\magnification=1200
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\refstyle A
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\topmatter
\title A survey on practical numbers \endtitle
\author Giuseppe Melfi \endauthor
\address Dipartimento di Matematica, Universit\`a di Pisa,
Via Buonarroti 2, 56127 Pisa, Italy \endaddress
\email melfi\@dm.unipi.it \endemail
\abstract
A positive integer $m$ is said to be practical if every integer $n\in(1,m)$
is a sum of distinct positive divisors of $m.$ In this paper we give an
equivalent definition of practical number, and describe some arithmetical
properties of practical numbers showing a remarkable analogy with primes.
We give an improvement of the estimate of the gap between consecutive
practical numbers and prove the existence of infinitely many practical
numbers in suitable binary recurrence sequences, including the sequences
of Fibonacci, Lucas and Pell.
\endabstract
\endtopmatter

\document

\subheading{1. Introduction}
\vskip 0.1cm

A positive integer $m$ is said to be {\sl practical} (see \cite{11}) if
every $n$ with $1\le n\le m$ is a sum of distinct positive divisors of $m.$
Several authors dealt with some aspects of the theory of practical numbers.
P. Erd\H os \cite3 in 1950 announced that practical numbers have zero
asymptotic density. B. M. Stewart \cite{12} proved the following structure
theorem: an integer $\,m\ge 2,$
$\,m=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k},$
with primes $p_1<p_2<\dots<p_k$ and integers $\alpha_i\ge 1,$
is practical if and only if $\,p_1=2\,$ and, for $i=2,3,\dots,k,$
$$p_i\le\sigma\!\left(p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_{i-1}
^{\alpha_{i-1}}\right)+1$$
where $\sigma(n)$ denotes the sum of the positive divisors of $n.$
 
Let $P(x)$ be the counting function of practical numbers:
$$P(x)=\sum\Sb m\le x\\ m\text{ practical}\endSb 1\,.$$
M. Hausman and H. N. Shapiro \cite5 showed in 1984 that
$$P(x)\ll\frac x{(\log x)^\beta}$$ for any
$\beta<\frac12(1-1/\log 2)^2\simeq 0.\,0979.\,$ M. Margenstern 
(\cite6, \cite7)
proved that $$P(x)\gg\frac x{\exp\left\{\frac1{2\log 2}(\log\log x)^2
+3\log\log x\right\}}.$$ 
Gerald Tenenbaum (\cite{13}, \cite{14}) improved the above upper
and lower bounds as follows:
$$\frac{x}{\log x}(\log\log x)^{-5/3-\varepsilon}\ll_\varepsilon
P(x)\ll\frac{x}{\log x}\log\log x\log\log\log x.$$
Moreover, Margenstern conjectured that:
$$P(x)\sim\lambda\,\frac x{\log x}$$ with $\lambda\simeq 1.\,341,$
in analogy with the asymptotic behaviour of primes.

The author \cite8 recently proved two Goldbach-type conjectures for
practical numbers first stated in \cite6: (i) every even positive integer
is a sum of two practical numbers; (ii) there exist infinitely many
practical numbers $m$ such that $m-2$ and $m+2$ are also practical.

The purpose of the present paper is to survey some of the above results
and to give some new contributions to the theory of practical numbers.

Sierpi\'nski \cite{10} and Stewart \cite{12} independently remarked that a
positive integer $m$ is practical if and only if every integer $n$ with
$1\le n\le\sigma(m)$ is a sum of distinct positive divisors of $m.$
Here we give an alternative proof of this equivalence.

We also give an improved version of \cite{8, Lemma 2}, which yields a
slightly simpler proof of the Goldbach-type result (i) mentioned above.

We study the gap between consecutive practical numbers, improving upon
a result of Hausman and Shapiro \cite5.

Finally we prove that some binary recurrence sequences, including the
classical sequences of Fibonacci, Lucas and Pell, contain infinitely
many practical numbers. We incidentally note that it is unknown whether the
Fibonacci sequence $\{1,1,2,3,5,\dots\}$ and the Lucas sequence
$\{1,3,4,7,11,\dots\}$ contain infinitely many prime numbers. Dubner and Keller
\cite2 recently announced the primality of some ``titanic'' (i.e. having more
than 1000 digits) Fibonacci and Lucas numbers, such as $F_{9311},$
$\,F_{5387},$ $\,L_{14449},$ $\,L_{7741},$ $\,L_{5851},$
$\,L_{4793},$ $\,L_{4787}.$

\vskip 0.5cm
\subheading{2. An arithmetical result}
\vskip 0.1cm

In this section we give an equivalent definition of practical number.
We begin with the following lemma:

\proclaim{Lemma 1}
Let $m$ be a positive integer, and let $\,d_1=1<d_2<\dots<d_r=m\,$ be the
positive divisors of $m.$ Let $\,d_h$ be the least divisor such that
$\,d_h\ge\sqrt m.$ Then $\,d_1+d_2+\cdots+d_{h-1}+1\le m.$
\endproclaim
\demo{Proof}
The lemma is true for $m=1,2,3,4.$ Let $m>4;$ since $\,d_{h-1}<\sqrt m\,$
we have $$\spreadlines{0.1cm}\align
d_1+d_2+\cdots+d_{h-1}+1\ &\le\ 1+2+3+\cdots+[\sqrt m\,]+1\\
&=\ \frac{[\sqrt m\,]([\sqrt m\,]+1)}2+1\\
&\le\ \frac{\sqrt m\,(\sqrt m+1)}2+1\\
&<\ m\,.\qed
\endalign$$
\enddemo

\proclaim{Lemma 2} \rom{(Margenstern)}.
Let $m$ be a positive integer, and let $\,d_1,\dots,d_h,\dots,d_r$ be as in
Lemma 1. Then $m$ is such that every $n$ with $1\le n\le\sigma(m)$ is a sum
of distinct positive divisors of $m,$ if and only if
$\,d_{j+1}\le d_1+\cdots+d_j+1$ for every $\,j=1,\dots,\,h-1.$
\endproclaim
\demo{Proof}
For the proof see Margenstern's paper \cite7.\qed
\enddemo

\proclaim{Proposition 3}
A positive integer $m$ is practical if and only if every $n$ with
$1~\le~n~\le~\sigma(m)$ is a sum of distinct positive divisors of $m.$
\endproclaim
\demo{Proof}
Since $\sigma(m)\ge m,$ if $m$ is such that every $n$ with
$1\le n\le\sigma(m)$ is a sum of distinct positive divisors of $m,$
{\sl a fortiori\/} $m$ is a practical number.

Let $m$ be practical, i.e. \!every $n$ with $1\le n\le m$ is a sum of
distinct positive divisors of $m.$ Let $\,d_1,\dots,d_h,\dots,d_r$
be as in the preceding lemmas. For any $j$ satisfying $1\le j\le h-1$
we have $\,d_1+\cdots+d_j+1\le m$ by Lemma 1. Hence $\,d_1+\cdots+d_j+1$
is a sum of distinct divisors of $m,$ of which at least one must be
$\ge d_{j+1}.$ It follows that $\,d_{j+1}\le d_1+\cdots+d_j+1,$ whence,
by Lemma 2, every $n$ with $1\le n\le\sigma(m)$ is a sum of distinct
positive divisors of $m.$\qed
\enddemo
 
\vskip 0.5cm
\subheading{3. The Goldbach problem for practical numbers}
\vskip 0.1cm
 
In this section we prove that every even positive integer is a sum of two
practical numbers.
 
\proclaim{Lemma 4}
If $m$ is a practical number and $n$ is an integer such that
$1\le n\le\sigma(m)+1,$ then $mn$ is a practical number. In particular, for
$\,1\le n\le 2m,\,$ $\,mn$ is practical.
\endproclaim
\demo{Proof}
The first assertion easily follows from Stewart's structure theorem; see also
\cite{7, p. 6}. Since $\,m-1\,$ is a sum of distinct divisors of $m,$ we have
$\,m+(m-1)\le\sigma(m),$ i.e. $2m\le\sigma(m)+1,\,$ and this proves the
second assertion.\qed
\enddemo
 
The author \cite{8, Lemma 2} proved that if $m$ and $m+2$ are practical
numbers then every even integer $2n\in\left[m^2, 3m^2\right]$ is a sum of
two practical numbers. This can be improved as follows:

\proclaim{Lemma 5}
If $m$ and $m+2$ are two practical numbers, then every even integer $2n$
with $\frac12m^2\le2n\le\frac72m^2$ is a sum of two practical numbers.
\endproclaim
\demo{Proof}
We split up the interval $\,\left[\frac12m^2,\,\frac72m^2\right]\,$
into the union of three subintervals:
\roster
\item"{(i)}" $\ \left[\frac12m^2,\,m^2\right[\,;$
\item"{(ii)}" $\ \left[m^2,\,3m^2\right]\,;$
\item"{(iii)}" $\ \left]3m^2,\,\frac72m^2\right]\,.$
\endroster

(i) If $\,m=2,\,$ the only even number contained in the interval
$\,\left[\frac12m^2,\,m^2\right[\,$ is 2, which is a sum of two practical
numbers $(2=1+1).$ Suppose $\,m>2\,$ and let
$\,2n\in\left[\frac12m^2,\,m^2\right[\,.$ If $\,2n=\frac12m^2$ or
$\,2n=\frac12m^2+m,\,$ we use the decompositions
$$\align
\tfrac12m^2 &=m\left(\tfrac12m-1\right)+m,\\
\tfrac12m^2+m &=m\left(\tfrac12m-1\right)+2m.
\endalign$$
Otherwise we can represent $2n$ as $\,\frac12m^2+km+2j\,$ with
$0\le k<\frac12m,$ $\,1\le j\le\frac12m,$
$\,(k,j)\ne\left(0,\frac12m\right).$ Then
$$2n=\tfrac12m^2+km+2j=m\left(\tfrac12m+k-j\right)+(m+2)\,j.$$
By Lemma 4, $\,2n$ is a sum of two practical numbers.

