# Perfectly packing a square by squares of sidelength $f(n)^{-t}$

Keiju Sono

###### Abstract

In this paper, we prove that for any $1/2<t<1$, there exists a positive
integer $N_{0}$ depending on $t$ such that for any $n_{0}\geq N_{0}$, squares
of sidelength $f(n)^{-t}$ for $n\geq n_{0}$ can be packed with disjoint
interiors into a square of area $\sum_{n=n_{0}}^{\infty}f(n)^{-2t}$, if the
function $f$ satisfies some suitable conditions. The main theorem (Theorem
1.1) is a generalization of Tao’s theorem in [15], which argued the case
$f(n)=n$. As corollaries, we prove that there are such packings of squares
when $f(n)$ represents the $n$th element of either an arithmetic progression
or the set of prime numbers. In these cases, we give effective lower bounds
for $N_{0}$ with respect to $t$. Furthermore, we consider the case that $f(n)$
represents the $n$th element of the set of twin primes and prove that squares
of sidelength $f(n)^{-t}$ for $n\geq n_{0}$ can be packed with disjoint
interiors into a slightly larger square than theoretically expected.

0002020 Mathematics Subject Classification. 52C15 000Key Words and Phrases.
square packing in a square, perfectly packing, Meir-Moser conjecture

##  1 Introduction

Let $\Omega$ be a region in $\mathbb{R}^{n}$. It is said that a finite or
countably infinite number of squares in $\mathbb{R}^{n}$ are packed into
$\Omega$ if they are included in $\Omega$ with disjoint interiors. If the
region $\Omega$ is covered by these squares up to null subsets, then we say
that $\Omega$ is packed perfectly by these squares. We also call this
situation a perfectly packing of $\Omega$ by squares.

Meir and Moser [12] presented the problem of whether rectangles of dimension
$\frac{1}{n}\times\frac{1}{n+1}$ for $n\geq 1$ can be packed perfectly into a
square of area $1$. They also posed the question whether squares of sidelength
$n^{-1}$ for $n\geq 2$ can be packed perfectly into a square of area
$\frac{\pi^{2}}{6}-1$. Currently these questions are still unsolved. One of
the best current results on these problems is the work of Paulhus [13], who
obtained an algorithm for packing squares of sidelength
$\frac{1}{n}\times\frac{1}{n+1}$ for $n\geq 1$ with disjoint interiors into a
square of area $1+\frac{1}{10^{9}+1}$, and that for packing squares of
sidelength $n^{-1}$ for $n\geq 2$ with disjoint interiors into a square of
area $\frac{\pi^{2}}{6}-1+\frac{1}{1244918662}$. There were some incorrect
lemmas in Paulhus’ paper. However, Grzegorek and Januszewski [3] modified
Paulhus’ argument and gave completely correct proofs.

Another approach for the second question of Meir and Moser is to consider
whether squares of sidelength $n^{-t}$ for $n\geq 2$ can be packed with
disjoint interiors into a square of area $\zeta(2t)-1$ for $t>1/2$. The goal
is to prove that there exists such a packing for $t=1$. Januszewski and
Zielonka [7] proved that there is a proper packing for $1/2<t\leq 2/3$ by
combining ideas in [2], [8], [16] and Paulhus’ algorithm [13] for packing
squares. They also considered the 3-dimensional case, i.e., the problem of
perfectly packing a cube by cubes. Furthermore, there are several papers
considering perfect packing of the $d$-dimensional cubes of sidelength
$1^{-t}$, $2^{-t}$, $3^{-t},\ldots$ for special values of $t$ (see [6] and
[9], for example).

Recently Tao [15] conditionally improved their work. He proved that for any
$1/2<t<1$, there exists a large positive integer $N_{0}$ depending on $t$ such
that for any integer $n_{0}\geq N_{0}$, squares of sidelength $n^{-t}$ for
$n\geq n_{0}$ can be packed perfectly into a square of area
$\sum_{n=n_{0}}^{\infty}n^{-2t}$. His new idea is to arrange the squares in
near-lattice formations, which enables us to take $t$ arbitrarily close to
$1$. He also remarked that his method will also be applicable to the problem
of packing squares of dimension $\frac{1}{n^{t}}\times\frac{1}{(n+1)^{t}}$
into a square of area $\sum_{n=n_{0}}^{\infty}\frac{1}{n^{t}(n+1)^{t}}$. More
recently, McClenagan [11] obtained the 3-dimensional analogue of Tao’s result.
His theorem is also an improvement of the result of Januszewski and Zielonka
[7] mentioned above.

In this paper, for $q>r\geq 0$, we call the sequence $\\{(qn+r)^{-1}\\}$ an
AP-harmonic sequence, and call the sequence of reciprocals of prime numbers
the P-harmonic sequence. Furthermore, we call the sequence of reciprocals of
twin primes the TP-harmonic sequence.

We use a non-standard notation to estimate the size of functions effectively.
When two functions $F$, $G$ satisfy $|F(x)|\leq G(x)$ over some specified
range, we denote this relation by $F(x)=O_{1}(G(x))$.

The purpose of this paper is to generalize Tao’s theorem to the problem of
packing of squares of sidelength $f(n)^{-t}$ for $1/2<t<1$ and give some
effective lower bounds for the size of $N_{0}$ depending on $t$ in some
specific cases. The result includes (an effective version of) Tao’s theorem.
We will prove the main theorem below in Section 2. In Section 3, as a
corollary of Theorem 1.1, we will prove a result on perfectly packing a square
by squares of nearly AP-harmonic sidelength. In Section 4, we will prove a
similar result in the case of packing of squares of nearly P-harmonic
sidelength. Finally, in Section 5, we will give an ineffective version of
Theorem 1.1 and give a packing of squares of nearly TP-harmonic sidelength
into a slightly larger square than the expected size.

######  Theorem 1.1.

Let $1/2<t<1$ and put $\delta:=1-t$. Suppose that there exist positive
constants $c_{i},d_{i}(i=1,2),c_{\lambda,\nu},l,K$, functions
$\nu_{i},\lambda_{i}:\mathbb{N}\to\mathbb{R}_{>0}$ $(i=1,2)$, positive
functions $\xi_{i}$, $\eta_{i}$ on the open interval $(1/2,1)$ $(i=1,2)$,
positive integers $M$, $N_{0}$, $N_{1}$ $(N_{1}<N_{0})$ and a smooth
increasing function $f:\mathbb{R}_{\geq 1}\to\mathbb{R}_{>0}$ such that for
any positive integers $n_{0}\geq N_{0}$, $n_{1}\geq N_{1}$, the following ten
conditions hold.

$f(n_{0})\geq Kf((1+l)n_{0}),\quad K\geq(2/3)^{\frac{1}{2}},$ (1.1)

$c_{1}\frac{\xi_{1}(t)\nu_{1}(n_{1})}{f(n_{1})^{t+\delta
t}}\leq\sum_{n=1}^{n_{1}-1}\frac{1}{f(n)^{t+\delta t}}\leq
c_{2}\frac{\xi_{2}(t)\nu_{2}(n_{1})}{f(n_{1})^{t+\delta t}},$ (1.2)

$d_{1}\frac{\eta_{1}(t)\lambda_{1}(n_{1})}{f(n_{1})^{2t}}\leq\sum_{n=n_{1}}^{\infty}\frac{1}{f(n)^{2t}}\leq
d_{2}\frac{\eta_{2}(t)\lambda_{2}(n_{1})}{f(n_{1})^{2t}},$ (1.3)

$\lambda_{1}(n_{0})\geq c_{\lambda,\nu}\nu_{2}(n_{0}),$ (1.4)

$\lambda_{2}(n_{0})\leq d_{2}^{-1}\eta_{2}(t)^{-1}f(n_{0})^{2t},$ (1.5)

$M\geq\max\left\\{4^{\frac{1}{1-t}}\left(\frac{c_{2}\xi_{2}(t)}{c_{\lambda,\nu}d_{1}\eta_{1}(t)}\right)^{\frac{2}{t(1-t)}},\left(\frac{50}{K^{t(2-t)}}\right)^{\frac{2}{1-t}}\right\\},$
(1.6)

$\underset{x\in[n_{0},n_{0}+9M^{2}]}{\max}f^{{}^{\prime}}(x)\leq\frac{1}{4752}M^{-4}f(n_{0}),$
(1.7)

$f(n_{0}+9M^{2})\leq(264M^{2})^{\frac{1}{t}}f(n_{0}),$ (1.8)

$9Mf(n_{0})^{t}\leq ln_{0},$ (1.9)

$\nu_{1}(n_{0})\lambda_{2}(n_{0})^{-\frac{1+\delta}{2}}\geq\frac{(d_{2}\eta_{2}(t))^{\frac{1+\delta}{2}}}{c_{1}\xi_{1}(t)}M^{1-\frac{\delta}{2}}.$
(1.10)

Then, for any $n_{0}\geq N_{0}$, squares of sidelength $f(n)^{-t}$ for $n\geq
n_{0}$ can be packed perfectly into a square of area
$\sum_{n=n_{0}}^{\infty}f(n)^{-2t}$.

##  2 Proof of the main theorem

Given a rectangle $R$, we define the width $w(R)$ of $R$ to be the smaller of
two sidelengths of $R$, and define the ${\it height}$ $h(R)$ of $R$ to be the
larger one. If $R$ is a square, then $w(R)=h(R)$. The area of $R$ is equal to
$w(R)h(R)$. Given a family of rectangles ${\cal R}$, we define the total area
of ${\cal R}$ by

$\mathrm{area}({\cal R})=\sum_{R\in{\cal R}}w(R)h(R),$

and define its unweighted total perimeter by

$\mathrm{perim}({\cal R})=2\sum_{R\in{\cal R}}(w(R)+h(R)).$

###  2.1 Efficiently packing a small rectangle of bounded eccentricity

######  Proposition 2.1.

Let $1/2<t<1$ and $M$ be a positive integer. Suppose that a smooth increasing
function $f:\mathbb{R}_{\geq 1}\to\mathbb{R}_{>0}$ and any positive integer
$n_{0}\geq N_{0}$ satisfy (1.7) and (1.8). Let $R$ be a rectangle with

$Mf(n_{0})^{-t}\leq w(R)\leq h(R)\leq 3Mf(n_{0})^{-t}.$ (2.1)

Then, there exists a positive integer $n_{0}^{{}^{\prime}}$ satisfying
$M^{2}\leq n_{0}^{{}^{\prime}}-n_{0}\leq 9M^{2}$ such that $R$ is perfectly
packed by squares of sidelength $f(n)^{-t}$ $(n_{0}\leq
n<n_{0}^{{}^{\prime}})$ and a family of rectangles ${\cal R}$ with disjoint
interiors whose total perimeter satisfies

$\mathrm{perim}({\cal R})\leq 25Mf(n_{0})^{-t}$ (2.2)

and the width of each rectangle is at most $2f(n_{0})^{-t}$.

