only if \(8(B_{n}^{t})^{2}+8B_{n}^{t}(1+t)+(2t+1)^{2}\) is a perfect square and that \(b_{n}^{t}\) is a t-cobalancing number if and only if \(8(b_{n}^{t})^{2}+8tb_{n}^{t}+1\) is a perfect square. So
are integers, which are called the nth Lucas t-balancing and the nth Lucas t-cobalancing number (for further details, see [14]).
$$ C_{n}^{t}=\sqrt{8\bigl(B_{n}^{t} \bigr)^{2}+8B_{n}^{t}(1+t)+(2t+1)^{2}} \quad\text{and}\quad c_{n}^{t}=\sqrt{8 \bigl(b_{n}^{t}\bigr)^{2}+8tb_{n}^{t}+1} $$
(8)
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Santana and Diaz-Barrero [15], setting a sequence in Lemma 2 as
proved that it was a generalized Fibonacci sequence given by \(a_{n+1}=6a_{n} -a_{n-1}\) for \(n\geq 1\), with initial values \(a_{0}=1\) and \(a_{1}=7\). Since \(P_{n}=\frac{\alpha ^{n}-\beta ^{n}}{\alpha -\beta} \) for \(\alpha =1+\sqrt{2} \) and \(\beta =1-\sqrt{2}\), we get
for \(n\geq 0\). We can easily see that the sum of the first n nonzero terms of \(a_{n}\) is
$$ a_{n}=P_{2n}+P_{2n+1}, $$
(9)
$$ a_{n}=\frac{\alpha ^{2n+1}+\beta ^{2n+1}}{2} $$
(10)
$$ \sum_{i=1}^{n}a_{i}= \frac{5a_{n}-a_{n-1}-6}{4}. $$
(11)
It is proved in [15, Thm. 1] that the sum of the first \(4n+1\) nonzero terms of Pell numbers is
$$\begin{aligned} \sum_{i=1}^{4n+1}P_{i}= \biggl( \frac{\alpha ^{2n+1}+\beta ^{2n+1}}{2} \biggr) ^{2}. \end{aligned}$$
We conclude that the sum of the first \(4n+1\) nonzero terms of Pell numbers is
$$\begin{aligned} \sum_{i=1}^{4n+1}P_{i}=a_{n}^{2}. \end{aligned}$$
Moreover, Santana and Diaz-Barrero [15] proved that the sum of the first \(4n+1\) nonzero terms of Pell numbers is a perfect square:
$$\begin{aligned} \sum_{i=1}^{4n+1}P_{i}=\left ( \sum_{r=0}^{n} \left ( \begin{matrix} 2n+1 \\ 2r \end{matrix} \right ) 2^{r} \right ) ^{2}. \end{aligned}$$
Later, Tekcan and Tayat [16] proved that the sum of the first \(2n+1\) nonzero terms of Pell numbers is a perfect square if n is even or half of a perfect square if n is odd. They proved that the sum of the first \(2n+1\) nonzero terms of Pell numbers is
where \(\alpha =1+\sqrt{2}\) and \(\beta =1-\sqrt{2}\). Considering (12) and setting two integer sequences
for \(n\geq 0\), they proved that the sum of the first \(4n+1\) nonzero terms of Pell numbers is
for \(n\geq 1\).
$$ \sum_{i=1}^{2n+1}P_{i}= \textstyle\begin{cases} ( \frac{\alpha ^{n+1}+\beta ^{n+1}}{2} ) ^{2} & \mbox{for even }n\geq 2, \\ \frac{ ( \frac{\alpha ^{n+1}-\beta ^{n+1}}{\sqrt{2}} ) ^{2}}{2} & \mbox{for odd }n\geq 1,\end{cases} $$
(12)
$$ X_{n}=\frac{\alpha ^{n+1}+\beta ^{n+1}}{2}\quad \mbox{and}\quad Y_{n}= \frac{\alpha ^{n+1}-\beta ^{n+1}}{\sqrt{2}} $$
(13)
$$\begin{aligned} \sum_{i=1}^{4n+1}P_{i}= \bigl[2X_{n}^{2}-2X_{n}Y_{n-1}+(-1)^{n+1} \bigr]^{2} \end{aligned}$$
In this paper, we give several results on the sequences \(a_{n}\), \(X_{n}\), \(Y_{n}\), \(B_{n}\), \(b_{n}\), \(Q_{n}\), including sums, divisibility properties, perfect squares, and integer solutions of some specific Pell equations.
2 Results and discussion
In this section, we derive our main results.
2.1 Sums and divisibility properties
In this subsection, we deal with the sums and divisibility properties of numbers mentioned. First, we reformulate (11) in terms of Pell and Pell-Lucas numbers as follows.
Theorem 1
The sum of the first
n
nonzero terms of
\(a_{n}\)
is
$$\begin{aligned} \sum_{i=1}^{n}a_{i}= \textstyle\begin{cases} \frac{17P_{2n-1}+7P_{2n-2}-3}{2}, \\ \frac{5Q_{2n}+2Q_{2n-1}-6}{4}. \end{cases}\displaystyle \end{aligned}$$
Proof
Since \(a_{1}+a_{2}+\cdots +a_{n}=\frac{5a_{n}-a_{n-1}-6}{4}\) by (11) and since \(a_{n}=P_{2n}+P_{2n+1}\) by (9), we deduce that
The second result can be proved similarly. □
$$\begin{aligned} \sum_{i=1}^{n}a_{i} =& \frac{5a_{n}-a_{n-1}-6}{4} \\ =&\frac{5(P_{2n}+P_{2n+1})-(P_{2n-2}+P_{2n-1})-6}{4} \\ =&\frac{5P_{2n}+5(2P_{2n}+P_{2n-1})-P_{2n-2}-P_{2n-1}-6}{4} \\ =&\frac{15(2P_{2n-1}+P_{2n-2})+4P_{2n-1}-P_{2n-2}-6}{4} \\ =&\frac{17P_{2n-1}+7P_{2n-2}-3}{2}. \end{aligned}$$
Theorem 2
For the sequences
\(a_{n}\), \(X_{n}\), \(Y_{n}\), \(B_{n}\), \(b_{n}\), \(Q_{n}\), and
\(P_{n}\), we have:
(1)
\(X_{n}=P_{n+1}+P_{n}\)
for
\(n\geq 0\), and the sum of the first
n
nonzero terms of
\(X_{n}\)
is
Moreover, \(Y_{n}=P_{n+2}-P_{n}\)
for
\(n\geq 0\), and the sum of the first
n
nonzero terms of
\(Y_{n}\)
is
$$\begin{aligned} \sum_{i=1}^{n}X_{i}=P_{n}+2P_{n+1}-2. \end{aligned}$$
$$\begin{aligned} \sum_{i=1}^{n}Y_{i}=P_{n}+3P_{n+1}-3. \end{aligned}$$
(2)
\(a_{n}=X_{2n}\)
for
\(n\geq 0\)
or
\(a_{n}=X_{2n-1}+Y_{2n-1}\)
for
\(n\geq 1\); \(Q_{n+1}=2X_{n}\)
and
\(Q_{n+2}-Q_{n+1}=2Y_{n}\)
for
\(n\geq 0\); \(a_{n}=\frac{Q_{2n+1}}{2}\)
and
\(B_{n}=\frac{P_{n}Q_{n}}{2}\)
for
\(n\geq 0\), and
\(b_{n}=\frac{P_{n}Q_{n-1}+P_{n-1}Q_{n}-2}{4}\)
for
\(n\geq 1\).