(ii) For the interval $\,\left[m^2,\,3m^2\right]\,$ see \cite{8, Lemma 2}.

(iii) If $m=2,$ the only even number contained in the interval
$\,\left]3m^2,\,\frac72m^2\right]\,$ is 14, which is a sum of two practical
numbers $(14=6+8).$ Suppose $m>2$ and let
$\,2n\in\,\left]3m^2,\,\frac72m^2\right].$
We can represent $\,2n\,$ as $\,\frac72m^2-km+2j\,$ with
$\,1\le k\le\frac12m,$ $\,1\le j\le\frac12m.$ Then
$$2n=\tfrac72m^2-km+2j=m(2m-k-j-3)+(m+2)\left(\tfrac32m+j\right),$$
which is a sum of two practical numbers by Lemma 4.\qed
\enddemo
 
\proclaim{Theorem 6}
Every even positive integer is a sum of two practical numbers.
\endproclaim
\demo{Proof}
Since $(2,4),\,(4,6),\,(6,8)$ are pairs of twin practical numbers,
by Lemma 5 every $2n\le 126$ is a sum of two practical numbers.
Suppose we have a sequence $\{m_n\}$ such that
\roster
\item"{(i)}" $m_1=16$
\endroster
and for every $n$
\roster
\item"{(ii)}" $m_n$ is practical
\item"{(iii)}" $m_n+2$ is practical
\item"{(iv)}" $1<m_{n+1}/m_n<\sqrt7.$
\endroster
 
Since, by (iv), the intervals
$\,\left[\frac12m_n^2,\,\frac72m_n^2\right]\,$ and
$\,\left[\frac12m_{n+1}^2,\,\frac72m_{n+1}^2\right]\,$ overlap, every even
positive integer $2n\ge128$ is a sum of two practical numbers by Lemma 5.
We shall construct a sequence $\{m_n\}$ satisfying (i), (ii), (iii) and a
condition slightly stronger than (iv), i.e. $1<m_{n+1}/m_n<2.$
 
Let $S_0=\{16,\,30,\,54,\,88,\,160\}.$ For every $r\in S_0,$ $\,r$ and $r+2$
are practical numbers. Denote $\,S_0=\{r_{0,1},\,r_{0,2},\dots,\,r_{0,5}\}\,$
with $\,r_{0,1}<r_{0,2}<\dots<r_{0,5}.$ Note that $\,r_{0,i}<2r_{0,i-1}$
$\,(i=2,3,4,5)\,$ and $\,r_{0,5}=\frac12r_{0,1}^2+2r_{0,1}.$ Let $h_0=5$ 
and, for $k=1,2,\dots,$ define
$$\align
S_k &=\left\{\tfrac12r_{k-1,i}^2+2r_{k-1,i}\ ,
\ r_{k-1,i}^2+3r_{k-1,i}\ \big\vert\ i=1,2,\dots,h_{k-1}\right\}\\
\vspace{0.1cm}
&=\left\{r_{k,1},\,r_{k,2},\dots,\,r_{k,h_k}\right\}
\endalign$$
with $\,r_{k,1}<r_{k,2}<\dots<r_{k,h_k}.$ Further let
$\,S=\bigcup_{k=0}^{\infty}S_k.$ If we write $\,S=\{m_n\},$ with
$\,m_n<m_{n+1}\,$ for every $n,$ one can see that $\{m_n\}$ satisfies (i),
(ii), (iii) and $\,m_{n+1}<2m_n.\,$ The proof of this is similar to the
argument given in \cite{8, Theorem~1}.\qed
\enddemo

\vskip 0.5cm
\subheading{4. k-tuples of twin practical numbers}
\vskip 0.1cm

It is easy to find infinitely many pairs $(m,\,m+2)$ of twin practical
numbers (see the proof of Theorem 6 above). The following was conjectured
in \cite6 and \cite7:

\proclaim{Theorem 7}
There exist infinitely many practical numbers $m$ such that $m-2$ and $m+2$
are also practical.
\endproclaim
\demo{Proof}
For the proof see \cite{8, Theorem 2}.\qed
\enddemo