Proof of Proposition 2.1.
We may assume that $R$ is the rectangle $[0,w(R)]\times[0,h(R)]$. Then by
(2.1), there exist positive integers $M_{1},M_{2}$ $(M\leq M_{1}\leq M_{2}\leq
3M)$ satisfying

$M_{1}f(n_{0})^{-t}\leq w(R)<(M_{1}+1)f(n_{0})^{-t},\;M_{2}f(n_{0})^{-t}\leq
h(R)<(M_{2}+1)f(n_{0})^{-t}.$

Put $n_{0}^{{}^{\prime}}:=n_{0}+M_{1}M_{2}$. Then $M^{2}\leq
n_{0}^{{}^{\prime}}-n_{0}\leq 9M^{2}$. Rewrite the set $\\{n\;|\;n_{0}\leq
n<n_{0}+M_{1}M_{2}\\}$ as $\\{n_{i,j}\;|\;0\leq i<M_{1},0\leq j<M_{2}\\}$,
where

$n_{i,j}:=n_{0}+jM_{1}+i.$

Let $S_{i,j}$ be a square of sidelength $f(n_{i,j})^{-t}$. Following the idea
in [15], we arrange this square by

$S_{i,j}=[x_{i,j},x_{i,j}+f(n_{i,j})^{-t}]\times[y_{i,j},y_{i,j}+f(n_{i,j})^{-t}],$
(2.3)

where

$x_{i,j}:=w(R)-\sum_{i^{{}^{\prime}}=i}^{M_{1}}f(n_{i^{{}^{\prime}},j})^{-t},$
(2.4)
$y_{i,j}:=\sum_{j^{{}^{\prime}}=0}^{j-1}f(n_{i,j^{{}^{\prime}}})^{-t}\;(j\geq
1),\quad y_{i,0}:=0$ (2.5)

for $1\leq i<M_{1},1\leq j<M_{2}$. We need some analysis on $f$. By mean value
theorem,

$f(n_{i,j})=f(n_{0}+jM_{1}+i)=f(n_{0})+f^{{}^{\prime}}(c)(jM_{1}+i)$

for some $c\in(n_{0},n_{0}+jM_{1}+i)$. Notice that $n_{0}+jM_{1}+i\leq
n_{0}+9M^{2}$. We denote the maximum value of $f^{{}^{\prime}}(x)$ in
$[n_{0},n_{0}+9M^{2}]$ by ${\cal D}_{n_{0},M}$. Then

$\displaystyle f(n_{i,j})=$ $\displaystyle f(n_{0})+O_{1}(9M^{2}{\cal
D}_{n_{0},M})$ $\displaystyle=$ $\displaystyle
f(n_{0})\left(1+O_{1}\left(\frac{9M^{2}{\cal
D}_{n_{0},M}}{f(n_{0})}\right)\right).$

By (1.7), we have

$\frac{9M^{2}{\cal D}_{n_{0},M}}{f(n_{0})}<\frac{1}{100}.$ (2.6)

Hence

$f(n_{i,j})^{-t}=f(n_{0})^{-t}\left(1+O_{1}\left(\frac{9M^{2}{\cal
D}_{n_{0},M}}{f(n_{0})}\right)\right)$

uniformly for $1/2<t<1$. Then by (2.4) and (2.5), it follows that

$x_{i,j}=w(R)-f(n_{0})^{-t}\left(M_{1}-i+1+O_{1}\left(\frac{27M^{3}{\cal
D}_{n_{0},M}}{f(n_{0})}\right)\right),$ (2.7)
$y_{i,j}=f(n_{0})^{-t}\left(j+O_{1}\left(\frac{27M^{3}{\cal
D}_{n_{0},M}}{f(n_{0})}\right)\right)$ (2.8)

for $0\leq i<M_{1},0\leq j<M_{2}$.

######  Lemma 2.2.

$(1)$ $S_{i,j}(0\leq i<M_{1},0\leq j<M_{2})$ are included in $R$.
$(2)$ If $(i,j)\neq(i^{{}^{\prime}},j^{{}^{\prime}})$, the interiors of
$S_{i,j}$ and $S_{i^{{}^{\prime}},j^{{}^{\prime}}}$ are disjoint.

###### Proof.

(1) Since $f$ is increasing, we see that

$0\leq w(R)-M_{1}f(n_{0})^{-t}\leq x_{i,j}\leq x_{i,j}+f(n_{i,j})^{-t}\leq
w(R),$ $0\leq y_{i,j}\leq y_{i,j}+f(n_{i,j})^{-t}\leq M_{2}f(n_{0})^{-t}\leq
h(R).$

Therefore, $S_{i,j}$ is included in $R$.
(2) If $i\geq i^{{}^{\prime}}$, $j<j^{{}^{\prime}}$, then
$y_{i,j}+f(n_{i,j})^{-t}\leq y_{i,j^{{}^{\prime}}}\leq
y_{i^{{}^{\prime}},j^{{}^{\prime}}}$. Hence the interior of $S_{i,j}$ lies
bottom to the interior of $S_{i^{{}^{\prime}},j^{{}^{\prime}}}$. Similarly if
$i^{{}^{\prime}}\geq i$, $j^{{}^{\prime}}<j$.
If $i<i^{{}^{\prime}}$, $j\leq j^{{}^{\prime}}$, then
$x_{i,j}+f(n_{i,j})^{-t}\leq x_{i^{{}^{\prime}},j}\leq
x_{i^{{}^{\prime}},j^{{}^{\prime}}}$. Hence the interior of $S_{i,j}$ lies
left to the interior of $S_{i^{{}^{\prime}},j^{{}^{\prime}}}$. Similarly if
$i^{{}^{\prime}}<i$, $j^{{}^{\prime}}\leq j$. ∎

######  Lemma 2.3.

Suppose

$9M^{2}\underset{0\leq x\leq
9M^{2}}{\max}f^{{}^{\prime}}(n_{0}+x)f(n_{0}+x)^{-t-1}\leq
f(n_{0}+9M^{2})^{-t}.$ (2.9)

Then, for $0\leq i<M_{1}-1$ and $0\leq j<M_{2}-1$, the squares $S_{i,j}$,
$S_{i+1,j}$, $S_{i,j+1}$ and $S_{i+1,j+1}$ surround a rectangle

$[x_{i+1,j},x_{i+1,j+1}]\times[y_{i+1,j+1},y_{i,j+1}].$ (2.10)

(0,0)($w(R)$,0)(0,$h(R)$)($w(R)$,$h(R)$)$S_{i,j}$$S_{i+1,j}$$S_{i+1,j+1}$$S_{i,j+1}$

###### Proof.

This fact follows from the following equalities and inequalities.

$\displaystyle x_{i+1,j}=x_{i,j}+f(n_{i,j})^{-t},$
$\displaystyle\underline{y_{i,j}<y_{i+1,j}+f(n_{i+1,j})^{-t}}<y_{i,j}+f(n_{i,j})^{-t},$
$\displaystyle x_{i,j}<\underline{x_{i,j+1}<x_{i,j}+f(n_{i,j})^{-t}}$
$\displaystyle y_{i,j+1}=y_{i,j}+f(n_{i,j})^{-t}$ $\displaystyle
x_{i+1,j+1}=x_{i,j+1}+f(n_{i,j+1})^{-t},$ $\displaystyle
y_{i+1,j+1}<\underline{y_{i,j+1}<y_{i+1,j+1}+f(n_{i+1,j+1})^{-t}},$
$\displaystyle x_{i+1,j}<\underline{x_{i+1,j+1}<x_{i+1,j}+f(n_{i+1,j})^{-t}},$
$\displaystyle y_{i+1,j+1}=y_{i+1,j}+f(n_{i+1,j})^{-t}.$

Among these relations, the underlined four (essentially two) inequalities are
nontrivial, and others easily follow from the definitions of $x_{i,j}$ and
$y_{i,j}$. By definition,

$y_{i,j}-y_{i+1,j}=\sum_{j^{{}^{\prime}}=0}^{j-1}\left(f(n_{0}+j^{{}^{\prime}}M_{1}+i)^{-t}-f(n_{0}+j^{{}^{\prime}}M_{1}+i+1)^{-t}\right).$

By mean value theorem, we have

$\displaystyle
f(n_{0}+j^{{}^{\prime}}M_{1}+i)^{-t}-f(n_{0}+j^{{}^{\prime}}M_{1}+i+1)^{-t}$
$\displaystyle\quad=-\frac{\partial}{\partial
u}f(n_{0}+j^{{}^{\prime}}M_{1}+u)^{-t}|_{u=c_{i,j^{{}^{\prime}}}}$
$\displaystyle\quad=tf^{{}^{\prime}}(n_{0}+j^{{}^{\prime}}M_{1}+c_{i,j^{{}^{\prime}}})f(n_{0}+j^{{}^{\prime}}M_{1}+c_{i,j^{{}^{\prime}}})^{-t-1}$

for some $c_{i,j^{{}^{\prime}}}\in(i,i+1)$. Hence

$\displaystyle y_{i,j}-y_{i+1,j}=$ $\displaystyle
t\sum_{j^{{}^{\prime}}=0}^{j-1}f^{{}^{\prime}}(n_{0}+j^{{}^{\prime}}M_{1}+c_{i,j^{{}^{\prime}}})f(n_{0}+j^{{}^{\prime}}M_{1}+c_{i,j^{{}^{\prime}}})^{-t-1}$
$\displaystyle\leq 3M\underset{0\leq x\leq
9M^{2}}{\max}f^{{}^{\prime}}(n_{0}+x)f(n_{0}+x)^{-t-1}.$

On the other hand, since $f$ is increasing, $f(n_{i+1,j})^{-t}$ is equal or
larger than
$f(n_{0}+9M^{2})^{-t}$. Hence if

$3M\underset{0\leq x\leq
9M^{2}}{{\max}}f^{{}^{\prime}}(n_{0}+x)f(n_{0}+x)^{-t-1}\leq
f(n_{0}+9M^{2})^{-t}$ (2.11)

holds, then the first and third underlined inequalities hold. Moreover, the
condition (2.11) is weaker than (2.9). Hence the first and third underlined
inequalities follow from (2.9). We now prove the second and fourth
inequalities assuming (2.9). Again by definition,

$x_{i,j+1}-x_{i,j}=\sum_{i^{{}^{\prime}}=i}^{M_{1}}\left(f(n_{0}+jM_{1}+i^{{}^{\prime}})^{-t}-f(n_{0}+(j+1)M_{1}+i^{{}^{\prime}})^{-t}\right).$

By mean value theorem,

$\displaystyle
f(n_{0}+jM_{1}+i^{{}^{\prime}})^{-t}-f(n_{0}+(j+1)M_{1}+i^{{}^{\prime}})^{-t}$
$\displaystyle\quad=-\frac{\partial}{\partial
u}f(n_{0}+uM_{1}+i^{{}^{\prime}})^{-t}|_{u=c_{i^{{}^{\prime}},j}}$
$\displaystyle\quad=tM_{1}f^{{}^{\prime}}(n_{0}+c_{i^{{}^{\prime}},j}M_{1}+i^{{}^{\prime}})f(n_{0}+c_{i^{{}^{\prime}},j}M_{1}+i^{{}^{\prime}})^{-t-1}.$

for some $c_{i^{{}^{\prime}},j}\in(j,j+1)$. Hence by (2.9),

$\displaystyle x_{i,j+1}-x_{i,j}=$ $\displaystyle
tM_{1}\sum_{i=i^{{}^{\prime}}}^{M_{1}}f^{{}^{\prime}}(n_{0}+c_{i^{{}^{\prime}},j}M_{1}+i^{{}^{\prime}})f(n_{0}+c_{i^{{}^{\prime}},j}M_{1}+i^{{}^{\prime}})^{-t-1}$
$\displaystyle\leq$ $\displaystyle 9M^{2}\underset{0\leq x\leq
9M^{2}}{{\max}}f^{{}^{\prime}}(n_{0}+x)f(n_{0}+x)^{-t-1}\leq
f(n_{0}+9M^{2})^{-t}.$

On the other hand, $f(n_{i,j})^{-t}$ is equal or larger than
$f(n_{0}+9M^{2})^{-t}$. Thus we obtain the second and fourth underlined
inequalities under the assumption of (2.9). ∎