Proof
(1) Let
for some \(T_{1}\) and \(T_{2}\). If we take \(n=0\) and \(n=1\), then we have the system of equations \(T_{1}+T_{2}=1\) and \(T_{1}\alpha +T_{2}\beta =3 \). This system of equations has the solution \(T_{1}=\frac{\alpha +1}{\alpha -\beta} \) and \(T_{2}=\frac{-\beta -1}{\alpha -\beta} \). So (14) becomes
as we wanted. Since \(X_{n}=P_{n}+P_{n+1}\) and \(P_{1}+P_{2}+\cdots +P_{n}=\frac{P_{n}+P_{n+1}-1}{2}\), we easily deduce that
Similarly, it can be shown that \(Y_{n}=P_{n+2}-P_{n}\) for \(n\geq 0\) and \(Y_{1}+Y_{2}+\cdots +Y_{n}=P_{n}+3P_{n+1}-3\).
$$ X_{n}=T_{1}\alpha ^{n}+T_{2}\beta ^{n} $$
(14)
$$\begin{aligned} X_{n}& =T_{1}\alpha ^{n}+T_{2}\beta ^{n} \\ & = \biggl( \frac{\alpha +1}{\alpha -\beta} \biggr) \alpha ^{n}+ \biggl( \frac{-\beta -1}{\alpha -\beta} \biggr) \beta ^{n} \\ & =\frac{(\alpha +1)\alpha ^{n}-(\beta +1)\beta ^{n}}{\alpha -\beta} \\ & =\frac{\alpha ^{n+1}-\beta ^{n+1}}{\alpha -\beta} +\frac{\alpha ^{n}-\beta ^{n}}{\alpha -\beta} \\ & =P_{n+1}+P_{n}, \end{aligned}$$
$$\begin{aligned} \sum_{i=1}^{n}X_{i}=\sum _{i=1}^{n}(P_{i}+P_{i+1})=P_{n}+2P_{n+1}-2. \end{aligned}$$
(2) Since \(a_{n}=\frac{\alpha ^{2n+1}+\beta ^{2n+1}}{2}\) and \(X_{n}=\frac{\alpha ^{n+1}+\beta ^{n+1}}{2}\), we get
Similarly, since \(X_{n}=P_{n+1}+P_{n}\) and \(Y_{n}=P_{n+2}-P_{n}\), we get \(X_{n}+Y_{n}=P_{n+1}+P_{n+2}\), and hence
The remaining cases can be proved similarly. □
$$\begin{aligned} X_{2n}=\frac{\alpha ^{2n+1}+\beta ^{2n+1}}{2}=a_{n}. \end{aligned}$$
$$\begin{aligned} X_{2n-1}+Y_{2n-1}=P_{2n}+P_{2n+1}=a_{n}. \end{aligned}$$
Theorem 3
Let
\(P_{n}\)
denote the
nth Pell number.
(1)
If
\(n\geq 2\)
is even, then
and if
\(n\geq 1\)
is odd, then
$$\begin{aligned} 1+\sum_{i=1}^{2n-1}{P_{i}} =& \textstyle\begin{cases} (\frac{Q_{n}}{2})^{2}, \\ (2B_{\frac{n}{2}}+2b_{\frac{n}{2}}+1)^{2},\end{cases}\displaystyle \end{aligned}$$
$$\begin{aligned} \sum_{i=1}^{2n-1}{P_{i}} =& \textstyle\begin{cases} (\frac{Q_{n}}{2})^{2}, \\ (b_{\frac{n+3}{2}}-B_{\frac{n+1}{2}}-B_{\frac{n-1}{2}}-b_{\frac{n-1}{2}})^{2}.\end{cases}\displaystyle \end{aligned}$$
(2)
If
\(n\geq 2\)
is even, then
and if
\(n\geq 1\)
is odd, then
$$\begin{aligned} \sum_{i=1}^{4n+1}{P_{i}}= \bigl(2P_{n+1}^{2}-2P_{n}^{2}-1 \bigr)^{2}, \end{aligned}$$
$$\begin{aligned} \sum_{i=1}^{4n+1}{P_{i}}= \bigl(2P_{n+1}^{2}-2P_{n}^{2}+1 \bigr)^{2}. \end{aligned}$$
Proof
(1) Let n be even, say \(n=2k\) for some positive integer k. Since \(P_{1}+P_{2}+\cdots +P_{n}=\frac{P_{n}+P_{n+1}-1}{2}\), we easily get
The other cases can be proved similarly. □