It is shown in \cite7 that for any even $m>2,$ at least one of $m,$ $\,m+2,$
$\,m+4,$ $\,m+6$ is not practical. However, we state the following

\proclaim{Conjecture 8}
There exist infinitely many $5$-tuples of practical numbers of the form
$\,(m-6,\,m-2,\,m,\,m+2,\,m+6).$
\endproclaim
 
\vskip 0.5cm
\subheading{5. Gaps between practical numbers}
\vskip 0.1cm

Here we give an estimate of the gap between consecutive practical numbers.
The same problem for primes has been extensively studied. If $\{p_n\}$ is
the sequence of primes, R. C. Baker and G. Harman \cite1 recently proved that
$$p_{n+1}-p_n\ll p_n^{0.535},$$ the exponent $0.\,535$ being of course
replaced by $\frac12+\varepsilon$ under the Riemann Hypothesis. If
$\{s_n\}$ is the sequence of practical numbers, Hausman and Shapiro
\cite5 proved that $$s_{n+1}-s_n\le2s_n^{1/2}.$$ We can improve this
inequality as follows:

\proclaim{Theorem 9}
Let $\{s_n\}$ be the sequence of practical numbers and let
$A>4e^{-\gamma/2},$ where $\gamma$ is the Euler-Mascheroni constant.
For any sufficiently large $n$ we have
$$s_{n+1}-s_n<A\,\frac{s_n^{1/2}}{(\log\log s_n)^{1/2}}.$$
\endproclaim
\demo{Proof}
Let $\delta>0$ and $c<e^\gamma$ be such that
$4c^{-1/2}(1+\delta)(1-\delta)^{-1/2}<A.$ Let $N_k=\prod_{p\le e^k}p^k,$
where $p$ denotes a prime. By \cite{4, \S 22.9} we have
$$\lim_{k\to\infty}\frac{\sigma(N_k)}{N_k\log\log N_k}=e^\gamma.\tag1$$

For every $k,$ let $m^{(k)}$ be any integer such that $N_{k-1}|m^{(k)},$
$\,m^{(k)}|N_k.$ It is easy to see, by induction on $k,$ that $N_k$ is
practical for all $k\ge1,$ and if $k\ge3$ then $m^{(k)}$ is also practical.
To prove this, note that $N_1=2$ and $N_2=2^2\cdot3^2\cdot5^2\cdot7^2$
are practical, and $m^{(k)}/N_{k-1}$ is a product of primes not exceeding
$e^k.$ Since $e^k\le2N_{k-1}$ for $k\ge3,$ $\,m^{(k)}$ and hence $N_k$
are practical by repeated application of Lemma 4.

Since $n|m$ easily implies $\sigma(n)/n\le\sigma(m)/m,$ we get
$$\frac{\sigma(N_{k-1})}{N_{k-1}\log\log N_k}\le
\frac{\sigma\!\left(m^{(k)}\right)}{m^{(k)}\log\log m^{(k)}}\le
\frac{\sigma(N_k)}{N_k\log\log N_{k-1}}.$$
Clearly $$\log\log N_{k-1}\sim\log\log N_k,$$ whence, by (1),
$$\lim_{k\to\infty}\frac{\sigma\!\left(m^{(k)}\right)}
{m^{(k)}\log\log m^{(k)}}=e^\gamma.$$
Thus there exists an integer $k_0$ such that for any $k\ge k_0$
$$\min\Sb m\\ N_{k-1}|m\\ m|N_k\endSb \frac{\sigma(m)}{m\log\log m}>c.\tag2$$
Let $s_n$ be a practical number such that $s_n>c\,N_{k_0}^2\log\log N_{k_0}$
and let $\kappa$ be the least positive integer such that
$$N_\kappa\ge\frac{s_n}{c\,N_\kappa\log\log N_\kappa}.$$
Further, let
$$m_1^{(\kappa)}=N_{\kappa-1}<m_2^{(\kappa)}<\cdots<m_{\lambda}^{(\kappa)}
=N_\kappa$$ be all the integers satisfying $N_{\kappa-1}|m_i^{(\kappa)},$
$\,m_i^{(\kappa)}|N_\kappa,\,$ and let $\nu$ be such that $$m_\nu^{(\kappa)}<
\frac{s_n}{c\,m_\nu^{(\kappa)}\log\log m_\nu^{(\kappa)}}\tag3$$ and $$m_{\nu+1}
^{(\kappa)}\ge\frac{s_n}{c\,m_{\nu+1}^{(\kappa)}\log\log m_{\nu+1}^{(\kappa)}}.
\tag4$$ Let $\vartheta$ and $\tau$ be defined by $m_\nu^{(\kappa)}=\vartheta
N_{\kappa-1},$ $\,N_\kappa=\tau m_\nu^{(\kappa)}.$ Clearly $\tau>1.$ Let
$p''$ be the least prime factor of $\tau,$ and let $p'$ be the greatest prime
$<p''$ (if $p''=2,$ we let $p'=1).$ By Bertrand's postulate we have
$p''\le 2p'.$ Since $N_\kappa=\vartheta\tau N_{\kappa-1},$ we have
$$\vartheta\tau=\left(\prod_{p\le e^{\kappa-1}}p\right)
\left(\prod_{e^{\kappa-1}<p\le e^\kappa}p^\kappa\right),$$
whence $\,p'|\vartheta\tau,$ $\,p'|\vartheta,\,$ and $\,p'|m_\nu^{(\kappa)}.$
Therefore $$p''\cdot\frac{m_\nu^{(\kappa)}}{p'}=p''\cdot\frac{\vartheta}{p'}
\cdot N_{\kappa-1}$$ is a multiple of $N_{\kappa-1}.$ Moreover
$$N_\kappa=\tau m_\nu^{(\kappa)}=p'\cdot\frac{\tau}{p''}\cdot p''\cdot
\frac{m_\nu^{(\kappa)}}{p'}$$ is a multiple of $\,p''m_\nu^{(\kappa)}/p'.$
Hence $$p''\cdot\frac{m_\nu^{(\kappa)}}{p'}=m_i^{(\kappa)}$$ for some $i>\nu,$
since $p''>p'.$ It follows that
$$m_{\nu+1}^{(\kappa)}\le p''\cdot\frac{m_\nu^{(\kappa)}}{p'}\le2m_\nu^
{(\kappa)}.\tag5$$
 