By (1.7), the left hand side of (2.9) is at most $(1/264M^{2})f(n_{0})^{-t}$.
Hence (2.9) is satisfied if (1.8) holds. Summing up, the rectangle $R$ is
perfectly packed by the following five types of rectangles:
Type I. The squares $S_{i,j}$ $(0\leq i<M_{1},0\leq j<M_{2})$.
Type II. The rectangles (2.10) surrounded by Type I squares. By (2.7) and
(2.8), the perimeters of these surrounded rectangles are
$O_{1}\left(\frac{216M^{3}{\cal D}_{n_{0},M}}{f(n_{0})^{1+t}}\right)$ and the
number of these rectangles is at most $9M^{2}$.
Type III. The rectangles $[0,x_{0,j}]\times[y_{0,j},y_{0,j}+f(n_{0,j})^{-t}]$
$(0\leq j<M_{2})$. By (2.7), the perimeters of these rectangles are at most
$4f(n_{0})^{-t}+\frac{54M^{3}{\cal D}_{n_{0},M}}{f(n_{0})^{t+1}}$ and the
number of these rectangles is at most $3M$.
Type IV. The rectangles
$[x_{i,M_{2}-1},x_{i,M_{2}-1}+f(n_{i,M_{2}-1})^{-t}]\times[y_{i,M_{2}-1}+f(n_{i,M_{2}-1})^{-t},h(R)]$
$(0\leq i<M_{1})$. The perimeters of these rectangles are at most
$4f(n_{0})^{-t}+\frac{54M^{3}{\cal D}_{n_{0},M}}{f(n_{0})^{t+1}}$, and the
number of these rectangles is at most $3M$.
Type V. The rectangle
$[0,x_{0,M_{2}-1}]\times[y_{0,M_{2}-1}+f(n_{0,M_{2}-1})^{-t},h(R)]$. The
perimeter of this rectangle is at most $4f(n_{0})^{-t}+\frac{108M^{3}{\cal
D}_{n_{0},M}}{f(n_{0})^{1+t}}$.
Put ${\cal R}:=R\backslash\\{S_{i,j}\\}_{i,j}$. By the above argument, it
follows that the widths of the rectangles contained in ${\cal R}$ exceed
neither $54M^{3}{\cal D}_{n_{0},M}f(n_{0})^{-t-1}$ nor
$f(n_{0})^{-t}+27M^{3}{\cal D}_{n_{0},M}f(n_{0})^{-t-1}$, so the widths are at
most $2f(n_{0})^{-t}$ by (1.7). Moreover, we see that the total perimeter of
${\cal R}$ is

$\displaystyle\mathrm{perim}({\cal R})\leq$ $\displaystyle\frac{216M^{3}{\cal
D}_{n_{0},M}}{f(n_{0})^{1+t}}\times
9M^{2}+\left(4f(n_{0})^{-t}+\frac{54M^{3}{\cal
D}_{n_{0},M}}{f(n_{0})^{1+t}}\right)\times 6M$ (2.12)
$\displaystyle\quad+\left(4f(n_{0})^{-t}+\frac{108M^{3}{\cal
D}_{n_{0},M}}{f(n_{0})^{1+t}}\right)$ $\displaystyle\leq$
$\displaystyle\frac{2376M^{5}{\cal
D}_{n_{0},M}}{f(n_{0})^{t+1}}+\frac{49}{2}Mf(n_{0})^{-t}$ $\displaystyle\leq$
$\displaystyle 25Mf(n_{0})^{-t}.$

Thus the proof of Proposition 2.1 is completed. $\Box$

###  2.2 Perfectly packing some families of rectangles

Put $\delta=1-t$ and define the weighted total perimeter of ${\cal R}$ by

$\mathrm{perim}_{\delta}({\cal R})=\sum_{R\in{\cal R}}w(R)^{\delta}h(R).$

######  Proposition 2.4.

Suppose that $M$, $f$, $K$, $l$ and $1/2<t<1$ satisfy conditions (1.1), (1.2),
(1.3), (1.4) and (1.6) for any $n_{0}\geq N_{0}$, $n_{1}\geq N_{1}$. Fix
positive integers $n_{\max}\geq n_{0}(\geq N_{0})$. Let ${\cal R}$ be a family
of finite number of rectangles with disjoint interiors satisfying

$\mathrm{area}({\cal R})=\sum_{n=n_{0}}^{\infty}\frac{1}{f(n)^{2t}},$ (2.13)
$\mathrm{perim}_{\delta}({\cal R})\leq
M^{-1+\frac{\delta}{2}}\sum_{n=1}^{n_{0}-1}\frac{1}{f(n)^{t+\delta t}}$ (2.14)

and

$\sup_{R\in{\cal R}}h(R)\leq 1.$ (2.15)

Then, squares of sidelength $f(n)^{-t}$ $(n_{0}\leq n<n_{\max})$ can be packed
into $\bigcup_{R\in{\cal R}}R$.

###### Proof.

Fix $n_{\max}$ and prove the statement by a downward induction on $n_{0}$. The
above proposition is trivial if $n_{\max}=n_{0}$. So we suppose that
$n_{\max}>n_{0}$ and that the claim has already been proved for larger
$n_{0}$. Then, by (2.14) and (1.2), we have

$\sum_{R\in{\cal R}}w(R)^{\delta}h(R)\leq
M^{-1+\frac{\delta}{2}}\frac{c_{2}\xi_{2}(t)\nu_{2}(n_{0})}{f(n_{0})^{t+\delta
t}}.$ (2.16)

On the other hand, by (2.13) and (1.3), we have

$\sum_{R\in{\cal
R}}w(R)h(R)\geq\frac{d_{1}\eta_{1}(t)\lambda_{1}(n_{0})}{f(n_{0})^{2t}}.$
(2.17)

Hence by the pigeonhole principle, there exists a rectangle $R\in{\cal R}$
that satisfies

$w(R)^{1-\delta}\geq\frac{d_{1}\eta_{1}(t)\lambda_{1}(n_{0})f(n_{0})^{-2t}}{M^{-1+\frac{\delta}{2}}c_{2}\xi_{2}(t)\nu_{2}(n_{0})f(n_{0})^{-t-\delta
t}}.$

Therefore, we have

$\displaystyle w(R)$
$\displaystyle\geq(M^{1-\frac{\delta}{2}})^{\frac{1}{1-\delta}}\left(\frac{d_{1}\eta_{1}(t)\lambda_{1}(n_{0})}{c_{2}\xi_{2}(t)\nu_{2}(n_{0})}\right)^{\frac{1}{1-\delta}}f(n_{0})^{-t}$
$\displaystyle\geq
M^{1+\frac{\delta}{2}}\left(\frac{d_{1}\eta_{1}(t)\lambda_{1}(n_{0})}{c_{2}\xi_{2}(t)\nu_{2}(n_{0})}\right)^{\frac{1}{t}}f(n_{0})^{-t},$

since $(1-\frac{\delta}{2})/(1-\delta)>1+\frac{\delta}{2}$. Since
$\lambda_{1}$ and $\nu_{2}$ satisfy (1.4), we have

$w(R)\geq
M^{1+\frac{\delta}{2}}\left(\frac{c_{\lambda,\nu}d_{1}\eta_{1}(t)}{c_{2}\xi_{2}(t)}\right)^{\frac{1}{t}}f(n_{0})^{-t}.$

Since $M$ satisfies (1.6), it follows that

$M^{\frac{\delta}{2}}\left(\frac{c_{\lambda,\nu}d_{1}\eta_{1}(t)}{c_{2}\xi_{2}(t)}\right)^{\frac{1}{t}}\geq
2.$

Therefore,

$h(R)\geq w(R)\geq 2Mf(n_{0})^{-t}.$ (2.18)

We cut the rectangle $R$ into a rectangle $R_{0}$ of dimension
$(w(R)-Mf(n_{0})^{-t})\times h(R)$ and a rectangle $R_{*}$ of dimension
$Mf(n_{0})^{-t}\times h(R)$. Then cut off the squares of sidelength
$Mf(n_{0})^{-t}$ from $R_{*}$ until the height of the remaining rectangle is
below $2Mf(n_{0})^{-t}$. Thus $R_{*}$ is decomposed into rectangles
$R_{1},\ldots,R_{m}$. These rectangles are of dimension $Mf(n_{0})^{-t}\times
h(R_{i})$, where

$Mf(n_{0})^{-t}\leq h(R_{i})<2Mf(n_{0})^{-t}\quad(i=1,\ldots,m)$ (2.19)

with disjoint interiors, and

$\sum_{i=1}^{m}h(R_{i})=h(R).$

Since $h(R)\leq 1$, it follows that

$m\leq M^{-1}f(n_{0})^{t}.$ (2.20)

This gives a perfectly packing of $R$

$R=R_{1}\cup R_{2}\cup\cdots\cup R_{m}.$

We adapt Proposition 2.1 $m$ times. Then there exists a sequence of positive
integers
$n_{0}=n_{0}^{{}^{\prime}}<n_{1}^{{}^{\prime}}<\cdots<n_{m}^{{}^{\prime}}$
such that

$M^{2}\leq n_{i+1}^{{}^{\prime}}-n_{i}^{{}^{\prime}}\leq 9M^{2}\quad(0\leq
i\leq m-1)$ (2.21)

and each $R_{i}$ is perfectly packed by squares of sidelength $f(n)^{-t}$
$(n_{i-1}^{{}^{\prime}}\leq n<n_{i}^{{}^{\prime}})$ and a family ${\cal
R}_{i}$ of finite number of rectangles with disjoint interiors satisfying

$\mathrm{perim}({\cal R}_{i})\leq 25Mf(n_{0})^{-t}$ (2.22)

and the width of each rectangle in ${\cal R}_{i}$ is at most $2f(n_{0})^{-t}$,
provided that the height of $R_{i}$ satisfies

$Mf(n_{i-1}^{{}^{\prime}})^{-t}\leq h(R_{i})\leq
3Mf(n_{i-1}^{{}^{\prime}})^{-t}$ (2.23)

for $i=1,\ldots,m$. Let us confirm that (2.23) is satisfied. Since
$Mf(n_{i-1}^{{}^{\prime}})^{-t}\leq Mf(n_{0})^{-t}$,
$Mf(n_{i-1}^{{}^{\prime}})^{-t}\leq h(R_{i})$ is satisfied. By (2.20) and
(2.21),

$n_{i}^{{}^{\prime}}\leq n_{0}+9Mf(n_{0})^{t}.$

Combining this with (1.9), we have

$n_{0}\leq n_{i}^{{}^{\prime}}\leq(1+l)n_{0}$ (2.24)

for $i=1,\ldots,m$. Since $h(R_{i})$ satisfies (2.19), to prove $h(R_{i})\leq
3Mf(n_{i-1}^{{}^{\prime}})^{-t}$, it suffices to see $2Mf(n_{0})^{-t}\leq
3Mf(n_{i-1}^{{}^{\prime}})^{-t}$, which is equivalent to
$f(n_{0})\geq(2/3)^{\frac{1}{t}}f(n_{i-1}^{{}^{\prime}})$. For $i=1,\ldots,m$,
these inequalities are valid because of (1.1) and (2.24). Hence (2.23) holds
and the inductive argument can be applied to the rectangles
$R_{1},\ldots,R_{m}$. Put

${\cal R}^{{}^{\prime}}:=({\cal
R}\backslash\\{R\\})\cup\\{R_{0}\\}\cup\bigcup_{i=1}^{m}{\cal R}_{i}.$

By the above argument, $\bigcup_{R^{{}^{\prime}}\in{\cal R}}R^{{}^{\prime}}$
is perfectly packed by squares of sidelength $f(n)^{-t}$, $n_{0}\leq
n<n_{m}^{{}^{\prime}}$ and rectangles in ${\cal R}^{{}^{\prime}}$. If
$n_{m}^{{}^{\prime}}\geq n_{\max}$, the induction process is completed. So we
assume $n_{m}^{{}^{\prime}}<n_{\max}$. Before the evaluation of
$\textrm{perim}_{\delta}({\cal R}^{{}^{\prime}})$, we see that for
$i=1,\ldots,m$,

$f(n_{0})^{-t-\delta t}\leq M^{-2}K^{-t-\delta
t}\sum_{n=n_{i-1}^{{}^{\prime}}}^{n_{i}^{{}^{\prime}}-1}f(n)^{-t-\delta t}$
(2.25)

holds, because $f(n_{0})\geq Kf(n)$ for $n_{i-1}^{{}^{\prime}}\leq
n<n_{i}^{{}^{\prime}}-1$. By (2.22), (2.14), (2.25) and
$w(R^{{}^{\prime}})\leq 2f(n_{0})^{-t}$ for $R^{{}^{\prime}}\in{\cal R}_{i}$,
it follows that