$$\begin{aligned} 1+\sum_{i=1}^{2n-1}{P_{i}} =&1+ \frac{P_{2n-1}+P_{2n}-1}{2} \\ =&\frac{P_{2n-1}+P_{2n}+1}{2} \\ =&\frac{\frac{\alpha ^{4k-1}-\beta ^{4k-1}+\alpha ^{4k}-\beta ^{4k}}{2\sqrt{2}}+1}{2} \\ =&\frac{\alpha ^{4k}(\alpha ^{-1}+1)+\beta ^{4k}(-\beta ^{-1}-1)+2\sqrt{2}}{4\sqrt{2}} \\ =&\frac{\alpha ^{4k}+\beta ^{4k}+2(\alpha \beta )^{2k}}{4} \\ =& \biggl( \frac{\alpha ^{2k}+\beta ^{2k}}{2} \biggr) ^{2} \\ =&\biggl(\frac{Q_{n}}{2}\biggr)^{2}. \end{aligned}$$
Theorem 4
For the sequences mentioned, we have
$$\begin{aligned}& \sum_{i=1}^{2n+1}{a_{i}}={a_{n}} {a_{n+1}},\qquad \sum_{i=1}^{2n}{a_{i}}=4B_{n+1}P_{2n} , \\& \sum_{i=1}^{2n+1}{B_{i}}={a_{n}} {B_{n+1}},\qquad \sum_{i=0}^{2n}{Y_{2i+1}}=2a_{n}P_{2n+2}, \\& \sum_{i=1}^{2n+1}{Q_{i}}={2(P_{2n+2}-1)}, \qquad \sum_{i=1}^{2n+1}{P_{i}}= \textstyle\begin{cases} (a_{\frac{n}{2}})^{2} , & n \textit{ even}, \\ Y_{n}P_{n+1}, & n \textit{ odd}, \end{cases}\displaystyle \\& \sum_{i=1}^{2n}{P_{2i}}={a_{n}} {P_{2n}}, \qquad \sum_{i=1}^{2n}{P_{i}}= \textstyle\begin{cases} 2P_{n}P_{n+1} , & n \textit{ even}, \\ a_{\frac{n-1}{2}}X_{n}, & n \textit{ odd}, \end{cases}\displaystyle \\& \sum_{i=1}^{2n}{X_{2i}}={8X_{n}} {P_{n+1}} {B_{n}}, \qquad \sum_{i=1}^{2n}{B_{i}}=a_{n}B_{n}, \\& \sum_{i=1}^{2n}{Q_{i}}={4b_{n+1}}, \qquad \sum_{i=1}^{2n}{B_{2i}}=P_{2n}P_{2n+1}X_{2n}C_{n}, \\& \sum_{i=1}^{2n}{Y_{2i-1}}={2a_{n}} {P_{2n}}, \qquad \sum_{i=1}^{2n}{a_{2i-1}}=a_{2n}P_{2n}X_{2n-1}, \\& \sum_{i=0}^{2n}{P_{2i+1}}={a_{n}} {P_{2n+1}}, \qquad \sum_{i=1}^{2n}{X_{2i-1}}=P_{2n+1}Y_{2n-1}, \\& \sum_{i=1}^{2n}{P_{2i-1}}={X_{2n-1}} {P_{2n}}, \qquad \sum_{i=0}^{2n}{a_{2i+1}}=a_{n}P_{2n+1}X_{4n+2}, \\& \sum_{i=1}^{2n}{a_{2i}}={B_{2n}} {a_{2n+1}}, \qquad \sum_{i=1}^{2n}{Q_{2i}}=2P_{2n+1}Y_{2n-1}. \end{aligned}$$
Proof
Since \(\sum_{i=1}^{2n+1}{a_{i}}=\frac{ 29a_{2n}-5a_{2n-1} - 6}{ 4} \) by (11), we get
The other cases can be proved similarly. □
$$\begin{aligned} \sum_{i=1}^{2n+1}a_{i} =& \frac{29 ( \frac{\alpha ^{4n+1}+\beta ^{4n+1}}{2 } ) -5 ( \frac{\alpha ^{4n-1}+\beta ^{4n-1}}{2} ) -6}{4} \\ =&\frac{\alpha ^{4n}(29\alpha -5\alpha ^{-1})+\beta ^{4n}(29\beta -5\beta ^{-1})-12}{8} \\ =&\frac{\alpha ^{4n}(17+12\sqrt{2})+\beta ^{4n}(17-12\sqrt{2})-6}{4} \\ =&\frac{\alpha ^{4n+4}+\beta ^{4n+4}-6}{4} \\ =&\frac{\alpha ^{4n+4}+\beta ^{4n+4}-(\alpha \beta )^{2n+1}(\beta ^{2}+\alpha ^{2})}{4} \\ =& \biggl( \frac{\alpha ^{2n+1}+\beta ^{2n+1}}{2} \biggr) \biggl( \frac{\alpha ^{2n+3}+\beta ^{2n+3}}{2} \biggr) \\ =&a_{n}a_{n+1}. \end{aligned}$$
From Theorem 4 we have the following result.
Theorem 5
For the divisibility properties, we have
(1)
If
n
is even, then
\(a_{\frac{n}{2}}\vert \sum_{i=1}^{2n+1}{P_{i}}\), \(P_{n}\vert \sum_{i=1}^{2n}{P_{i}}\), and
\(P_{n+1}\vert \sum_{i=1}^{2n}{P_{i}} \), and if
n
is odd, then
\(Y_{n}\vert \sum_{i=1}^{2n+1}{P_{i}} \), \(P_{n+1}\vert \sum_{i=1}^{2n+1}{P_{i}} \), \(a_{\frac{n-1}{2}}\vert \sum_{i=1}^{2n}{P_{i}} \), and
\(X_{n}\vert \sum_{i=1}^{2n}{P_{i}} \).