Let $\,q=\left[s_n/{m_{\nu+1}^{(\kappa)}}\right]+1.$ By (2) and (4) we have
$$\spreadlines{0.1cm}\align
q\ &\le\ \frac{s_n}{m_{\nu+1}^{(\kappa)}}+1\\
&\le\ c\,m_{\nu+1}^{(\kappa)}\log\log m_{\nu+1}^{(\kappa)}+1\\
&<\ \sigma\!\left(m_{\nu+1}^{(\kappa)}\right)+1,\endalign$$
whence, by Lemma 4, $\,r=q\,m_{\nu+1}^{(\kappa)}\,$ is a practical number.
Further
$$r-s_n=m_{\nu+1}^{(\kappa)}\left(\left[\frac{s_n}{m_{\nu+1}^{(\kappa)}}
\right]+1\right)-s_n>0,$$ whence, by (3) and (5),
$$\spreadlines{0.1cm}\align
s_{n+1}-s_n\ &\le\ r-s_n\\
&=\ m_{\nu+1}^{(\kappa)}\left(1-\left\{\frac{s_n}{m_{\nu+1}^{(\kappa)}}
\right\}\right)\\
&<\ 2\,\frac{s_n}{c\,m_\nu^{(\kappa)}\log\log m_\nu^{(\kappa)}}.\endalign$$
For any $\varepsilon>0$ and any sufficiently large $n$ we have, by (3),
(4) and (5),
$${m_{\nu+1}^{(\kappa)}}^2\ge\frac{s_n}{c\log\log m_{\nu+1}^{(\kappa)}}
\ge s_n^{1-\varepsilon}\tag6$$ and
$$\spreadlines{0.1cm}\aligned
s_{n+1}-s_n\ &<\ 2\,\frac{m_{\nu+1}^{(\kappa)}}{m_\nu^{(\kappa)}}\cdot
\frac{s_n}{c\,m_{\nu+1}^{(\kappa)}\log\log m_\nu^{(\kappa)}}\\
&\le\ 4\,\frac{c^{1/2}\left(\log\log m_{\nu+1}^{(\kappa)}\right)^{1/2}}
{s_n^{1/2}}\cdot\frac{s_n}{c\log\log m_\nu^{(\kappa)}}\\
&\le\ 4c^{-1/2}(1+\delta)\,\frac{s_n^{1/2}}{\left(\log\log m_\nu^{(\kappa)}
\right)^{1/2}}.\endaligned\tag7$$
Since, by (5) and (6),
$$m_\nu^{(\kappa)}=\frac{m_\nu^{(\kappa)}}{m_{\nu+1}^{(\kappa)}}
\,m_{\nu+1}^{(\kappa)}\ge\tfrac12s_n^{(1-\varepsilon)/2},$$ we get
$$\log\log m_\nu^{(\kappa)}\ge\log\left(\frac{1-\varepsilon}2
\log s_n-\log2\right)\ge(1-\delta)\log\log s_n,$$ whence, by (7),
$$s_{n+1}-s_n<A\,\frac{s_n^{1/2}}{(\log\log s_n)^{1/2}}.\qed$$
\enddemo