$\displaystyle\mathrm{perim}_{\delta}({\cal R}^{{}^{\prime}})$
$\displaystyle=\mathrm{perim}_{\delta}({\cal
R})-w(R)^{\delta}h(R)+w(R_{0})^{\delta}h(R)+\sum_{i=1}^{m}\sum_{R^{{}^{\prime}}\in{\cal
R}_{i}}w(R^{{}^{\prime}})^{\delta}h(R^{{}^{\prime}})$
$\displaystyle\leq\mathrm{perim}_{\delta}({\cal
R})+\sum_{i=1}^{m}O_{1}(2^{\delta}f(n_{0})^{-\delta t}\mathrm{perim}({\cal
R}_{i}))$ $\displaystyle\leq\mathrm{perim}_{\delta}({\cal R})+2^{\delta}\cdot
25M\sum_{i=1}^{m}O_{1}(f(n_{0})^{-t-\delta t})$
$\displaystyle\leq\mathrm{perim}_{\delta}({\cal R})+\frac{50}{K^{t+\delta
t}M}\sum_{i=1}^{m}O_{1}\left(\sum_{n=n_{i-1}^{{}^{\prime}}}^{n_{i}^{{}^{\prime}}-1}f(n)^{-t-\delta
t}\right)$ $\displaystyle\leq
M^{-1+\frac{\delta}{2}}\sum_{n=1}^{n_{0}-1}\frac{1}{f(n)^{t+\delta
t}}+\frac{50}{K^{t+\delta
t}M}\sum_{n=n_{0}}^{n_{m}^{{}^{\prime}}-1}\frac{1}{f(n)^{t+\delta t}}.$

Since $M\geq(50/K^{t(2-t)})^{\frac{2}{1-t}}$ by (1.6), we have

$\mathrm{perim}_{\delta}({\cal R}^{{}^{\prime}})\leq
M^{-1+\frac{\delta}{2}}\sum_{n=1}^{n_{m}^{{}^{\prime}}-1}\frac{1}{f(n)^{t+\delta
t}}.$

Therefore, ${\cal R}^{{}^{\prime}}$ satisfies (2.14) with $n_{0}$ replaced
with $n_{m}^{{}^{\prime}}$. Moreover,

$\mathrm{area}({\cal R}^{{}^{\prime}})=\mathrm{area}({\cal
R})-\sum_{n=n_{0}}^{n_{m}^{{}^{\prime}}-1}\frac{1}{f(n)^{2t}}=\sum_{n=n_{m}^{{}^{\prime}}}^{\infty}\frac{1}{f(n)^{2t}}.$

Hence the identity (2.13) with $n_{0}$ replaced with $n_{m}^{{}^{\prime}}$ is
obtained. Finally, the heights of the rectangles in ${\cal R}^{{}^{\prime}}$
are at most $1$. Therefore, by induction hypothesis, squares of sidelength
$f(n)^{-t}$ with $n_{m}^{{}^{\prime}}\leq n<n_{\max}$ can be packed into
$\bigcup_{R^{{}^{\prime}}\in{\cal R}^{{}^{\prime}}}R^{{}^{\prime}}$. On the
other hand, squares of sidelength $f(n)^{-t}$ for $n_{0}\leq
n<n_{m}^{{}^{\prime}}$ have already been packed into ${\cal R}\backslash{\cal
R}^{{}^{\prime}}$. Hence the induction is now completed. ∎

###  2.3 Proof of Theorem 1.1

Suppose that $f$, $t$, $K$, $l$, $M$ and $N_{0}$ satisfy all the conditions in
the statement of Theorem 1.1. Let $S$ be a square of area
$\sum_{n=n_{0}}^{\infty}f(n)^{-2t}$. By (1.3), the sidelength of $S$ is at
most
$d_{2}^{\frac{1}{2}}\eta_{2}(t)^{\frac{1}{2}}\lambda_{2}(n_{0})^{\frac{1}{2}}f(n_{0})^{-t}$.
Therefore,

$\displaystyle\mathrm{perim}_{\delta}(S)$
$\displaystyle\leq(d_{2}^{\frac{1}{2}}\eta_{2}(t)^{\frac{1}{2}}\lambda_{2}(n_{0})^{\frac{1}{2}}f(n_{0})^{-t})^{1+\delta}=(d_{2}\eta_{2}(t)\lambda_{2}(n_{0}))^{\frac{1+\delta}{2}}f(n_{0})^{-(1+\delta)t}.$

On the other hand, by (1.2),

$\sum_{n=1}^{n_{0}-1}\frac{1}{f(n)^{t+\delta t}}\geq
c_{1}\xi_{1}(t)\nu_{1}(n_{0})f(n_{0})^{-(1+\delta)t}.$

Hence the condition (2.14) is satisfied if

$(d_{2}\eta_{2}(t)\lambda_{2}(n_{0}))^{\frac{1+\delta}{2}}f(n_{0})^{-(1+\delta)t}\leq
c_{1}M^{-1+\frac{\delta}{2}}\xi_{1}(t)\nu_{1}(n_{0})f(n_{0})^{-(1+\delta)t},$

and the above inequality is equivalent to (1.10). Furthermore, the height of
$S$ is at most $1$ provided that (1.5) is satisfied. Consequently, if the
functions and parameters satisfy these conditions, one can apply Proposition
2.4 to ${\cal R}=S$ and see that for any $n_{\max}>n_{0}\;(\geq N_{0})$, the
squares of sidelength $f(n)^{-t}$ $(n_{0}\leq n<n_{\max})$ can be packed into
$S$. Since $n_{\max}$ is arbitrary, by sending $n_{\max}\to\infty$ and using a
standard compactness argument (see e.g., [10]), we see that there is a
perfectly packing of squares of sidelength $f(n)^{-t}$ for $n\geq n_{0}$ into
$S$, under the assumptions of Theorem 1.1. Thus the proof of Theorem 1.1 is
completed. $\Box$

##  3 Perfectly packing a square by squares of nearly AP-harmonic sidelength

In this section, we apply Theorem 1.1 to the problem of perfectly packing a
square by squares of sidelength $(qn+r)^{-t}$ for $q>r\geq 0$, $1/2<t<1$.

######  Corollary 3.1.

Let $1/2<t<1$ and $q>r\geq 0$ be fixed constants and put $\delta=1-t$. Suppose
that $M$ and $N_{0}$ are positive integers satisfying

$M\geq\max\left\\{4^{\frac{1}{1-t}}\left(\frac{2(2t-1)}{1-t-\delta
t}\right)^{\frac{2}{t(1-t)}},\left(50\left(\frac{11}{10}\right)^{t(2-t)}\right)^{\frac{2}{1-t}}\right\\},$
$\displaystyle N_{0}\geq\max\biggl{\\{}$ $\displaystyle
4752M^{4},\;\frac{2r}{q((264M^{2})^{\frac{1}{t}}-2)},\;(10q(2q)^{t}M)^{\frac{1}{1-t}},\;2^{\frac{1}{1-t-\delta
t}}\frac{q+r}{q},$ $\displaystyle\quad q^{-1}\left(\frac{q(1-t-\delta
t)}{(q+r)^{t+\delta t}}\right)^{\frac{1}{1-t-\delta
t}},\;\frac{1}{1-2^{-\frac{1}{2t-1}}},$ $\displaystyle q^{-1}(2q(1-t-\delta
t))^{\frac{2}{1-\delta}}\left(\frac{2}{q(2t-1)}\right)^{\frac{1+\delta}{1-\delta}}M^{\frac{2-\delta}{1-\delta}},\;q^{-1}\left(\frac{2}{q(2t-1)}\right)^{\frac{1}{2t-1}}\biggr{\\}}.$

Then, for any positive integer $n_{0}\geq N_{0}$, squares of sidelength
$(qn+r)^{-t}$ for $n\geq n_{0}$ can be packed perfectly into a square of area
$\sum_{n=n_{0}}^{\infty}(qn+r)^{-2t}$.

######  Remark 3.2 (Effective lower bound for $N_{0}$ in the case of nearly
harmonic sidelength).

In the case of $f(n)=n$ (hence $q=1,r=0$), a specific lower bound for $N_{0}$
is given by the following table.

Table 1: Lower bound for $N_{0}$ in the case $f(n)=n$ $t$ | 0.55 | 0.6 | 0.7 | 0.8 | 0.9 | 0.95 | 0.99 | 0.999
---|---|---|---|---|---|---|---|---
$\log_{10}N_{0}$ | 35 | 38 | 50 | 114 | 563 | 2673 | 92863 | 13216295

As $t\nearrow 1$, a sufficient condition for the value of $N_{0}$ is

$N_{0}\geq 4^{\frac{1}{(1-t)^{2}}}(2^{t}\cdot
10)^{\frac{1}{1-t}}\left(\frac{2(2t-1)}{(1-t)^{2}}\right)^{\frac{2}{t(1-t)^{2}}}.$

Roughly speaking, as $t\nearrow 1$, the perfectly packing is possible if
$N_{0}\geq\left(3/(1-t)^{2}\right)^{3/(1-t)^{2}}$.

###### Proof.

All we have to do is to identify when the function $f(x)=qx+r$ satisfies the
whole conditions of Theorem 1.1. The condition (1.7) is

$q\leq\frac{1}{4752M^{4}}(qn_{0}+r),$

which is satisfied if

$n_{0}\geq 4752M^{4}.$ (3.1)

The condition (1.8) is

$(q(n_{0}+9M^{2})+r)\leq(264M^{2})^{\frac{1}{t}}(qn_{0}+r)$

If $n_{0}$ satisfies (3.1), then $n_{0}+9M^{2}\leq 2n_{0}$. Therefore, the
left hand side of the above inequality is at most $2(qn_{0}+r)$. Hence a
sufficient condition for (1.8) is

$2(qn_{0}+r)\leq(264M^{2})^{\frac{1}{t}}(qn_{0}+r),$

which is satisfied when

$n_{0}\geq\frac{2r}{q((264M^{2})^{\frac{1}{t}}-2)}.$ (3.2)

The condition (1.1) is

$qn_{0}+r\geq K(q(1+l)n_{0}+r),$

which holds when

$K\leq\frac{1}{1+l}.$ (3.3)

We choose $l=1/10,\;K=10/11$. Then $K\geq(2/3)^{\frac{1}{2}}$ is also
satisfied. The condition (1.9) is

$9M(qn_{0}+r)^{t}\leq ln_{0}.$

The left hand side is at most $9(2q)^{t}Mn_{0}^{t}$. Therefore, (1.9) holds
for

$n_{0}\geq\left(\frac{9(2q)^{t}M}{l}\right)^{\frac{1}{1-t}}.$ (3.4)

Next, we consider the conditions (1.2), (1.3). Evaluating the summation by
integrals, we have

$\int_{1}^{n_{1}}\frac{dx}{(qx+r)^{t+\delta
t}}\leq\sum_{n=1}^{n_{1}-1}\frac{1}{(qn+r)^{t+\delta
t}}\leq\frac{1}{(q+r)^{t+\delta
t}}+\int_{1}^{n_{1}-1}\frac{dx}{(qx+r)^{t+\delta t}}.$ (3.5)

The left hand side equals

$\frac{1}{q(1-t-\delta t)}\frac{qn_{1}+r}{(qn_{1}+r)^{t+\delta
t}}-\frac{1}{q(1-t-\delta t)}\frac{q+r}{(q+r)^{t+\delta t}},$

which is equal to or larger than

$\frac{1}{2}\frac{1}{q(1-t-\delta t)}\frac{qn_{1}+r}{(qn_{1}+r)^{t+\delta t}}$

if $n_{1}$ satisfies

$n_{1}\geq 2^{\frac{1}{1-t-\delta t}}\frac{q+r}{q}.$ (3.6)

Therefore, for $n_{1}$ satisfying the condition (3.6), the former inequality
of (1.2) holds with

$c_{1}=\frac{1}{2},\quad\xi_{1}(t)=\frac{1}{q(1-t-\delta
t)},\quad\nu_{1}(n_{1})=qn_{1}+r.$

On the other hand, the right hand side of (3.5) is at most

$\frac{1}{(q+r)^{t+\delta t}}+\frac{1}{q(1-t-\delta
t)}\frac{qn_{1}+r}{(qn_{1}+r)^{t+\delta t}}.$

This is equal to or less than

$\frac{2}{q(1-t-\delta t)}\frac{qn_{1}+r}{(qn_{1}+r)^{t+\delta t}}$

if $n_{1}$ satisfies

$n_{1}\geq q^{-1}\left(\frac{q(1-t-\delta t)}{(q+r)^{t+\delta
t}}\right)^{\frac{1}{1-t-\delta t}}.$ (3.7)

Therefore, for $n_{1}$ satisfying the condition (3.7), the latter inequality
of (1.2) holds with