(2)
For every
\(n\geq 1\),
$$\begin{aligned}& a_{n}\Bigg\vert \sum_{i=0}^{2n}{a_{2i+1}}, \qquad P_{2n+1}\Bigg\vert \sum_{i=0}^{2n}{a_{2i+1}}, \qquad X_{4n+2}\Bigg\vert \sum_{i=0}^{2n}{a_{2i+1}}, \qquad a_{2n}\Bigg\vert \sum_{i=1}^{2n}{a_{2i-1}}, \\& a_{n}\Bigg\vert \sum_{i=1}^{2n+1}{a_{i}}, \qquad a_{n+1}\Bigg\vert \sum_{i=1}^{2n+1}{a_{i}}, \qquad a_{n}\Bigg\vert \sum_{i=1}^{2n+1}{B_{i}}, \qquad B_{n+1}\Bigg\vert \sum_{i=1}^{2n+1}{B_{i}}, \\& a_{n}\Bigg\vert \sum_{i=1}^{2n}{B_{i}}, \qquad B_{n+1}\Bigg\vert \sum_{i=1}^{2n}{a_{i}}, \qquad P_{2n}\Bigg\vert \sum_{i=1}^{2n}{a_{i}}, \qquad B_{n}\Bigg\vert \sum_{i=1}^{2n}{B_{i}}, \\& a_{2n+1}\Bigg\vert \sum_{i=1}^{2n}{a_{2i}}, \qquad b_{n+1}\Bigg\vert \sum_{i=1}^{2n}{Q_{i}}, \qquad a_{n}\Bigg\vert \sum_{i=0}^{2n}{X_{2i+1}}, \qquad P_{2n}\Bigg\vert \sum_{i=1}^{2n}{a_{2i-1}}, \\& X_{2n-1}\Bigg\vert \sum_{i=1}^{2n}{a_{2i-1}}, \qquad a_{n}\Bigg\vert \sum_{i=1}^{2n}{P_{2i}}, \qquad P_{2n}\Bigg\vert \sum_{i=1}^{2n}{P_{2i}}, \qquad B_{2n}\Bigg\vert \sum_{i=1}^{2n}{a_{2i}}, \\& P_{2n}\Bigg\vert \sum_{i=1}^{2n}{Q_{2i-1}}, \qquad a_{n}\Bigg\vert \sum_{i=0}^{2n}{Q_{2i+1}}, \qquad P_{2n+1}\Bigg\vert \sum_{i=1}^{2n}{X_{2i-1}}, \qquad P_{n}\Bigg\vert \sum_{i=1}^{2n}{B_{2i-1}}, \\& Y_{2n-1}\Bigg\vert \sum_{i=1}^{2n}{X_{2i-1}}, \qquad a_{n}\Bigg\vert \sum_{i=0}^{2n}{Y_{2i+1}}, \qquad P_{2n+2}\Bigg\vert \sum_{i=0}^{2n}{Y_{2i+1}}, \qquad a_{n}\Bigg\vert \sum_{i=1}^{2n}{Y_{2i-1}}, \\& P_{2n}\Bigg\vert \sum_{i=1}^{2n}{Y_{2i-1}}, \qquad P_{2n+1}\Bigg\vert \sum_{i=0}^{2n}{B_{2i+1}}, \qquad a_{n}\Bigg\vert \sum_{i=0}^{2n}{B_{2i+1}}, \qquad X_{n-1}\Bigg\vert \sum_{i=1}^{2n}{B_{2i-1}}, \\& P_{2n+1}\Bigg\vert \sum_{i=1}^{2n}{B_{2i}}, \qquad P_{2n+1}\Bigg\vert \sum_{i=1}^{2n}{Q_{2i}}, \qquad Y_{2n-1}\Bigg\vert \sum_{i=1}^{2n}{Q_{2i}}, \qquad P_{2n}\Bigg\vert \sum_{i=1}^{2n}{B_{2i}}, \\& X_{n}\Bigg\vert \sum_{i=1}^{2n}{X_{2i}}, \qquad P_{n+1}\Bigg\vert \sum_{i=1}^{2n}{X_{2i}}, \qquad B_{n}\Bigg\vert \sum_{i=1}^{2n}{X_{2i}}, \qquad X_{2n}\Bigg\vert \sum_{i=1}^{2n}{B_{2i}}. \end{aligned}$$
Finally, we give the following result.
Theorem 6
For the sequences
\(a_{n}\), \(X_{n}\), \(Y_{n}\), \(B_{n}\), \(Q_{n}\), and
\(P_{n}\), we have
$$\begin{aligned}& \frac{ \sum_{i=0}^{2n}{a_{2i+1}} }{ \sum_{i=0}^{2n}{P_{2i+1}} }=X_{4n+2}, \qquad \frac{ \sum_{i=1}^{2n}{a_{2i-1}} }{ \sum_{i=1}^{2n}{P_{2i-1}} }=a_{2n}, \qquad \frac{ \sum_{i=0}^{2n}{B_{2i+1}} }{ \sum_{i=0}^{2n}{Q_{2i+1}} }=\frac{P_{2n+1}^{2}}{2}, \\& \frac{ \sum_{i=1}^{2n}{Q_{2i}} }{ \sum_{i=1}^{2n}{X_{2i-1}} }=2, \qquad \frac{ \sum_{i=1}^{2n}{Y_{2i-1}}}{ \sum_{i=1}^{2n}{P_{2i}} }=2. \end{aligned}$$
2.2 Perfect squares
We see in (5) that \(B_{n}\) is a balancing number if and only if \(8B_{n}^{2}+1\) is a perfect square and that \(b_{n}\) is a cobalancing number if and only if \(8b_{n}^{2}+8b_{n}+1\) is a perfect square. Similarly, we can give the following result.
Theorem 7
For every
\(n\geq 1\),
(1)
\(\frac{Q_{4n+2}-2}{4}\)
is a perfect square and
\(\sqrt{\frac{ Q_{4n+2}-2}{4}}=a_{n}\);
(2)
\(2P_{2n-1}^{2}-1\)
is a perfect square, and
\(\sqrt{2P_{2n-1}^{2}-1} =a_{n-1}\);
(3)
\(2P_{2n}^{2}+1\)
is a perfect square, and
\(\sqrt{2P_{2n}^{2}+1} =C_{n}\);
(4)
\(P_{2n+1}^{2}+P_{2n}P_{2n+2}\)
is a perfect square, and
\(\sqrt{ P_{2n+1}^{2}+P_{2n}P_{2n+2}}=X_{2n}\);
(5)
\(P_{2n}^{2}+P_{2n-1}^{2}+P_{4n}-1\)
is a perfect square, and
\(\sqrt{ P_{2n}^{2}+P_{2n-1}^{2}+P_{4n}-1}=Y_{2n-1}\).
Proof
(1) Applying the Binet formulas, we deduce that
as claimed. The other cases can be proved similarly. □
$$\begin{aligned} \sqrt{\frac{Q_{4n+2}-2}{4}} =&\sqrt{\frac{\alpha ^{4n+2}+\beta ^{4n+2}-2}{4}} =\sqrt{ \frac{\alpha ^{4n+2}+\beta ^{4n+2}+2(\alpha \beta )^{2n+1}}{4}} \\ =&\sqrt{ \biggl( \frac{\alpha ^{2n+1}+\beta ^{2n+1}}{2} \biggr) ^{2}}= \frac{ \alpha ^{2n+1}+\beta ^{2n+1}}{2}=a_{n}, \end{aligned}$$
As in Theorem 7, we can give the following result.