\remark{Remark}
By Gronwall's theorem \cite{4, Theorem 323} we have
$$\limsup_{n\to\infty}\frac{\sigma(n)}{n\log\log n}=e^\gamma,$$
which justifies the choice of the sequence $N_k$ in our proof of Theorem 9.
\endremark

\vskip 0.5cm
\subheading{6. Binary recurrence sequences}
\vskip 0.1cm
 
Let $P,$ $Q$ be non-zero integers; a pair of Lucas sequences $\{u_n(P,Q)\}$,
$\{v_n(P,Q)\}$ is a pair of binary recurrence sequences defined as
$$\left\{\aligned
u_0(P,Q)&=0\\
u_1(P,Q)&=1\\
u_n(P,Q)&=P\,u_{n-1}(P,Q)-Q\,u_{n-2}(P,Q)\quad\text{for}\ n\ge2
\endaligned\right.$$ and
$$\left\{\aligned
v_0(P,Q)&=2\\
v_1(P,Q)&=P\\
v_n(P,Q)&=P\,v_{n-1}(P,Q)-Q\,v_{n-2}(P,Q)\quad\text{for}\ n\ge2.
\endaligned\right.$$
 
The sequence $\,\{u_n(P,Q)\}\,$ is also called a {\sl \ fundamental Lucas
sequence\ } and $\{v_n(P,Q)\}\,$~its {\sl \ companion sequence\/}.
 
Suppose $P^2-4Q\ne0$ and let $\alpha,$ $\beta$ be the distinct roots of
the polynomial $$x^2-Px+Q.$$ We have
$$u_n(P,Q)=\frac{\alpha^n-\beta^n}{\alpha-\beta}$$ and
$$v_n(P,Q)=\alpha^n+\beta^n.$$ Using a shorter notation, we shall write
$u_n$ and $v_n$ instead of $u_n(P,Q)$ and $v_n(P,Q).$ For $(P,Q)=(1,-1)$,
$u_n$ and $v_n$ are the sequence of Fibonacci numbers and the sequence of
Lucas numbers, respectively; for $(P,Q)=(2,-1)$, $u_n$ is the sequence of
Pell numbers \cite{9, p. 56}.

\proclaim{Theorem 10}
Let $\{u_n(P,Q)\}$ be a fundamental Lucas sequence. If $P^2-4Q>~\!\!0$ and
$PQ+P$ is even, then the sequence $\{|u_n(P,Q)|\}$ contains infinitely
many practical numbers.
\endproclaim
\demo{Proof}
We shall prove that, for sufficiently large $k,$ $|u_{3\cdot2^k}|$ is a
practical number. Let $\{v_n\}$ be the companion sequence of $\{u_n\}.$
Since $u_{2m}=u_mv_m$ for every $m,$ we have, for $k>0,$
$$u_{3\cdot2^k}=u_3\cdot\prod_{h=0}^{k-1}v_{3\cdot2^h}.$$ Also, $P^2-4Q>0$
implies $u_3=P^2-Q>0.$ Note that $v_3=P(P^2-3Q),$ whence
$\text{sgn}\,v_3=\text{sgn}\,P.$ Since $P^2-4Q>0,$ we have
$\alpha,\,\beta\in\Bbb R,$ whence
$v_n=\alpha^n+\beta^n$ is positive for $n$ even. Therefore
$$|u_{3\cdot2^k}|=u_3\,|v_3|\cdot\prod_{h=1}^{k-1}v_{3\cdot2^h}.$$
Since $PQ+P$ is even, $v_{3m}$ is even for all $m.$ Denoting
$v'_{3m}=v_{3m}/2,$ we have
$$|u_{3\cdot2^k}|=2^k\,u_3\,|v'_3|\cdot\prod_{h=1}^{k-1}v'_{3\cdot2^h}.$$
Let $2^{k+1}\ge\max\{u_3,|v'_3|\},$ and define
$u_j^\ast=2^k\,u_3\,|v'_3|\cdot\prod_{h=1}^{j-1}v'_{3\cdot2^h}.$ We show,
by induction on $j,$ that $u_j^\ast$ is practical for $j=1,\dots,k.$
For $j=1$ this follows from Lemma 4 applied twice, since $2^k$ is practical
and $u_3,|v'_3|\le2^{k+1}.$ Let $1\le j\le k-1,$ and assume that $u_j^\ast$
is practical. We have $$u_j^\ast=2^{k-j}\,|u_{3\cdot2^j}|$$ and
$$u_{j+1}^\ast=u_j^\ast\,v'_{3\cdot2^j},$$ where $$v'_{3\cdot2^j}=
\tfrac12v_{3\cdot2^j}=\tfrac12\left(\alpha^{2^j}+\beta^{2^j}\right)
\left(\alpha^{2^{j+1}}-\alpha^{2^j}\beta^{2^j}+\beta^{2^{j+1}}\right).$$ Note
that $$\alpha^{2^j}+\beta^{2^j}=v_{2^j}$$ and $$\alpha^{2^{j+1}}-\alpha^{2^j}
\beta^{2^j}+\beta^{2^{j+1}}=v_{2^{j+1}}-Q^{2^j}$$ are positive integers (not
both odd). In order to prove that $u_{j+1}^\ast$ is practical, by Lemma 4
applied twice it suffices to show that $$M=\max\left\{\alpha^{2^j}+\beta^{2^j},
\ \alpha^{2^{j+1}}-\alpha^{2^j}\beta^{2^j}+\beta^{2^{j+1}}\right\}<u_j^\ast.$$
Since $\,x+y\le x^2-xy+y^2+1\,$ for all $x,y\in\Bbb R,$ we have $$\align
M&\le\alpha^{2^{j+1}}-\alpha^{2^j}\beta^{2^j}+\beta^{2^{j+1}}+1=
v_{2^{j+1}}-Q^{2^j}+1\\ \vspace{0.1cm}
&<v_{2^{j+1}}+Q^{2^j}=
\alpha^{2^{j+1}}+\alpha^{2^j}\beta^{2^j}+\beta^{2^{j+1}}.\endalign$$ From
$P^2-4Q>0$ and $P=\alpha+\beta\ne0$ it follows that $\alpha\ne\pm\beta.$
Therefore $$u_{2^j}=\frac{\alpha^{2^j}-\beta^{2^j}}{\alpha-\beta}\ne0,$$ i.e.
$|u_{2^j}|\ge1.$ Hence $$\align
M&\le|u_{2^j}|\left(\alpha^{2^{j+1}}+\alpha^{2^j}\beta^{2^j}+\beta^{2^{j+1}}
\right)\\ &=\left|\frac{\alpha^{3\cdot2^j}-\beta^{3\cdot2^j}}{\alpha-\beta}
\right|=|u_{3\cdot2^j}|<2^{k-j}\,|u_{3\cdot2^j}|=u_j^\ast.\qed\endalign$$
\enddemo

