Internat. J. Math. & Math. Sci.
131
(1987) 131-134
Vol. I0 No.
GAPS IN THE SEQUENCE n2
(mod 1)
VLADIMIR DROBOT
Department of Mathematics
Santa Clara University
Santa Clara, CA 95053
(Received September 16, 1985)
ABSTRACT:
Let
be an irrational number and let
$
For each N let
t.
[0,I] by
I0, I,
{k25}
the points
these intervals can assume.
case of the sequence
{nS},
KEY WORDS AND PHRASES.
I
k
{t}
denote the fractional part of
be the intervals resulting from the partition of
N
1,2,...,N.
Let T(N) be the number of distinct lengths
It is shown that T(N)
where T(N) <_ 3.
This is in contrast to the
=o.
Uniform distribution mod I.
980 AMS SUBJECT CLASSIFICATION CODE.
Primary 1OK31
INTRODUCTION.
1.
Let
({t}
t
fixed
N
5 be an irrational number and let {t} denote the fractional part of t
(mod I)
t
It], where [.] is the greatest integer function). For each
the points
{NS}
partition on the interval [0,,] into
It is well known that the lengths of these intervals can assume
N+I subintervals.
only 3 values:
{9}, {29}, {3
a,
and
a+.
The values of a and
can be actually given explicitly
in terms of N and the continued fraction expansion of 9.
conjecture and it was first proved by Swierczkowski in
exposition of all this, see [2].
This is known as Steinhaus
[|].
For an excellent
In this note we investigate the analogous problem
{n25}.
It turns out that in this case the number of different
lengths these subintervals can assume, is unbounded. More precisely we have the
for the sequence
following results.
2.
MAIN RESULTS.
I be the
For each integer N let I0,_I|,
N
N.
k
1,2
N+I subintervals resulting from partition of [0, I] by the points
Let T(N) be the number of distinct lengths these subintervals assume. Then for each
Theorem
Let 5 be an irrational.
{k25}
> O,
T(N) > Nexp {-( +E) in
22
In N
In In N
for N >
N(E)
(2 I)
V. DROBOT
132
for every 6 > 0 and N > N(6).
T(N) >N
In particular
In what follows
$ > 0 is some fixed irrational.
We need the following four
simple lemmas.
LEMMA I.
For any integers r,s
{r$}
{(r+s)$}
E
where
+
m
(2.2)
;.
0 or
We have
PROOF.
{(r+s)6}
(r + s)$
{r$}
{r$}
{r$}
Thus, if 0 <
<
and if
{s$}
LEMMA 2.
+
+ [(r+s)$]
{uS}
{s$}
<
then (2) holds with E
O,
+
{s}
< 2
then (2) holds with E
I.
{x$}
If
m
If
y < x, let x
then
0
then
e
+
{k$}
{y6}
<
{x$}
{y&}
+
{k&}
{x$}
>
{y$}
{y$}.
f
{x$}
If
{(x+y)$}
+
Then by Lena
x + k.
contrary to hypothesis, so that E
+
0 and (2.3) holds.
Again, by Lena
m
-;
so that E
{y$}
{x$}
(2.3)
y < x
and (2.3) holds again.
For any two non-negative integers x,y,
3.
PROOF.
Then
E
y + k, k > 0.
{x&}
E
{(x-y)$}
{x6}
<
x < y
Suppose x < y so that y
{y6}
If
{x6}
+ integer
(i f }
{(y-x)$}
PROOF.
{s6}
+
+
Suppose x,y are integers,
{y$}
{r$}
+ [r] + [s$]
then by
m
m
{y$}
{x$} #
{y$}
ee=a
0 or
contradicting the fact that $ is irrational.
Suppose
L 4.
{y]$} -{x,$}>0
A
If A
B then
x;
II:
x
x>
III:
IV:
< yl, x <
2
y,
y;,
y;,
<
x]
x
y
>
x
2,
Y2
are non-negative integers and let
{y26} -{x25}>0
B
x2
Y2
We will use Leman 2 and 3 and consider 4 cases
PROOF.
I:
x;, y,
x
x
x
2
2
2
Y2
>
Y2
<
Y2
>
Y2
In case I we get from Lena 2
A
so A
{(y]
B implies
x])$}
Y
xl
B
Y2
{(Y2 x2)6}
x2
In case II, by Lena 2 we get
A
{(yl
x])$}
B
{(x 2
Y2)6}
GAPS IN THE SEQUENCE
so A
B cannot hold by Lemma 3.
Similarly, A
Y2
133
x2
B cannot hold in case III, and A
x;
in case IV.
We are now ready to prove the Theorem I.
partition of [0,1] by the points
{029}
with a collection A(N) of N intervals.
[{x9}, {y9}]
2
by Lemma 4.
2
2
xl
{N29}.
some
x),
[{x2|9},
If two of these intervals
If we
we are left
{y219 }]
and
equal length then
are of
Y;
Let N be fixed and consider the
{129}, {229}, {329}
interval [{x29},I] for
0,
exclude the right most interval (i.e. the
assume.
y;
B implies
2
(2.4)
x2
Y2
Let T(N) be the number of distinct lengths these intervals from A(N) can
The collection A(N) is then divided into T(N) subsets, any two intervals from
one subset are of equal length.
Thus, by (2 4)
k
One of these subsets must contain N/T(N) intervals.
< k < N 2 such that the equation
there exists an integer k,
y
2
x
2
(y-x)(y+x)
(2.5)
has N/T(N) solutions in integers x,y,
distinct divisors of k.
(xl,y I)
(x2,Y2),
2
If
then
2
2
< x < y < N
2
Yl xl Y2 x2’
Yl xl Y2 x2 and y|
2
Each such solution produces 2
<- xi
+
<
x
< N
Yi
Y2
+
2
for
,2 and
Thus
x2"
N/T(N) < d(k)
(2.6)
where d(z) is the number of divisors of d.
d(z) <
exp{(l+)
in 2
In z
in
in
%0( z)
z
This was first proved by Wigert in [3]
It is well known that for each e > 0
see also
for
z > z(e)
[4], Satz 5.2
Since k < N
2
we get
from (2.6)
2N/T(N) < %0(e,k) < 0(e,N 2)
exp
{(I+g)
in 2
< exp
{(+6)
in 2
2 InN
(2 7)
in 2 + in in N
2
in N
in in N
for N >
N(6)
Solving this inequality for T(N) gives (2.1).
The argument carries over almost without any change to the sequence
integer p > I.
THEOREM 2.
I0, I],
points
I
{k29},
{nPg}
for any
The corresponding estimate is then as follows.
N
k
Let 9 be an irrational and p >
be the
N+|
T (N) > N
p
For each integer N let
[0,] by the
Let T (N) be the number of distinct lengths these
P
subintervals resulting from partition of
1,2,...,N.
intervals can assume.
an integer.
Then for each
exp{-(l+e)
In 2 p
> 0
in N
In In N
for
N > N()
V. DROBOT
13
REFERENCES
SWIERCZKOWSKI, S. On successive settings of an arc on the circumference of a
circle, Fund. Math. 46 (1958), 187-189.
2. SLATER, N.B. Gaps and steps for the sequence n$ mod I, Proc. Cambridge Philos.
Soc. 63 (1967), Ill5-1123.
3. WIGERT, S. Sur l’ordre de grandeur du numbre des diviseurs d’un entier, Arkiv for
Math., Astr., och Fhsik 3, No. 18 (1907), I-9.
4. PRACHAR, K.
Primzahlverteilung, Springer Verlag, Gottingen, 1957.