Theorem 8
For the sequences
\(a_{n}\), \(X_{n}\), \(B_{n}\), \(b_{n}\), \(Q_{n}\), and
\(P_{n}\), we have
$$\begin{aligned}& \sqrt{\sum_{i=0}^{2n}{B_{2i+1}}}=a_{n}P_{2n+1}, \qquad \sqrt{\sum_{i=1}^{2n}{B_{2i-1}}}=2C_{n}X_{n-1}P_{n}, \\& \sqrt{\frac{ \sum_{i=0}^{2n}{Q_{2i+1}} }{2}}=a_{n},\qquad \sqrt{\sum _{i=1}^{2n}{Q_{2i-1}}}=2P_{2n}. \end{aligned}$$
Proof
Since \(B_{n}=\frac{\alpha ^{2n}-\beta ^{2n}}{4\sqrt{2}}\), \(P_{n}=\frac{\alpha ^{n}-\beta ^{n}}{2\sqrt{2}}\), and \(a_{n}=\frac{\alpha ^{2n+1}+\beta^{2n+1}}{2}\), we get
The other cases can be proved similarly. □
$$\begin{aligned} \sqrt{\sum_{i=0}^{2n}{B_{2i+1}}} =&\sqrt{B_{1}+B_{3}+\cdots +B_{4n+1}} \\ =&\sqrt{\frac{\alpha ^{2}-\beta ^{2}}{4\sqrt{2}}+\frac{\alpha ^{6}-\beta ^{6} }{4\sqrt{2}}+\cdots +\frac{\alpha ^{8n+2}-\beta ^{8n+2}}{4\sqrt{2}}} \\ =&\sqrt{\frac{1}{4\sqrt{2}} \biggl( \frac{\alpha ^{8n+6}-\alpha ^{2}}{\alpha ^{4}-1}-\frac{\beta ^{8n+6}-\beta ^{2}}{\beta ^{4}-1} \biggr) } \\ =&\sqrt{\frac{\alpha ^{8n+4}-1}{32}+\frac{\beta ^{8n+4}-1}{32}} \\ =&\sqrt{ \biggl( \frac{\alpha ^{4n+2}+\beta ^{4n+2}-2(\alpha \beta )^{2n+1}}{8} \biggr) \biggl( \frac{\alpha ^{4n+2}+\beta ^{4n+2}+2(\alpha \beta )^{2n+1}}{4} \biggr) } \\ =&\sqrt{ \biggl( \frac{\alpha ^{2n+1}+\beta ^{2n+1}}{2} \biggr) ^{2} \biggl( \frac{\alpha ^{2n+1}-\beta ^{2n+1}}{2\sqrt{2}} \biggr) ^{2}} \\ =&a_{n}P_{2n+1}. \end{aligned}$$
2.3 Continued fraction expansion
Theorem 9
The continued fraction expansion of
\(\frac{a_{n+1}}{a_{n}}\)
is
for
\(n\geq 1\) (here
\((x)_{k}\)
means that there are
k
successive terms ‘x’).
$$\begin{aligned} \frac{a_{n+1}}{a_{n}}=\bigl[5;(1,4)_{n-1},1,6\bigr] \end{aligned}$$
Proof
Let \(n=1\). Then
Let us assume that it is satisfied for \(n-1\), that is, \(\frac{a_{n}}{a_{n-1}} =[5;(1,4)_{n-2},1,6]\). Then we get
So it is true for all \(n\geq 1\) since \(a_{n+1}=6a_{n}-a_{n-1}\). □
$$\begin{aligned} \frac{a_{2}}{a_{1}}=\frac{41}{7}=5+\frac{1}{1+\frac{1}{6}}=[5;1,6]. \end{aligned}$$
$$\begin{aligned} \frac{a_{n+1}}{a_{n}} =&\bigl[5;(1,4)_{n-1},1,6\bigr]=5+ \frac{1}{1+\frac{1}{4+\frac{1}{ \textstyle\begin{array}{c} 1+ \\ \cdots \\ 1+\frac{1}{6} \end{array}\displaystyle }}} \\ =&5+\frac{1}{1+\frac{1}{-1+5+\frac{1}{ \textstyle\begin{array}{c} 1+ \\ \cdots \\ 1+\frac{1}{6} \end{array}\displaystyle }}} =5+\frac{1}{1+\frac{1}{-1+\frac{a_{n}}{a_{n-1}}}} \\ =&5+\frac{1}{1+\frac{a_{n-1}}{a_{n}-a_{n-1}}}=5+\frac{a_{n}-a_{n-1}}{a_{n}}= \frac{6a_{n}-a_{n-1}}{a_{n}}. \end{aligned}$$
2.4 Companion matrix
The companion matrix for Pell numbers is \(\big[ {\scriptsize\begin{matrix}{} 2 & 1 \cr 1 & 0\end{matrix}} \big] \). It is known that
So \(P_{n+1}P_{n-1}-P_{n}^{2}=(-1)^{n}\), which known as the Cassini identity, is an immediate consequence of the matrix formula [17]. If we take the nth power of the matrix in the left side of (15), then we can give the following theorem, which can be proved by induction on n.
$$ \left [ \begin{matrix} P_{n+1} & P_{n} \\ P_{n} & P_{n-1}\end{matrix} \right ] =\left [ \begin{matrix} 2 &1 \\ 1 &0\end{matrix} \right ] ^{n}. $$
(15)
Theorem 10
For the Pell numbers
\(P_{n}\), we have
for
\(n\geq 1\).
$$\begin{aligned} \left [ \begin{matrix} P_{n+1} & P_{n} \\ P_{n} & P_{n-1} \end{matrix} \right ] ^{n}= \left [ \begin{matrix} P_{n^{2}+1} & P_{n^{2}} \\ P_{n^{2}} & P_{n^{2}-1} \end{matrix} \right ] \end{aligned}$$
2.5 Pythagorean triples
It is known that the Pell numbers \(P_{n}\) have a close connection with square triangular numbers, that is,
Note that the left side of (16) describes a square number, whereas the right side describes a triangular number, so the result is a square triangular number (see [18]). Notice that if a right triangle has integer side lengths a, b, c (necessarily satisfying the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\)), then \((a,b,c)\) is known as a Pythagorean triple. As Martin [19] described, Pell numbers can be used to form Pythagorean triples in which a and b are one unit apart, corresponding to right triangles that are nearly isosceles. For instance,
is a Pythagorean triple. Now we can give the following theorem related to Pythagorean triples.