\proclaim{Theorem 11}
Let $\{v_n(P,Q)\}$ be a companion Lucas sequence with $Q=-1$ and $P>0.$
If there exists a positive integer $t$ such that $v_{35t}$ is practical,
then $\{v_n\} $ contains infinitely many practical numbers.
\endproclaim
\demo{Proof}
We shall prove by induction that, for every $k\ge0,$ $v_{3^k35t}$ is practical.
For $k=0$ this is true by assumption. Suppose that $v_{3^k35t}$ is practical
for some $k.$ Since $v_n=\alpha^n+\beta^n,$ where $\alpha$ and $\beta$ are
the roots of the polynomial $x^2-Px+Q,$ we have
$$v_{3^{k+1}35t}=v_{3^k35t}\left(\alpha^{3^k70t}-
\alpha^{3^k35t}\beta^{3^k35t}+\beta^{3^k70t}\right).$$
Define $$\Phi_d(x,y)\,=\left\{\alignedat2
&x^{\varphi(d)}\phi_d(y/x)\quad&&\text{if }x\ne0\\ \vspace{0.1cm}
&0&&\text{if }x=y=0\\ &y^{\varphi(d)}\phi_d(x/y)&&\text{if }y\ne0,
\endalignedat\right.$$ where
$\phi_d$ is the $d$-th cyclotomic polynomial and $\varphi$ is the Euler totient
function. Note that $\,x^{\varphi(d)}\phi_d(y/x)=y^{\varphi(d)}\phi_d(x/y)\,$
if $x\ne0$ and $y\ne0.$