$c_{2}=2,\quad\xi_{2}(t)=\frac{1}{q(1-t-\delta
t)},\quad\nu_{2}(n_{1})=qn_{1}+r.$

Next, we consider (1.3). We have

$\int_{n_{1}}^{\infty}\frac{dx}{(qx+r)^{2t}}\leq\sum_{n=n_{1}}^{\infty}\frac{1}{(qn+r)^{2t}}\leq\int_{n_{1}-1}^{\infty}\frac{dx}{(qx+r)^{2t}}.$

The left hand side equals

$\frac{qn_{1}+r}{q(2t-1)(qn_{1}+r)^{2t}}.$

Hence the former inequality of (1.3) holds with

$d_{1}=1,\quad\eta_{1}(t)=\frac{1}{q(2t-1)},\quad\lambda_{1}(n_{1})=qn_{1}+r.$

The right hand side is

$\frac{1}{q(2t-1)}\frac{q(n_{1}-1)+r}{(q(n_{1}-1)+r)^{2t}},$

which becomes smaller than

$\frac{2}{q(2t-1)}\frac{qn_{1}+r}{(qn_{1}+r)^{2t}}$

when

$n_{1}\geq\frac{1}{1-2^{-\frac{1}{2t-1}}}.$ (3.8)

Therefore, if $n_{1}$ satisfies the condition (3.8), then the latter
inequality of (1.3) holds with

$d_{2}=2,\quad\eta_{2}(t)=\frac{1}{q(2t-1)},\quad\lambda_{2}(n_{1})=qn_{1}+r.$

We assume (3.6), (3.7) and (3.8). Then, (1.4) holds with $c_{\lambda,\nu}=1$.
The condition (1.10) is

$(qn_{0}+r)^{1-\frac{1+\delta}{2}}\geq 2q(1-t-\delta
t)\left(\frac{2}{q(2t-1)}\right)^{\frac{1+\delta}{2}}M^{1-\frac{\delta}{2}},$

which holds for

$n_{0}\geq q^{-1}(2q(1-t-\delta
t))^{\frac{2}{1-\delta}}\left(\frac{2}{q(2t-1)}\right)^{\frac{1+\delta}{1-\delta}}M^{\frac{2-\delta}{1-\delta}}.$
(3.9)

Finally, the condition (1.5) is

$qn_{0}+r\leq\frac{1}{2}q(2t-1)(qn_{0}+r)^{2t},$

which holds for

$n_{0}\geq q^{-1}\left(\frac{2}{q(2t-1)}\right)^{\frac{1}{2t-1}}.$ (3.10)

We choose a positive integer $N_{0}$ that is larger than the largest value of
the right hand sides of (3.1), (3.2), (3.4), (3.6) to (3.10). Then for
$1/2<t<1$ and for any $n_{0}\geq N_{0}$, the squares of sidelength
$(qn+r)^{-t}$ for $n\geq n_{0}$ are packed perfectly into a square of area
$\sum_{n=n_{0}}^{\infty}(qn+r)^{-2t}$. ∎

##  4 Perfectly packing a square by squares of nearly P-harmonic sidelength

In this section, for $1/2<t<1$, we give a perfectly packing a square by
squares of sidelength $p_{n}^{-t}$ for $n\geq n_{0}$ for any $n_{0}\geq
N_{0}$, where $p_{n}$ denotes the $n$th prime number and $N_{0}$ is a
sufficiently large positive integer depending on $t$, which will be explicitly
given in Corollary 4.6. Before the statement of the theorem, we prepare some
basic facts on primes. First, it is known that the sequence $\\{p_{n}\\}$
satisfies

$p_{n}>n\log n$ (4.1)

for any positive integer $n$, and

$p_{n}<n(\log n+\log\log n)$ (4.2)

for any $n\geq 6$ (see for example [14]). We first give an upper bound for
$\sum_{n\leq x}1/p_{n}^{t}$ for any fixed $1/2<t<1$. By decomposing the
interval $[1,x]$ into $[1,\sqrt[4]{x^{3}}]\cup(\sqrt[4]{x^{3}},x]$ and using
(4.1), we have

$\displaystyle\sum_{n\leq x}\frac{1}{p_{n}^{t}}\leq$
$\displaystyle\frac{1}{2^{t}}+\frac{1}{(\log 2)^{t}}\sum_{2\leq
n\leq\sqrt[4]{x^{3}}}\frac{1}{n^{t}}+\frac{1}{(\log\sqrt[4]{x^{3}})^{t}}\sum_{\sqrt[4]{x^{3}}<n\leq
x}\frac{1}{n^{t}}$ (4.3) $\displaystyle\leq$
$\displaystyle\frac{1}{2^{t}}+\frac{1}{(\log
2)^{t}}\int_{1}^{\sqrt[4]{x^{3}}}\frac{du}{u^{t}}+\frac{\left(\frac{4}{3}\right)^{t}}{(\log
x)^{t}}\int_{2}^{x}\frac{du}{u^{t}}$ $\displaystyle\leq$
$\displaystyle\frac{1}{2^{t}}+\frac{x^{\frac{3}{4}(1-t)}}{(1-t)(\log
2)^{t}}+\frac{\left(\frac{4}{3}\right)^{t}x^{1-t}}{(1-t)(\log x)^{t}}.$

In the right hand side of (4.3), the inequality

$\max\left\\{\frac{1}{2^{t}},\;\frac{x^{\frac{3}{4}(1-t)}}{(1-t)(\log
2)^{t}}\right\\}\leq\frac{x^{1-t}}{(1-t)(\log x)^{t}}$

holds for

$x\geq e^{\frac{16}{(1-t)^{2}}}.$ (4.4)

Furthermore,

$\frac{\left(\frac{4}{3}\right)^{t}x^{1-t}}{(1-t)(\log
x)^{t}}\leq\frac{4}{3}\frac{x^{1-t}}{(1-t)(\log x)^{t}}$

for any $x>1$. Therefore, by (4.3) and the above estimates, we have

$\sum_{n\leq x}\frac{1}{p_{n}^{t}}\leq\frac{10}{3}\frac{x^{1-t}}{(1-t)(\log
x)^{t}}$ (4.5)

for $x$ satisfying (4.4). On the other hand, since $1/(1+(\log\log n/\log
n))>7/10$ for any $n\geq 6$, by (4.2) we have

$\displaystyle\sum_{n\leq x}\frac{1}{p_{n}^{t}}$
$\displaystyle>\frac{7}{10}\sum_{6\leq n\leq x}\frac{1}{(n\log
n)^{t}}\geq\frac{7}{10}\frac{1}{(\log x)^{t}}\sum_{6\leq n\leq
x}\frac{1}{n^{t}}\geq\frac{7}{10}\frac{1}{(\log
x)^{t}}\int_{6}^{x}\frac{du}{u^{t}}$ (4.6)
$\displaystyle\geq\frac{7(x^{1-t}-6^{1-t})}{10(1-t)(\log x)^{t}}$

for $x\geq 6$. If $x$ satisfies

$x\geq 6\cdot 2^{\frac{1}{1-t}},$ (4.7)

then $x^{1-t}-6^{1-t}\geq x^{1-t}/2$. Therefore, for $x$ satisfying the
condition (4.7), we have

$\sum_{n\leq x}\frac{1}{p_{n}^{t}}\geq\frac{7}{20}\frac{x^{1-t}}{(1-t)(\log
x)^{t}}.$ (4.8)

Finally, we note that $e^{\frac{16}{(1-t)^{2}}}>6\cdot 2^{\frac{1}{1-t}}$
holds for any $1/2<t<1$, we have the following conclusion.

######  Lemma 4.1.

For any $1/2<t<1$ and $x\geq e^{\frac{16}{(1-t)^{2}}}$, we have

$\frac{7}{20}\frac{x^{1-t}}{(1-t)(\log x)^{t}}<\sum_{n\leq
x}\frac{1}{p_{n}^{t}}<\frac{10}{3}\frac{x^{1-t}}{(1-t)(\log x)^{t}}.$ (4.9)

Replacing $x$ with $x-1$ in (4.9), we see that

$\frac{7}{20}\frac{(x-1)^{1-t}}{(1-t)(\log(x-1))^{t}}<\sum_{n\leq
x-1}\frac{1}{p_{n}^{t}}<\frac{10}{3}\frac{(x-1)^{1-t}}{(1-t)(\log(x-1))^{t}}$

for $x\geq 1+e^{\frac{16}{(1-t)^{2}}}$. In this region it follows that

$\frac{1}{2}\frac{x^{1-t}}{(1-t)(\log
x)^{t}}<\frac{(x-1)^{1-t}}{(1-t)(\log(x-1))^{t}}<\frac{2x^{1-t}}{(1-t)(\log
x)^{t}}.$

Combining (4.9) and the above inequality, we have the following result.

######  Lemma 4.2.

For $1/2<t<1$ and $x\geq 2e^{\frac{16}{(1-t)^{2}}}$, we have

$\frac{7}{40}\frac{x^{1-t}}{(1-t)(\log x)^{t}}<\sum_{n\leq
x-1}\frac{1}{p_{n}^{t}}<\frac{20}{3}\frac{x^{1-t}}{(1-t)(\log x)^{t}}.$ (4.10)

Next, we evaluate $\sum_{n\geq x}1/p_{n}^{t}$ for $t>1$. By (4.1), we have

$\sum_{n\geq x}\frac{1}{p_{n}^{t}}<\frac{1}{(\log x)^{t}}\sum_{n\geq
x}\frac{1}{n^{t}}\leq\frac{1}{(\log
x)^{t}}\int_{x-1}^{\infty}\frac{du}{u^{t}}=\frac{(x-1)^{1-t}}{(t-1)(\log
x)^{t}}.$

If $x\geq 1/(1-2^{-\frac{1}{t-1}})$, then $(x-1)^{1-t}\leq 2x^{1-t}$.
Therefore, for this $x$ we have

$\sum_{n\geq x}\frac{1}{p_{n}^{t}}\leq\frac{2x^{1-t}}{(t-1)(\log x)^{t}}.$
(4.11)

On the other hand, by (4.2), for $x\geq 6$ we have

$\displaystyle\sum_{n\geq x}\frac{1}{p_{n}^{t}}$ $\displaystyle\geq\sum_{n\geq
x}\frac{1}{(n\log n)^{t}\left(1+\frac{\log\log n}{\log
n}\right)^{t}}\geq\frac{1}{2^{t}}\sum_{n\geq x}\frac{1}{(n\log
n)^{t}}\geq\frac{1}{2^{t}}\frac{1}{(\log 2x)^{t}}\sum_{x\leq
n<2x}\frac{1}{n^{t}}$ $\displaystyle\geq\frac{1}{(2\log
2x)^{t}}\int_{x}^{2x}\frac{du}{u^{t}}=\frac{(1-2^{1-t})x^{1-t}}{(t-1)(2\log
2x)^{t}}\geq\frac{(1-2^{1-t})x^{1-t}}{2^{2t}(t-1)(\log x)^{t}}.$

Consequently,

######  Lemma 4.3.

For $t>1$ and $x\geq\max\\{1/(1-2^{-\frac{1}{t-1}}),6\\}$, we have

$\frac{(1-2^{1-t})x^{1-t}}{2^{2t}(t-1)(\log x)^{t}}\leq\sum_{n\geq
x}\frac{1}{p_{n}^{t}}\leq\frac{2x^{1-t}}{(t-1)(\log x)^{t}}.$ (4.12)

Finally, by (4.1) and (4.2), it follows that

$\frac{p_{n_{1}}}{2}<n_{1}\log n_{1}<p_{n_{1}}.$

Combining the above inequality and (4.10), (4.12), we obtain the following
conclusion.

######  Lemma 4.4.

We have

$\frac{7}{40}\frac{n_{1}}{(1-t)p_{n_{1}}^{t}}<\sum_{n\leq
n_{1}-1}\frac{1}{p_{n}^{t}}<\frac{20}{3}\frac{2^{t}n_{1}}{(1-t)p_{n_{1}}^{t}}$
(4.13)

for $1/2<t<1$, $n_{1}\geq 2e^{\frac{16}{(1-t)^{2}}}$, and

$\frac{(1-2^{1-t})n_{1}}{2^{2t}(t-1)p_{n_{1}}^{t}}<\sum_{n\geq
n_{1}}\frac{1}{p_{n_{1}}^{t}}<\frac{2^{1+t}n_{1}}{(t-1)p_{n_{1}}^{t}}$ (4.14)

for $t>1$, $n_{1}\geq\max\\{6,1/(1-2^{-\frac{1}{t-1}})\\}$.