$$ \bigl( (P_{k-1}+P_{k})P_{k} \bigr) ^{2}=\frac{(P_{k-1}+P_{k})^{2} ( (P_{k-1}+P_{k})^{2}-(-1)^{k} ) }{2}. $$
(16)
$$\begin{aligned} \bigl(2P_{n}P_{n+1},P_{n+1}^{2}-P_{n}^{2},P_{n+1}^{2}+P_{n}^{2} \bigr) \end{aligned}$$
Theorem 11
\(\sqrt{2}B_{2n},P_{2n-1}+P_{2n}\), and
\(B_{2n}+b_{2n}+1\)
form a Pythagorean triple, that is,
and
\((P_{n}Q_{n}+2b_{n}+1)^{2}-8B_{n}^{2},\sqrt{2}P_{2n}\), and
\(X_{2n-1}\)
form a Pythagorean triple, that is,
$$\begin{aligned} (\sqrt{2}B_{2n})^{2}+(P_{2n-1}+P_{2n})^{2}=(B_{2n}+b_{2n}+1)^{2}, \end{aligned}$$
$$\begin{aligned} {}\bigl[ (P_{n}Q_{n}+2b_{n}+1)^{2}-8B_{n}^{2} \bigr]^{2}+(\sqrt{2} P_{2n})^{2}=X_{2n-1}^{2}. \end{aligned}$$
Proof
Applying the Binet formulas, we deduce that
The other case can be proved similarly. □
$$\begin{aligned}& (\sqrt{2}B_{2n})^{2}+(P_{2n-1}+P_{2n})^{2} \\& \quad = \biggl[ \sqrt{2} \biggl( \frac{\alpha ^{4n}-\beta ^{4n}}{4\sqrt{2}} \biggr) \biggr] ^{2}+ \biggl( \frac{\alpha ^{2n-1}-\beta ^{2n-1}}{2\sqrt{2}}+\frac{ \alpha ^{2n}-\beta ^{2n}}{2\sqrt{2}} \biggr) ^{2} \\& \quad = \frac{\alpha ^{8n}-2(\alpha \beta )^{2n}+\beta ^{8n}}{16}+\frac{2\alpha ^{4n}+4(\alpha \beta )^{2n}+2\beta ^{4n}}{8} \\& \quad = \frac{(\alpha ^{4n}+\beta ^{4n})^{2}+4(\alpha ^{4n}+\beta ^{4n})+4}{16} \\& \quad = \biggl( \frac{\alpha ^{4n}(1+\alpha ^{-1})-\beta ^{4n}(1+\beta ^{-1})}{4 \sqrt{2}}+\frac{1}{2} \biggr) ^{2} \\& \quad = \biggl( \frac{\alpha ^{4n}-\beta ^{4n}}{4\sqrt{2}}+\frac{\alpha ^{4n-1}-\beta ^{4n-1}}{4\sqrt{2}}-\frac{1}{2}+1 \biggr) ^{2} \\& \quad = (B_{2n}+b_{2n}+1)^{2}. \end{aligned}$$
2.6 The Pell equation
Let d be any positive nonsquare integer, and let N be any fixed integer. Then the equation
is known as a Pell-type equation; \(x^{2}-dy^{2}=N\) is the positive Pell-type equation, and \(x^{2}-dy^{2}=-N\) is the negative Pell-type equation. It is named after John Pell (1611-1685), a mathematician who searched for integer solutions to equations of this type in the seventeenth century. Ironically, Pell was not the first to work on this problem, nor did he contribute to our knowledge for solving it. Euler (1707-1783), who brought us the ψ-function, accidentally named the equation after Pell, and the name stuck.
$$ x^{2}-dy^{2}=\pm N $$
(17)
For \(N=1\), the Pell equation \(x^{2}-dy^{2}=\pm 1\) is known as the classical Pell equation. The Pell equation \(x^{2}-dy^{2}=1\) was first studied by Brahmagupta (598-670) and Bhaskara (1114-1185). Its complete theory was worked out by Lagrange (1736-1813), not Pell. It is often said that Euler (1707-1783) mistakenly attributed Brouncker’s (1620-1684) work on this equation to Pell. However, the equation appears in a book by Rahn (1622-1676), which was certainly written with Pell’s help: some say that it is entirely written by Pell. Perhaps Euler knew what he was doing in naming the equation. In 1657, Fermat stated (without giving proof) that the positive Pell equation \(x^{2}-dy^{2}=1\) has an infinite number of solutions. If \((m,n) \) is a solution, that is, \(m^{2}-dn^{2}=1\), then \((m^{2}+dn^{2},2mn)\) is also a solution since
So the Pell equation \(x^{2}-dy^{2}=1\) has infinitely many integer solutions. Later, in 1766, Lagrange proved that the Pell equation \(x^{2}-dy^{2}=1\) has an infinite number of solutions if d is positive and nonsquare. The first nontrivial solution \((x_{1},y_{1})\neq (\pm 1,0)\) of this equation is called the fundamental solution from which all others are easily computed since \(x_{n}+y_{n}\sqrt{d}=(x_{1}+y_{1}\sqrt{d})^{n}\) for \(n\geq 1\) can be found using, for example, the cyclic method [20], known in India in the 12th century, or using the slightly less efficient but more regular English method [20] (17th century). There are other methods to compute this so-called fundamental solution, some of which are based on a continued fraction expansion of the square root of d given as follows. Let \(\sqrt{d}=[m_{0};\overline{m_{1},m_{2},\ldots ,m_{l}}]\) denote the continued fraction expansion of period length l. Set \(A_{-2}=0\), \(A_{-1}=1\), \(A_{k}=m_{k}A_{k-1}+A_{k-2}\) and \(B_{-2}=1\), \(B_{-1}=0\), \(B_{k}=m_{k}B_{k-1}+B_{k-2} \) for nonnegative integers k. Then \(C_{k}=\frac{A_{k}}{B_{k}}\) is the kth convergent of \(\sqrt{d}\), and the fundamental solution of \(x^{2}-dy^{2}=1\) is \((x_{1},y_{1})=(A_{l-1},B_{l-1})\) if l is even or \((A_{2l-1},B_{2l-1})\) if l is odd. Also, if l is odd, then the the fundamental solution of \(x^{2}-dy^{2}=-1\) is \((x_{1},y_{1})=(A_{l-1},B_{l-1})\) (for further details on Pell equations, see [21‐23]).
$$\begin{aligned} \bigl(m^{2}+dn^{2}\bigr)^{2}-d(2mn)^{2}= \bigl(m^{2}-dn^{2}\bigr)^{2}=1. \end{aligned}$$
It is known that there is a connection between integer sequences and Pell equations. For instance, Olajas [9] gave the integer solutions to \(x^{2}-5y^{2}=\pm 4\) as follows.