Since $\,x^{70}-x^{35}y^{35}+y^{70}=\Phi_6(x,y)\,\Phi_{30}(x,y)\,\Phi_{42}(x,y)
\,\Phi_{210}(x,y),\,$ we have
$$v_{3^{k+1}35t}=v_{3^k35t}\,\Phi_6\bigl(\alpha^{3^kt},\beta^{3^kt}\bigr)
\,\Phi_{30}\bigl(\alpha^{3^kt},\beta^{3^kt}\bigr)
\,\Phi_{42}\bigl(\alpha^{3^kt},\beta^{3^kt}\bigr)
\,\Phi_{210}\bigl(\alpha^{3^kt},\beta^{3^kt}\bigr).$$
Note that, since $Q=-1,$
$$\alignat2
&\Phi_6\bigl(\alpha^{3^kt},\beta^{3^kt}\bigr)&&=\ v_{3^k2t}-(-1)^t\\
&\Phi_{30}\bigl(\alpha^{3^kt},\beta^{3^kt}\bigr)&&=\ v_{3^k8t}
+(-1)^tv_{3^k6t}-(-1)^tv_{3^k2t}-1\\
&\Phi_{42}\bigl(\alpha^{3^kt},\beta^{3^kt}\bigr)&&=\ v_{3^k12t}
+(-1)^tv_{3^k10t}-(-1)^tv_{3^k6t}-v_{3^k4t}+1\\
&\Phi_{210}\bigl(\alpha^{3^kt},\beta^{3^kt}\bigr)\ &&=\ v_{3^k48t}
-(-1)^tv_{3^k46t}+v_{3^k44t}+(-1)^tv_{3^k38t}-v_{3^k36t}\\
&&&+2(-1)^tv_{3^k34t}-v_{3^k32t}+(-1)^tv_{3^k30t}+v_{3^k24t}
-(-1)^tv_{3^k22t}\\
&&&+v_{3^k20t}-(-1)^tv_{3^k18t}+v_{3^k16t}-(-1)^tv_{3^k14t}-v_{3^k8t}
-v_{3^k4t}-1.\endalignat$$
Since $P>0$ and $Q=-1,$ for every $n>0$ we have $v_n<v_{n+1},$ whence
$$\alignat2
0<v_{3^k2t}-1&\le\,\Phi_6\bigl(\alpha^{3^kt},\beta^{3^kt}\bigr)&&\le v_{3^k2t}
+1<v_{3^k35t},\\
0<v_{3^k8t}-v_{3^k6t}+v_{3^k2t}-1&\le\Phi_{30}\bigl(\alpha^{3^kt},
\beta^{3^kt}\bigr)&&\le v_{3^k8t}+v_{3^k6t}<v_{3^k35t},\\
0<v_{3^k12t}-v_{3^k10t}+v_{3^k6t}-v_{3^k4t}&\le
\Phi_{42}\bigl(\alpha^{3^kt},\beta^{3^kt}\bigr)&&\le
v_{3^k12t}+v_{3^k10t}+1<v_{3^k35t}.\endalignat$$
Since $v_{3^{k+1}35t},$ $v_{3^k35t},$ $\Phi_6\bigl(\alpha^{3^kt},
\beta^{3^kt}\bigr),$ $\Phi_{30}\bigl(\alpha^{3^kt},\beta^{3^kt}\bigr),$
$\Phi_{42}\bigl(\alpha^{3^kt},\beta^{3^kt}\bigr)$ are positive integers, we
have $\Phi_{210}\bigl(\alpha^{3^kt},\beta^{3^kt}\bigr)>0,$ and it is easy
to show that $\Phi_{210}\bigl(\alpha^{3^kt},\beta^{3^kt}\bigr)<2v_{3^k48t}.$
By Lemma 4, we have that $$m=v_{3^k35t}\,\Phi_6\bigl(\alpha^{3^kt},
\beta^{3^kt}\bigr)\,\Phi_{30}\bigl(\alpha^{3^kt},\beta^{3^kt}\bigr)
\,\Phi_{42}\bigl(\alpha^{3^kt},\beta^{3^kt}\bigr)$$ is a practical number.
Since $v_{3^{k+1}35t}=m\,\Phi_{210}\bigl(\alpha^{3^kt},\beta^{3^kt}\bigr),$
to complete the proof it suffices to show that $2v_{3^k48t}\le2m,$ and this
can be proved by straightforward and tedious calculations that we omit.\qed
\enddemo

The Fibonacci sequence $\{u_n(1,-1)\}$ and the Pell sequence $\{u_n(2,-1)\}$
satisfy the assumptions of Theorem 10. Since $L_{630}=v_{35\cdot18}(1,-1)$
is a practical number, the Lucas sequence $\{v_n(1,-1)\}$ satisfies the
assumptions of Theorem 11. Therefore there exist infinitely many practical
Fibonacci, Pell and Lucas numbers.

It is interesting to note that the first practical Fibonacci numbers are
$F_3,$ $F_6,$ $F_{12},$ $F_{24},$ $F_{30},$ $F_{36},$ $F_{42},$ $F_{48},$
which, except for $F_3,$ have practical subscripts. It is well known that
every prime Fibonacci number, except for $F_4,$ has a prime subscript \cite4,
but there exist some practical Fibonacci numbers with non-practical subscripts.
The smallest such number is $F_{444}.$ In fact, $444=2^2\cdot3\cdot37$ is not
practical, but $$\align
F_{444}=\ &2^4\cdot3^2\cdot73\cdot149\cdot443\cdot2221\cdot4441\cdot11987
\cdot1121101\cdot54018521\cdot55927129\\
&\cdot6870470209\cdot8336942267\cdot81143477963\cdot1459000305513721\endalign$$
is a practical number.
\bigskip
 
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\enddocument