The next purpose is to extend the sequence $\\{p_{n}\\}$ to a smooth
increasing function on $\mathbb{R}_{\geq 1}$.

######  Lemma 4.5.

There exists a smooth increasing function $f:\mathbb{R}_{\geq
1}\to\mathbb{R}_{\geq 1}$ which satisfies the following two conditions.
1\. For any positive integer $n$, $f(n)=p_{n}$.
2\. There exists an absolute positive constant $C_{P}<7$ such that for any
positive integer $n$ and for any $x\in[n,n+1]$,

$f^{{}^{\prime}}(x)\leq C_{p}(p_{n+1}-p_{n}).$ (4.15)

###### Proof.

Define a function $\varphi$ by

$\varphi(x)=\begin{cases}0&(x<-\frac{1}{6})\\\
\frac{1}{2}\exp\left(1-\frac{1}{1-36x^{2}}\right)&(-\frac{1}{6}\leq x\leq
0)\\\ 1-\frac{1}{2}\exp\left(1-\frac{1}{1-36x^{2}}\right)&(0\leq
x\leq\frac{1}{6})\\\ 1&(x>\frac{1}{6}).\end{cases}$

This is a smooth increasing function satisfying $\varphi(x)=0$ if $x<-1/6$,
$\varphi(x)=1$ if $x>1/6$ and $\max\varphi^{{}^{\prime}}(x)=6.511\ldots<7$. We
define the function $f(x)$ by

$f(x)=p_{1}+\sum_{k=1}^{\infty}\varphi(x-(k+1/2))(p_{k+1}-p_{k}).$ (4.16)

We fix a positive integer $n$ arbitrarily and suppose $x\in[n,n+1]$. If $k\geq
n+1$, then

$x-\left(k+\frac{1}{2}\right)\leq(n+1)-(n+1)-\frac{1}{2}=-\frac{1}{2}<-\frac{1}{6}$

Hence $\varphi(x-(k+1/2))=0$. If $k\leq n-1$, then

$x-\left(k+\frac{1}{2}\right)\geq
n-(n-1)-\frac{1}{2}=\frac{1}{2}>\frac{1}{6}.$

Hence $\varphi(x-(k+1/2))=1$. Therefore,

$\displaystyle f(x)=$ $\displaystyle
p_{1}+\sum_{k=1}^{n-1}(p_{k+1}-p_{k})+\varphi(x-(n+1/2))(p_{n+1}-p_{n})$
$\displaystyle=$ $\displaystyle p_{n}+\varphi(x-(n+1/2))(p_{n+1}-p_{n}).$

Hence $f$ satisfies the conditions of the lemma. ∎

Next, it is known that there exist some $0<\theta<1$ and a positive integer
$N_{\theta}$ for which

$p_{n+1}-p_{n}\leq p_{n}^{\theta}$ (4.17)

holds for any $n\geq N_{\theta}$. The best result currently known is due to
Baker, Harman and Pintz [1], according to which (4.17) holds for
$\theta=0.525$. By (4.15) and (4.17), we have

$f^{{}^{\prime}}(x)\leq C_{P}p_{n}^{\theta}$

for any $n\geq N_{\theta}$, $x\in[n,n+1]$. Combining this with $p_{n}<2n\log
n$ $(n\geq 6)$, we have

$f^{{}^{\prime}}(x)\leq 7(2\log n)^{\theta}$ (4.18)

for any $n\geq\max\\{6,N_{\theta}\\}$, $x\in[n,n+1]$. Now our theorem on
perfectly packing a square by squares of sidelength $p_{n}^{-t}$ is described
as follows.

######  Corollary 4.6.

Assume that (4.17) holds for any $n\geq N_{\theta}$ with some positive
constant $\theta<1$. for $1/2<t<1$, let $M$ be a positive integer satisfying

$M\geq\max\left\\{4^{\frac{1}{1-t}}\left(\frac{20\cdot
2^{6t-t^{2}}(2t-1)}{3(1-t)^{2}(1-2^{1-2t})}\right)^{\frac{2}{t(1-t)}},\left(50\left(\frac{6}{5}\right)^{t(2-t)}\right)^{\frac{2}{1-t}}\right\\},$

and $N_{0}$ be a positive integer satisfying

$\displaystyle N_{0}\geq$
$\displaystyle\max\biggl{\\{}N_{\theta},\;85536M^{5},\;3\times
10^{18},\;2e^{\frac{16}{(1-t)^{4}}},\;\frac{1}{1-2^{-\frac{1}{2t-1}}},$
$\displaystyle\quad\quad\quad
M^{\frac{1}{(1-t)^{2}}},\;\left(\frac{40}{7}(1-t)^{2}\right)^{\frac{2}{t}}\left(\frac{2^{1+2t}}{2t-1}\right)^{\frac{2-t}{t}}M^{\frac{1+t}{t}},\;e^{\frac{1}{(2t-1)(1-t)}}\biggr{\\}}.$

Then, for any positive integer $n_{0}\geq N_{0}$, there exists a perfectly
packing a square of area $\sum_{n=n_{0}}^{\infty}p_{n}^{-2t}$ by squares of
sidelength $p_{n}^{-t}$ for $n\geq n_{0}$.

######  Remark 4.7.

Roughly speaking, the perfectly packing exists as $t\nearrow 1$ if $N_{0}\geq
2\exp\left(\frac{16}{(1-t)^{4}}\right)$.

###### Proof.

From now on, we assume

$N_{0}\geq N_{\theta}.$ (4.19)

By (4.18), it follows that

$\max_{x\in[n_{0},n_{0}+9M^{2}]}f^{{}^{\prime}}(x)\leq
8(2\log(n_{0}+9M^{2}))^{\theta}$

for any $n_{0}\geq N_{\theta}$, the condition (1.7) is valid if

$8(2\log(n_{0}+9M^{2}))^{\theta}\leq\frac{1}{4752}M^{-4}n_{0}\log
n_{0}\quad(\forall n_{0}\geq N_{0}),$

which holds for

$N_{0}\geq 2\cdot 4752\cdot 9M^{5}.$ (4.20)

Since $n\log n<f(n)<2n\log n$ for any $n\geq 6$, the condition (1.8) is valid
if

$2(n_{0}+9M^{2})\log(n_{0}+9M^{2})\leq(264M^{2})^{\frac{1}{t}}n_{0}\log n_{0}$

holds for any $n_{0}\geq N_{0}$. Moreover, if

$N_{0}\geq 9M^{2},$ (4.21)

then the left hand side is at most $4n_{0}\log 2n_{0}$. Hence under the
assumption of (4.21), the condition (1.8) is satisfied if

$4n_{0}\log n_{0}\leq(264M^{2})^{\frac{1}{t}}n_{0}\log n_{0},$

and this inequality holds whenever $M$ satisfies (1.6). Consequently, if $M$
satisfies (1.6) and $N_{0}$ satisfies (4.21), then the condition (1.8) is also
satisfied. The condition (1.1) is valid if

$n_{0}\log n_{0}\geq K(1+l)n_{0}(\log(1+l)n_{0}+\log\log(1+l)n_{0})$

holds for any $n_{0}\geq N_{0}$, and by choosing

$l=\frac{1}{10},\;K=\frac{5}{6},$

the above inequality holds for

$N_{0}\geq 3\times 10^{18}.$ (4.22)

By replacing the parameter $t$ in (4.13) and (4.14) with $t+\delta t$
$(\delta:=1-t)$, $2t$, respectively, the conditions (1.2), (1.3) hold with

$c_{1}=\frac{7}{40},\quad
c_{2}=\frac{20}{3},\quad\xi_{1}(t)=\frac{1}{(1-t)^{2}},\quad\xi_{2}(t)=\frac{2^{2t-t^{2}}}{(1-t)^{2}},\quad\nu_{1}(n_{1})=\nu_{2}(n_{1})=n_{1},$
$d_{1}=d_{2}=1,\quad\eta_{1}(t)=\frac{1-2^{1-2t}}{2^{4t}(2t-1)},\quad\eta_{2}(t)=\frac{2^{1+2t}}{2t-1},\quad\lambda_{1}(n_{1})=\lambda_{2}(n_{1})=n_{1}.$

(We used $1-t-\delta t=(1-t)^{2}$.) Hence (1.4) holds with
$c_{\lambda,\nu}=1$. The condition (1.9) holds if

$9Mn_{0}^{t}(\log n_{0}+\log\log n_{0})^{t}\leq\frac{1}{10}n_{0}\quad(\forall
n_{0}\geq N_{0}),$

which is equivalent to

$n_{0}^{1-t}\geq 90M(\log n_{0}+\log\log n_{0})^{t}\quad(\forall n_{0}\geq
N_{0}).$

Since the right hand side is at most $180M\log n_{0}$, a sufficient condition
for (1.9) is that

$n_{0}^{1-t}\geq 180M\log n_{0}\quad(\forall n_{0}\geq N_{0}).$

This condition is satisfied if

$N_{0}\geq M^{\frac{1}{(1-t)^{2}}}.$ (4.23)

The condition (1.10) is

$n_{0}^{1-\frac{1+\delta}{2}}\geq\frac{\left(\frac{2^{1+2t}}{2t-1}\right)^{\frac{1+\delta}{2}}}{\frac{7}{40}\frac{1}{(1-t)^{2}}}M^{1-\frac{\delta}{2}}\quad(\forall
n_{0}\geq N_{0}),$

which is satisfied if

$N_{0}\geq\left(\frac{40}{7}(1-t)^{2}\right)^{\frac{2}{t}}\left(\frac{2^{1+2t}}{2t-1}\right)^{\frac{2-t}{t}}M^{\frac{1+t}{t}}.$
(4.24)

Finally, the condition (1.5) is satisfied if

$n_{0}\leq\frac{2t-1}{2^{1+2t}}(n_{0}\log n_{0})^{2t}\quad(\forall n_{0}\geq
N_{0}),$

and this inequality holds for

$N_{0}\geq e^{\frac{1}{(2t-1)(1-t)}}.$ (4.25)

Combining the lower bounds for $N_{0}$, we obtain the conclusion. ∎

##  5 Ineffective version of the perfectly packing theorem and an application
to the packing of squares of nearly TP-harmonic sidelength

###  5.1 Ineffective version of the perfectly packing theorem

Clearly, one cannot adapt Theorem 1.1 when $f(x)$ is larger than
$x^{1+\delta_{0}}$ for some constant $\delta_{0}>0$ up to constant multiples,
because the condition (1.9) fails. On the other hand, we have seen that one
can adapt Theorem 1.1 to the function with $f(n)=p_{n}$, where $p_{n}\sim
n\log n$. Hence one might wonder that how large the function $f(x)$ can be.
The following theorem partially answers this question.

######  Theorem 5.1.

Let $\kappa:\mathbb{R}_{>1}\to\mathbb{R}_{>1}$ be a smooth function for which
$f(x):=x\kappa(x)$ becomes a smooth increasing function. Suppose that for any
$\varepsilon>0$, there exists a positive integer $N_{\varepsilon}$ such that
for any $n\geq N_{\varepsilon}$,

$\max_{x\in[n,2n]}\kappa^{{}^{\prime}}(x)\leq\varepsilon\kappa(x),$
$(1-\varepsilon)\kappa(M(2n))\leq\kappa(n)\leq n^{\varepsilon},$
$c_{\mathfrak{m}}\kappa(n)\leq\kappa(\mathfrak{m}(n))\leq
C_{\mathfrak{m}}\kappa(n),$

where $M(2n)$ denotes one of the values $x\in[1,2n]$ such that $\kappa(x)$
takes its minimum in $[1,2n]$, $\mathfrak{m}(n)$ denotes one of the values
$x\in[n,\infty)$ such that $\kappa(x)$ takes its minimum in $[n,\infty)$, and
$C_{\mathfrak{m}}>c_{\mathfrak{m}}>0$ are positive constants. Moreover, we
assume that there exist constants $0<\alpha<1$ and $C_{\alpha}>1$ such that

$\kappa(n)\leq C_{\alpha}\kappa(m_{\alpha}(n))$

holds for any $n\geq N_{\varepsilon}$, where $m_{\alpha}(n)$ denotes one of
the values $x\in[n,\infty)$ such that $\kappa(x)$ takes its minimum in
$[n,\infty)$. Then, for any $1/2<t<1$, there exists a positive integer
$N_{0}=N_{0}(t)$ such that for any integer $n_{0}\geq N_{0}$, squares of
sidelength $(n\kappa(n))^{-t}$ for $n\geq n_{0}$ can be packed perfectly into
a square of area $\sum_{n=n_{0}}^{\infty}(n\kappa(n))^{-2t}$.