Theorem 12
([9, Thm. 2.17])
The only solutions of the equation
\(x^{2}-5y^{2}=\pm 4\)
are
\(x=\pm L_{m}\)
and
\(y=\pm F_{m}\), where
\(L_{m}\)
and
\(F_{m}\)
are the
mth terms of the Lucas and Fibonacci sequences, respectively.
For integers A and B such that \(A^{2}-4B\neq 0\) (to exclude a degenerate case), \(R=\{R_{i}\}_{i=0}^{\infty }=R(A,B,R_{0},R_{1})\) is a second-order linear recurrence if the recurrence relation for \(i\geq 2\)
holds for its terms and \(R_{0},R_{1}\) are fixed integers. For the Pell equation \(x^{2}-8y^{2}=1\), Liptai [24] proved the following:
$$ R_{i}=AR_{i-1}-BR_{i-2} $$
(18)
Theorem 13
([24, Thm. 1])
The terms of the second-order linear recurrence
\(R(6,-1,1,6)\)
are the solutions of the equation
\(z^{2}-8y^{2}=1\)
for some integer
z.
Now we can return to our main problem. We consider the integer solutions of the Pell equations
$$\begin{aligned} x^{2}-8y^{2}=1,\qquad x^{2}-2y^{2}=\pm 4,\quad \mbox{and}\quad x^{2}-8y^{2}=\pm 4. \end{aligned}$$
Theorem 14
(1)
For the positive Pell equation
\(x^{2}-8y^{2}=1\), we have:
for
\(n\geq 3\).
(a)
The integer solutions are
\((x_{n},y_{n})=(\frac{Q_{2n}}{2},\frac{P_{2n} }{2})\)
for
\(n\geq 1\).
(b)
The integer solutions satisfy the recurrence relation
$$\begin{aligned} x_{n}=\frac{5(Q_{2n-2}+Q_{2n-4})-Q_{2n-6}}{2}\quad \textit{and}\quad y_{n}= \frac{ 5(P_{2n-2}+P_{2n-4})-P_{2n-6}}{2} \end{aligned}$$
(2)
The negative Pell equation
\(x^{2}-8y^{2}=-1\)
has no integer solutions.
(3)
For the positive Pell equation
\(x^{2}-2y^{2}=4\), we have:
for
\(n\geq 4\).
(a)
The integer solutions are
\((x_{n},y_{n})=(Q_{2n},Y_{2n-1})\)
for
\(n\geq 1\).
(b)
The integer solutions
\((x_{n},y_{n})\)
satisfy the recurrence relation
$$\begin{aligned} x_{n}=5(Q_{2n-2}+Q_{2n-4})-Q_{2n-6}\quad \textit{and}\quad y_{n}=5(Y_{2n-3}+Y_{2n-5})-Y_{2n-7} \end{aligned}$$
(4)
For the negative Pell equation
\(x^{2}-2y^{2}=-4\), we have:
for
\(n\geq 4\).
(a)
The integer solutions are
\((x_{n},y_{n})=(Q_{2n-1},Y_{2n-2})\)
for
\(n\geq 1\).
(b)
The integer solutions
\((x_{n},y_{n})\)
satisfy the recurrence relation
$$\begin{aligned} x_{n}=5(Q_{2n-3}+Q_{2n-5})-Q_{2n-7}\quad \textit{and}\quad y_{n}=5(Y_{2n-4}+Y_{2n-6})-Y_{2n-8} \end{aligned}$$
(5)
For the positive Pell equation
\(x^{2}-8y^{2}=4\), we have:
for
\(n\geq 3\).
(a)
The integer solutions are
\((x_{n},y_{n})=(Q_{2n},P_{2n})\)
for
\(n\geq 1\) .
(b)
The integer solutions
\((x_{n},y_{n})\)
satisfy the recurrence relation
$$\begin{aligned} x_{n}=5(Q_{2n-2}+Q_{2n-4})-Q_{2n-6}\quad \textit{and}\quad y_{n}=5(P_{2n-2}+P_{2n-4})-P_{2n-6} \end{aligned}$$
(6)
For the negative Pell equation
\(x^{2}-8y^{2}=-4\), we have:
for
\(n\geq 4\).
(a)
The integer solutions are
\((x_{n},y_{n})=(Q_{2n-1},P_{2n-1})\)
for
\(n\geq 1\).
(b)
The integer solutions
\((x_{n},y_{n})\)
satisfy the recurrence relation
$$\begin{aligned} x_{n}=5(Q_{2n-3}+Q_{2n-5})-Q_{2n-7}\quad \textit{and}\quad y_{n}=5(P_{2n-3}+P_{2n-5})-P_{2n-7} \end{aligned}$$
Proof
(1a) Notice that \(Q_{n}=\alpha ^{n}+\beta ^{n}\) and \(P_{n}=\frac{\alpha ^{n}-\beta ^{n}}{2\sqrt{2}}\). So
since \(\alpha \beta =-1\).
$$\begin{aligned} x^{2}-8y^{2}&=\biggl(\frac{Q_{2n}}{2} \biggr)^{2}-8\biggl(\frac{P_{2n}}{2}\biggr)^{2}= \frac{(\alpha ^{2n}+\beta ^{2n})^{2}-8 ( \frac{\alpha ^{2n}-\beta ^{2n}}{2\sqrt{2}} ) ^{2}}{4} \\ &=\frac{\alpha ^{4n}+2(\alpha \beta )^{2n}+\beta ^{4n}-(\alpha ^{4n}-2(\alpha \beta )^{2n}+\beta ^{4n})}{4} \\ &\begin{aligned} &=\frac{4(\alpha \beta )^{2n}}{4} \\ &=1 \end{aligned} \end{aligned}$$
(1b)
Similarly, it can be shown that \(y_{n}=\frac{5(P_{2n-2}+P_{2n-4})-P_{2n-6}}{2} \).
$$\begin{aligned} \frac{5(Q_{2n-2}+Q_{2n-4})-Q_{2n-6}}{2} =&\frac{ 5Q_{2n-2}+5Q_{2n-4}-(2Q_{2n-5}+Q_{2n-6})+2Q_{2n-5}}{2} \\ =&\frac{5Q_{2n-2}+2(2Q_{2n-4}+Q_{2n-5})}{2} \\ =&\frac{2Q_{2n-1}+Q_{2n-2}}{2} \\ =&\frac{Q_{2n}}{2} \\ =&x_{n}. \end{aligned}$$
(2) The negative Pell equation \(x^{2}-8y^{2}=-1\) has no integer solutions since \(\sqrt{8}=[2;\overline{1,4}]\), that is, the length of 2, which is even.
(3a) Notice that \(Q_{n}=\alpha ^{n}+\beta ^{n}\) and \(Y_{n}=\frac{\alpha ^{n+1}-\beta ^{n+1}}{\sqrt{2}}\). So
$$\begin{aligned} x^{2}-2y^{2} =&Q_{2n}^{2}-2Y_{2n-1}^{2}= \bigl(\alpha ^{2n}+\beta ^{2n}\bigr)^{2}-2 \biggl( \frac{\alpha ^{2n}-\beta ^{2n}}{\sqrt{2}} \biggr) ^{2} \\ =& \alpha ^{4n}+2(\alpha \beta )^{2n}+\beta ^{4n}- \bigl(\alpha ^{4n}-2(\alpha \beta )^{2n}+\beta ^{4n} \bigr) \\ =& 4(\alpha \beta )^{2n} \\ =& 4. \end{aligned}$$
(3b) Similarly, we get
The other cases can be proved similarly. □
$$\begin{aligned} 5(Q_{2n-2}+Q_{2n-4})-Q_{2n-6} =& 5Q_{2n-2}+5Q_{2n-4}-(2Q_{2n-5}+Q_{2n-6})+2Q_{2n-5} \\ =& 5Q_{2n-2}+2(2Q_{2n-4}+Q_{2n-5}) \\ =& 4Q_{2n-2}+Q_{2n-2}+2Q_{2n-3} \\ =& 2(2Q_{2n-2}+Q_{2n-3})+Q_{2n-2} \\ =& 2Q_{2n-1}+Q_{2n-2} \\ =& Q_{2n} \\ =& x_{n} \end{aligned}$$
It is known that there are a number of applications of Pell and Fibonacci sequences on the theory of numbers. For instance, Sellers proved in [25, Thm. 2.1] that the number of domino tilings of the graph \(W_{4}\times P_{n-1}\) equals the product of the nth Fibonacci and Pell numbers for all \(n\geq 2\). Also, there has been a connection between Diophantine quadruples and Fibonacci numbers. Recall that a set of m positive integers \(\{a_{1},a_{2},\dots,a_{m}\}\) is called a Diophantine m-tuble if \(a_{i}a_{j}+1 \) is a perfect square for all \(1\leq i< j\leq m\) and is called a \(D(n)\)-m-tuble (or a Diophantine m-tuble with the property \(D(n)\)) if \(a_{i}a_{j}+n\) is a perfect square for all \(1\leq i< j\leq m\).
Cassini’s identity for Fibonacci number is \(F_{n}F_{n+2}+(-1)^{n}=F_{n+1}^{2} \) and is the basis for the construction of so-called Hoggatt-Bergum’s quadruple. Hoggatt and Bergum [26] proved that, for any positive integer k, the set
is a Diophantine quadruple. Later Morgado [27] and Horadam [28] generalized this result. The identity
is known as the Morgado identity.
$$\begin{aligned} \{F_{2k},F_{2k+2},F_{2k+4},4F_{2k+1}F_{2k+2}F_{2k+3} \} \end{aligned}$$
$$\begin{aligned} F_{k-3}F_{k-2}F_{k-1}F_{k+1}F_{k+2}F_{k+3}+L_{k}^{2}= \bigl[F_{k}\bigl(2F_{k-1}F_{k+1}-F_{k}^{2} \bigr)\bigr]^{2} \end{aligned}$$
Using Fibonacci numbers, Dujella [29] defined the elliptic curve (see [30])
and determined the integer points on it by terms of Fibonacci numbers when the rank of \(E_{k}(\mathbb{Q})\) is 1.
$$\begin{aligned} E_{k}:y^{2}=(F_{2k}x+1) (F_{2k+2}x+1) (F_{2k+4}x+1) \end{aligned}$$
It is known that there are several identities on Fibonacci and Lucas numbers. Some of them can be given as \(4F_{k-1}F_{k+1}+F_{k}^{2}=L_{k}^{2}\) and \(4F_{k-1}F_{k}^{2}F_{k+1}+1=(F_{k}^{2}+F_{k-1}F_{k+1})^{2}\). Using these, Dujella [31] obtained some quantities on \(D(F_{k}^{2})\)-quadruples
As in (19), the set
is a \(D(L_{k}^{2})\)-quadruple.
$$\begin{aligned} &\bigl\{ 2F_{k-1},2F_{k+1},2F_{k}^{3}F_{k+1}F_{k+2},2F_{k+1}F_{k+2}F_{k+3} \bigl(2F_{k+1}^{2}-F_{k}^{2}\bigr)\bigr\} , \\ &\bigl\{ F_{k-1},4F_{k+1},F_{k-2}F_{k-1}F_{k+1} \bigl(2F_{k}^{2}-F_{k-1}^{2} \bigr),F_{k}^{3}F_{k+2}F_{k+3}\bigr\} , \\ &\bigl\{ 4F_{k-1},F_{k+1},F_{k-2}F_{2k-2}F_{2k-1},F_{k}^{3}L_{k}L_{k+1} \bigr\} . \end{aligned}$$
(19)
$$\begin{aligned} \bigl\{ F_{k-3}F_{k-2}F_{k+1},F_{k-1}F_{k+2}F_{k+3},F_{k}L_{k}^{2},4F_{k-1}^{2}F_{k}F_{k+1}^{2} \bigl(2F_{k-1}F_{k+1}-F_{k}^{2}\bigr)\bigr\} \end{aligned}$$
Dujella and Ramasamy [32] considered the Fibonacci numbers and \(D(4)\)-quadruple. They proved that the set
is a \(D(4)\)-quadruple. Also, they considered integer solutions of the Pell equations by using a \(D(4)\)-quadruple.
$$\begin{aligned} \{F_{2k},5F_{2k},4F_{2k+2},4L_{2k}F_{4k+2} \} \end{aligned}$$
In the future work, we are planing to study \(D(n)\)-quadruples for the sequences mentioned for some n.
3 Conclusion
In this work, we deduced some new results on Pell, Pell-Lucas, and balancing numbers including sums, divisibility properties, perfect squares, and integer solutions of some specific Pell equations.
Acknowledgements
This work was supported by the Scientific Research Projects Coordination Unit of Istanbul University (No. BEK-2017-25878).
Competing interests
The author declares that she has no competing interests.
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