###### Proof.

In this proof we assume that $M$ satisfies (1.6) and that $N_{0}\geq 9M^{2}$.
The condition (1.7) becomes

$\max_{x\in[n_{0},2n_{0}]}(\kappa(x)+x\kappa^{{}^{\prime}}(x))\leq\frac{1}{4752}M^{-4}n_{0}\kappa(n_{0}).$

This inequality holds if both

$\kappa(M(2n_{0}))\leq\frac{1}{9504}M^{-4}n_{0}\kappa(n_{0})$ (5.1)

and

$\max_{x\in[n_{0},2n_{0}]}\kappa^{{}^{\prime}}(x)\leq\frac{1}{19008}M^{-4}\kappa(n_{0})$
(5.2)

are satisfied. These two inequalities are valid if the first two conditions in
the statement of the theorem are satisfied. The condition (1.8) is satisfied
if

$2n_{0}\kappa(M(2n_{0}))\leq(264M^{2})^{\frac{1}{t}}(n_{0}\kappa(n_{0})),$

which is equivalent to

$\kappa(M(2n_{0}))\leq\frac{1}{2}(264M^{2})^{\frac{1}{t}}\kappa(n_{0}).$

The right hand side is at least $(1/2)(264M^{2})^{\frac{1}{2}}\kappa(n_{0})$.
Therefore, (1.8) holds if

$\kappa(M(2n_{0}))\leq\sqrt{66}M\kappa(n_{0}).$ (5.3)

The condition (1.1) is

$n_{0}\kappa(n_{0})\geq K(1+l)n_{0}\kappa((1+l)n_{0}).$

Since the right hand side is at most $K(1+l)n_{0}\kappa(M(2n_{0}))$ for $l<1$,
the condition (1.1) is valid if

$n_{0}\kappa(n_{0})\geq K(1+l)n_{0}\kappa(M(2n_{0})).$

By taking $l=\varepsilon$, $K=(1-\varepsilon)/(1+\varepsilon)$, the above
condition becomes

$\kappa(n_{0})\geq(1-\varepsilon)\kappa(M(2n_{0})).$ (5.4)

Notice that if we assume (5.4), then (5.1) and (5.3) automatically hold for
any sufficiently large $n_{0}$. The condition (1.9) is

$9M(n_{0}\kappa(n_{0}))^{t}\leq ln_{0}.$

This condition is valid if

$\kappa(n_{0})\ll n_{0}^{\varepsilon}$ (5.5)

holds for any $\varepsilon>0$, where the implied constant above is dependent
only on $\varepsilon$. Next, we consider the condition (1.2). Since

$\frac{1}{f(n)^{t+\delta t}}\geq\frac{1}{n^{t+\delta
t}\kappa(M(n_{1}))^{t+\delta t}}$

for $1\leq n\leq n_{1}-1$, we have

$\displaystyle\sum_{n=1}^{n_{1}-1}\frac{1}{f(n)^{t+\delta t}}\geq$
$\displaystyle\frac{1}{\kappa(M(n_{1}))^{t+\delta
t}}\sum_{n=1}^{n_{1}-1}\frac{1}{n^{t+\delta
t}}\geq\frac{1}{\kappa(M(n_{1}))^{t+\delta
t}}\int_{1}^{n_{1}}\frac{du}{u^{t+\delta t}}$ $\displaystyle=$
$\displaystyle\frac{n_{1}^{1-t-\delta t}-1}{(1-t-\delta
t)\kappa(M(n_{1}))^{t+\delta t}}\geq\frac{(1-\varepsilon)n_{1}}{2(1-t-\delta
t)(\kappa(n_{1})n_{1})^{t+\delta t}}$

for any sufficiently large $n_{1}$. On the other hand, for any fixed
$0<\alpha<1$ and for any sufficiently large $n_{1}$, we have

$\displaystyle\sum_{n=1}^{n_{1}-1}\frac{1}{f(n)^{t+\delta t}}$
$\displaystyle\leq\sum_{1\leq n\leq
n_{1}^{\alpha}}\frac{1}{\kappa(n)^{t+\delta t}n^{t+\delta
t}}+\sum_{n_{1}^{\alpha}\leq n\leq n_{1}}\frac{1}{\kappa(n)^{t+\delta
t}n^{t+\delta t}}$ $\displaystyle\leq\sum_{1\leq n\leq
n_{1}^{\alpha}}\frac{1}{n^{t+\delta
t}}+\frac{1}{\kappa(m_{\alpha}(n_{1}))^{t+\delta
t}}\sum_{n=1}^{n_{1}}\frac{1}{n^{t+\delta t}}$
$\displaystyle\leq\frac{n_{1}^{\alpha(1-t-\delta t)}}{1-t-\delta
t}+\frac{n_{1}^{1-t-\delta t}}{(1-t-\delta
t)\kappa(m_{\alpha}(n_{1}))^{t+\delta t}}$
$\displaystyle\leq\frac{2n_{1}^{1-t-\delta t}}{(1-t-\delta
t)\kappa(m_{\alpha}(n_{1}))^{t+\delta
t}}=\frac{2n_{1}\left(\frac{\kappa(n_{1})}{\kappa(m_{\alpha}(n_{1}))}\right)^{t+\delta
t}}{(1-t-\delta t)(n_{1}\kappa(n_{1}))^{t+\delta t}},$

provided that

$\kappa(m_{\alpha}(n_{1}))^{t+\delta t}\leq n_{1}^{(1-\alpha)(1-t-\delta t)}.$

The above condition is satisfied if $\kappa$ satisfies $\kappa(n)\ll
n^{\varepsilon}$ for any $\varepsilon>0$. Combining these inequalities, for
any sufficiently large $n_{1}$ and any fixed $0<\alpha<1$, we have

$\frac{(1-\varepsilon)n_{1}}{2(1-t-\delta t)(\kappa(n_{1})n_{1})^{t+\delta
t}}\leq\sum_{n=1}^{n_{1}-1}\frac{1}{f(n)^{t+\delta
t}}\leq\frac{2n_{1}\left(\frac{\kappa(n_{1})}{\kappa(m_{\alpha}(n_{1}))}\right)^{t+\delta
t}}{(1-t-\delta t)(n_{1}\kappa(n_{1}))^{t+\delta t}}.$

Therefore, the condition (1.2) holds with

$c_{1}=\frac{1-\varepsilon}{2},\quad\xi_{1}(t)=\frac{1}{1-t-\delta
t},\quad\nu_{1}(n_{1})=n_{1},$ $c_{2}=2,\quad\xi_{2}(t)=\frac{1}{1-t-\delta
t},\quad\nu_{2}(n_{1})=\frac{n_{1}\kappa(n_{1})}{\kappa(m_{\alpha}(n_{1}))}.$

Next, we consider the condition (1.3). For $t>1/2$,

$\displaystyle\sum_{n=n_{1}}^{\infty}\frac{1}{f(n)^{2t}}$
$\displaystyle\geq\frac{1}{\kappa(M(2n_{1}))^{2t}}\sum_{n=n_{1}}^{2n_{1}}\frac{1}{n^{2t}}\geq\frac{1}{\kappa(M(2n_{1}))^{2t}}\int_{n_{1}}^{2n_{1}+1}\frac{du}{u^{2t}}$
$\displaystyle\quad=\frac{n_{1}^{1-2t}-(2n_{1}+1)^{1-2t}}{(2t-1)\kappa(M(2n_{1}))^{2t}}\geq\frac{n_{1}}{2(2t-1)\kappa(M(2n_{1}))^{2t}n_{1}^{2t}}$
$\displaystyle\quad\quad\geq\frac{n_{1}}{2(1-\varepsilon)(2t-1)(\kappa(n_{1})n_{1})^{2t}},$

provided that (5.4) holds. On the other hand,

$\displaystyle\sum_{n=n_{1}}^{\infty}\frac{1}{f(n)^{2t}}\leq\frac{1}{\mathfrak{m}(n_{1})^{2t}}\sum_{n=n_{1}}^{\infty}\frac{1}{n^{2t}}=\frac{n_{1}\left(\frac{\kappa(n_{1})}{\kappa(\mathfrak{m}(n_{1}))}\right)^{2t}}{(2t-1)(n_{1}\kappa(n_{1}))^{2t}}.$

Combining these inequalities, it follows that (1.3) holds with

$d_{1}=\frac{1}{2(1-\varepsilon)},\quad\eta_{1}(t)=\frac{1}{2t-1},\quad\lambda_{1}(n_{1})=n_{1},$
$d_{2}=1,\quad\eta_{2}(t)=\frac{1}{2t-1},\quad\lambda_{2}(n_{1})=n_{1}\left(\frac{\kappa(n_{1})}{\kappa(\mathfrak{m}(n_{1}))}\right)^{2}.$

The condition (1.4) is satisfied if $\kappa$ satisfies

$\kappa(n_{0})\leq C_{\alpha}\kappa(m_{\alpha}(n_{0}))$ (5.6)

for some $0<\alpha<1$, $C_{\alpha}>0$. The condition (1.10) becomes

$n_{0}^{\frac{1-\delta}{2}}\left(\frac{\kappa(n_{0})}{\kappa(\mathfrak{m}(n_{0}))}\right)^{-(1+\delta)}\geq\frac{2(1-t-\delta
t)}{1-\varepsilon}\left(\frac{1}{2t-1}\right)^{\frac{1+\delta}{2}}M^{1-\frac{\delta}{2}}.$

If $\kappa$ satisfies

$\kappa(\mathfrak{m}(n_{0}))\leq C_{\mathfrak{m}}\kappa(n_{0}),$ (5.7)

then the above inequality holds for any sufficiently large $n_{0}$. The
condition (1.5) becomes

$n_{0}\left(\frac{\kappa(n_{0})}{\kappa(\mathfrak{m}(n_{0}))}\right)^{2}\leq(2t-1)(n_{0}\kappa(n_{0}))^{2t}.$

If $\kappa$ satisfies

$\kappa(\mathfrak{m}(n_{0}))\geq c_{\mathfrak{m}}\kappa(n_{0}),$ (5.8)

then the above inequality holds for any sufficiently large $n_{0}$. Combining
these results we obtain the conclusion of the theorem. ∎

###  5.2 Packing of squares of nearly TP-harmonic sidelength

A prime number $p$ is called a twin prime if at least one of $p-2$ or $p+2$ is
also a prime number. We denote the $n$th twin prime by $\mathfrak{p}_{n}$.
Then
$\mathfrak{p}_{1}=3,\;\mathfrak{p}_{2}=5,\;\mathfrak{p}_{3}=7,\;\mathfrak{p}_{4}=11,\;\mathfrak{p}_{5}=13,\ldots$.
For $x\geq 2$, we denote the number of twin primes below $x$ by $\pi_{2}(x)$.
Then, it is known that for any sufficiently large $x$,

$\pi_{2}(x)\leq\frac{C\Pi_{2}x}{(\log x)^{2}}\left(1+c\frac{\log\log x}{\log
x}\right),$ (5.9)

where

$\Pi_{2}=\underset{p\geq
3}{\prod_{p:\mathrm{prime}}}\left(1-\frac{1}{(p-1)^{2}}\right)=0.6601618\ldots$
(5.10)

is the twin prime constant, and $c$ is some absolute constant. The coefficient
$C$ is also some absolute constant. Though the value of $C$ has not been known
correctly, according to the Hardy-Littlewood conjecture [5], the correct value
of $C$ is expected to be 2. Currently the best known upper bound is $C\leq
6.8325$ by Haugland [4].

######  Corollary 5.2.

Let ${\mathfrak{p}}_{n}$ be the $n$th twin prime, $\Pi_{2}$ be the twin prime
constant defined by (5.10) and $C$ be a constant in (5.9). Then, for any
$1/2<t<1$ and $C^{{}^{\prime}}>C$, there exists a positive integer $N_{0}$
depending on $t$ and $C^{{}^{\prime}}$ that for any $n_{0}\geq N_{0}$, squares
of sidelength ${\mathfrak{p}}_{n}^{-t}$ for $n\geq n_{0}$ can be packed into a
square of area $(C^{{}^{\prime}}\Pi_{2})^{2t}\sum_{n=n_{0}}^{\infty}(n(\log
n)^{2})^{-2t}$.

###### Proof.

Since $\pi_{2}(\mathfrak{p}_{n})=n$, the estimate (5.9) yields

$n\leq\frac{C\Pi_{2}\mathfrak{p}_{n}}{(\log\mathfrak{p}_{n})^{2}}\left(1+c\frac{\log\log\mathfrak{p}_{n}}{\log\mathfrak{p}_{n}}\right).$

From this, we easily see that for any $C^{{}^{\prime}}>C$,

$\mathfrak{p}_{n}\geq\frac{n}{C^{{}^{\prime}}\Pi_{2}}(\log n)^{2}$ (5.11)

holds for any $n\geq N(C^{{}^{\prime}})$, where $N(C^{{}^{\prime}})$ is some
positive integer depending on $C^{{}^{\prime}}$. We put

$f(x)=\frac{x}{C^{{}^{\prime}}\Pi_{2}}(\log
x)^{2},\quad\kappa(x)=\frac{1}{C^{{}^{\prime}}\Pi_{2}}(\log x)^{2}$

for $x>1$. By appropriately modifying the value of $\kappa(x)$ when $x$ is
small, $\kappa$ becomes a smooth function from $\mathbb{R}_{>1}$ to
$\mathbb{R}_{>1}$ and satisfies all the conditions of Theorem 5.1. Therefore,
for any $1/2<t<1$, there exists a positive integer $N_{0}(t)$ such that for
any positive integer $n_{0}\geq N_{0}(t)$, squares of sidelength
$((n/C^{{}^{\prime}}\Pi_{2})(\log n)^{2})^{-t}$ for $n\geq n_{0}$ can be
packed perfectly into a square of area
$\sum_{n=n_{0}}^{\infty}((n/C^{{}^{\prime}}\Pi_{2})(\log n)^{2})^{-2t}$. By
(5.11), a square of sidelength $\mathfrak{p}_{n}^{-t}$ for $n\geq
N(C^{{}^{\prime}})$ is contained in a square of sidelength
$((n/C^{{}^{\prime}}\Pi_{2})(\log n)^{2})^{-t}$. Hence by taking
$N_{0}:=\max\\{N_{0}(t),N(C^{{}^{\prime}})\\}$, we obtain the result. ∎

##  6 Acknowledgements

This work is partially supported by the JSPS, KAKENHI Grant Number 21K03204.
The author thanks Professor Yuta Suzuki for telling him how to construct a
smooth increasing function $f$ in Lemma 4.5. The author sincerely thanks to
the referee of this paper for reading the first manuscript very carefully and
giving a lot of valuable comments and suggestions.

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Kanto Gakuin University,
Kanazawa, Yokohama
Kanagawa, Japan
E-mail address<EMAIL_ADDRESS>


$\|\varphi_{s}(u)\|_{V}<\frac{1}{4}\frac{1-\alpha}{C_{\alpha}\tau^{1-\alpha}\max\\{1,e^{\omega\tau}\\}},\qquad\text{for
all }s\in[0,\tau],$ (B.4)

with $\alpha$, $C_{\alpha}$ and $\omega$ as in (B.2). Set
$O\subset\mathcal{D}^{\varphi}_{\tau}$ to be the neighborhood on which (B.1)
holds.

The derivative of the semiflow, $D\varphi_{t}(u)$, may be explicitly written
as (see Theorem 2.2 of [Wei79]),

$D\varphi_{t}(u)[y]=e^{A_{U}t}y+\int_{0}^{t}e^{A_{U}(t-s)}DB(\varphi_{s}(u))D\varphi_{s}(u)[y]ds,\qquad
t\in\mathcal{D}_{u}^{\varphi},\;y\in V,$

where $A_{U}$ and $B$ are as in (4.3) and (4.4); and $DB(u)[v]=B(u,v)+B(v,u)$.
We seek to estimate

$D\varphi_{t}(u)[y]-D\varphi_{t}(v)[y]=\int_{0}^{t}e^{A_{U}(t-s)}\Big{[}DB(\varphi_{s}(u))(D\varphi_{s}(u)-D\varphi_{s}(v))[y]\\\
+DB(\varphi_{s}(u)-\varphi_{s}(v))D\varphi_{s}(v)[y]\Big{]}ds$

for $v\in O$ and $y\in V$. Using (B.2),

$\left\|D\varphi_{t}(u)[y]-D\varphi_{t}(v)[y]\right\|_{V}\leq
2\int_{0}^{t}(t-s)^{-\alpha}C_{\alpha}e^{\omega(t-s)}\\\
\times\Big{[}\|\varphi_{s}(u)\|_{V}\|(D\varphi_{s}(u)-D\varphi_{s}(v))[y]\|_{V}+\|\varphi_{s}(u)-\varphi_{s}(v)\|_{V}\|D\varphi_{s}(v)[y]\|_{V}\Big{]}ds.$

Consequently,

$\left\|D\varphi_{t}(u)[y]-D\varphi_{t}(v)[y]\right\|_{V}\leq
2C_{\alpha}\frac{t^{1-\alpha}}{1-\alpha}\max\\{1,e^{\omega t}\\}\\\
\times\sup_{s\in[0,t]}\big{[}\|\varphi_{s}(u)\|_{V}\|(D\varphi_{s}(u)-D\varphi_{s}(v))[y]+\|\varphi_{s}(u)-\varphi_{s}(v)\|_{V}\|D\varphi_{s}(v)[y]\|_{V}\big{]}.$

Taking the supremum over $t\in[0,\tau]$, using (B.4) and (B.1), we obtain

$\sup_{t\in[0,\tau]}\left\|D\varphi_{t}(u)[y]-D\varphi_{t}(v)[y]\right\|_{V}\leq
C_{\tau,\alpha}\|u-v\|_{V}\sup_{t\in[0,\tau]}\|D\varphi_{t}(v)[y]\|_{V}$

It follows, by the uniform boundedness theorem, that

$\sup_{t\in[0,\tau]}\left\|D\varphi_{t}(u)-D\varphi_{t}(v)\right\|_{\mathcal{L}(V)}\leq
C_{\tau,\alpha}\|u-v\|_{V}.$

This concludes the case $i=1$; cases $i\geq 2$ follow trivially, as
$D^{i}\varphi_{t}(u)$ becomes independent of $u$ for $i\geq 2$. ∎

Suppose now that $u=0$ is a Lyapunov stable fixed point. Then, there exists a
neighborhood $O\subset V$ of $0$ on which the Lipschitz property (B.1)
continues to hold uniformly as $\tau\to\infty$. Moreover, (B.4) is trivially
satisfied for all $\tau\geq 0$. Via the above proof, we may hence conclude
that (B.3) also holds for all $\tau\geq 0$ over $O$, which, combined with the
boundedness of $D^{i}\varphi_{t}(0)$ for each $t\geq 0$ and $i\geq 0$, yields
that $\varphi_{t}$ is $C^{\infty}_{b}(O;X)$ for all $t\geq 0$ fixed.

The following Lemma establishes the sought-after joint continuity property.

######  Lemma B.2.

Let $\varphi$ denote the Navier-Stokes semiflow on $V$ (as in (4.2)), and let
$V_{0}$ be any closed linear subspace such that
$V_{0}\subset\mathrm{dom}(A_{U})$ (with $A_{U}$ as in (4.3)). Then, the map

$\displaystyle\mathbb{R}^{\geq}\times V_{0}\to\mathrm{M}_{i}(V_{0};V)$
$\displaystyle(t,v_{0})\mapsto D^{i}(\varphi_{t}\circ\imath_{0})(v_{0})$

is jointly continuous at $(0,0)$, for all $i\geq 0$ (here
$\imath_{0}:V_{0}\xhookrightarrow{}V$ is the inclusion).

###### Proof.

We first show the case $i=1$. Recall that $A_{U}$ is a sectorial operator. For
all $v_{0}\in V_{0}\subset\mathrm{dom}(A_{U})$, we have that

$\frac{e^{A_{U}t}v_{0}-v_{0}}{t}$

converges to $A_{U}v_{0}$ as $t\to 0$. Hence, the uniform boundedness
principle implies that, for any $\tau>0$,

$\|e^{A_{U}t}\imath_{0}-\imath_{0}\|_{\mathcal{L}(V_{0};V)}\leq Ct,\qquad
0\leq t\leq\tau.$

Supposed we are given $\varepsilon>0$. For $v_{0}\in V_{0}$, we estimate

$\displaystyle\|D(\varphi_{t}\circ\imath_{0})(v_{0})-\imath_{0}\|_{\mathcal{L}(V_{0};V)}$
$\displaystyle\leq\|D(\varphi_{t}\circ\imath_{0})(v_{0})-D(\varphi_{t}\circ\imath_{0})(0)\|_{\mathcal{L}(V_{0};V)}$
$\displaystyle\qquad+\|D(\varphi_{t}\circ\imath_{0})(0)-\imath_{0}\|_{\mathcal{L}(V_{0};V)}$
$\displaystyle\leq\|D\varphi_{t}(v_{0})\imath_{0}-e^{A_{U}t}\imath_{0}\|_{\mathcal{L}(V_{0};V)}+Ct.$
(B.5)

Choose $\tau<\varepsilon/(2C)$ and $O\subset\mathcal{D}^{\varphi}_{\tau}$
small enough so that (B.3) applies. Now taking
$\delta<\varepsilon/(2C_{\tau})$ (with $C_{\tau}$ as in (B.3), and such that
$B_{\delta}^{V_{0}}\subset O$) yields the claim, as
$D(\varphi\circ\tilde{\imath}_{0})\left([0,\tau)\times
B_{\delta}^{V_{0}}\right)\subset
B_{\varepsilon}^{\mathcal{L}(V_{0};V)}(\imath_{0})$.

The case $i=2$ is treated similarly. By Lemma B.1 and an argument analogous to
(B.5), it suffices to show $t\mapsto D^{2}(\varphi_{t}\circ\imath_{0})(0)$ is
norm-continuous at $t=0$. Observe that

$(t,s)\mapsto e^{A_{U}(t-s)}B(e^{A_{U}s}y_{0},e^{A_{U}s}\tilde{y}_{0})$

is continuous into $V$ for each fixed pair $y_{0},\tilde{y}_{0}\in V_{0}$ and
$t\geq s\geq 0$. Hence,

$\lim_{t\searrow
0}t^{-1}D^{2}(\varphi_{t}\circ\imath_{0})(0)[y_{0},\tilde{y}_{0}]=\lim_{t\searrow
0}t^{-1}\int_{0}^{t}2e^{A_{U}(t-s)}B(e^{A_{U}s}y_{0},e^{A_{U}s}\tilde{y}_{0})ds=2B(y_{0},\tilde{y}_{0}),$

for all $y_{0},\tilde{y}_{0}\in V_{0}$, and, by uniform boundedness once more,
$t\mapsto D^{2}(\varphi_{t}\circ\imath_{0})(0)$ is norm-continuous (at $0$).

The remainder ($i\neq 1,2$) follows trivially from Lemma B.1, since
$D^{i}\varphi_{t}(0)=0$ for all $t\geq 0$ and $i\neq 1,2$. ∎

## Acknowledgments

The author gratefully acknowledges the support of the Harding Foundation
through a PhD scholarship (https://www.hardingscholars.fund.cam.ac.uk).

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