Download binary digit distribution over naturally defined sequences

TRANSACTIONSOF THE
AMERICANMATHEMATICALSOCIETY
Volume 213. 1975
BINARYDIGIT DISTRIBUTIONOVER NATURALLY
DEFINED SEQUENCES
BY
D. J.NEWMAN AND MORTONSLATER
ABSTRACT. In a previous paper the first author showed that multiples
of 3 prefer to have an even number of ones in their binary digit expansion.
In this paper it is shown that in some general classes
of naturally defined
sequences,
the probability
that a member of a particular
sequence has an
even number of ones in its binary expansion
is A.
I. Binary digits contained
proved that in counting
in arithmetic
progressions.
In [4] Newman
the number of ones in the binary expansion
multiple
of 3 that count will come out even more often than odd.
striking
result
in that superfically
distinguishable
sequence
about multiples
would be expected
there seems
nothing
either
of 3 in the binary system.
to oscillate
in "odds"
of a
This is a
special
Certainly
or. "evens"
or
a random
being more
numerable.
In the case
of multiples
of 3, the probability
have an even number of digits.
odds is o(n), in fact about
A natural
question
which exhibit
proportion
That is, the preponderance
here would be:
Are there other natural
and further is there a natural
of evens over odds or vice versa
of the number of elements
in the sequence?
preponderance
of evens
sequence does
limN_#0O(E(N)/i4(N)) = 0 ?
In this paper
classes
of natural
Theorem.
ing function
of evens over
nlos3/1°«4.
this odd behavior
which the preponderance
is l/2 that the multiples
isa
sequences
sequence
positive
If E(N) gives
over odds and A(N) is the counting
in
function
the
of the
we shall prove that the above limit is zero over various
sequences.
Let A be the sequence
of this sequence.
{M • n\°°_x. Let A^(N)
be the count-
Then
F ( V)
lim -r^j-r-0
/V-»ooA.AN)
and
\EM(N) - %A iN)\ < C/Vl0«3/lo8 4.
Received by the editors June 4, 1974.
AMS(MOS)subject classifications (1970). Primary 10L10, 10H20, 10H30.
Key words and phrases.
abundant
Binary digits,
arithmetic
progressions,
sieving,
numbers.
Copyright © 1975, American Mathematical Society
71
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72
D. J. NEWMANAND M. SLATER
Proof.
Let 0(72) be the number of ones in the binary expansion
for n.
Let
y,v)=
Z
(-i)D<"-M'->.
0<y<N/M
Thus we want to prove that when N is a multiple
of M, |SM(iV)| < CNln3,/ln4.
As in [4] it follows that, if N = 2Kl + 2K2 + • • • + 2Ki, Kx > K2 > - -. >
K. > 0, then
SM{N)= S{2Kl-Pl)where
S{2K2 - p2) + ... + (- 1)/- lS{2Ki - p ) + (-1)'
p¿ is the small
positive
integer
so that
p¿ + 2 '+l + 2 «*2 + ••• +
2 J is divisible by M.
If it can be shown that S(2K'' - p.) < C2K'(ln3/I,l4)
-
j
y
then the result
yn3/ln4
V'V> < Ci Z 2K,0nVln 4) < Ci Z 2Kí
= CiNla3/'n "
¿=l
i=l
follows.
«2*-,)
This
= /,,
JLLu
.„ t.'»-^»
m. ok-b
\x\='A (j
_ xM)x2k-P
r
can be evaluated
,
w
M ..
,.
_
Z
.
.
by the method
.
£J • an Mth root of unity
of residues
n*-iu-a>f)
-iS~i-r—
2"-p+l
we shall
Jfe-1
max J] (1-22")|
and is equal to
fe-1
<max
II d-«?
- tú, l\
CO.
In the lemma which follows
dx.
+
+l
1
»2=0
l
>•
show that
,fcln3/ln4
<CX2*
H = 1 72=0
Thus pending proof of the lemma, the theorem is established.
Lemma.
i_
*-l
1
ax J]
Ün3/ln4
tl-*2 >")| <C12Asl
,= 0
The proof of this lemma is quite interesting.
idea of where the maximum occurs
the maximum
Although we have no
or what the value is, we can show that
is quite close to the value at z = e
Proof. If z = ei0, |l-z2"|
max
= 2|sin 2"-1ö|,
k-l
k-1
2k
n a-*2 ) = max
e
n=0
Ul=i B=0
,
so
II sin2"-1Ö
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73
BINARYDIGIT DISTRIBUTION
Substituting
26 for 6 to simplify
notation
we rewrite
the above product
as
follows:
(|sinl/30|
*n
|sin^32"ö
Since the max of the product
be treated separately:
sin1/3 2" +1Ö|j |sin2/3 2fe"10| .
< the product
of the maximums,
each term may
|sin1/30| |sin2/32A:- !0| < 1.
If
/(0) = sin2/32"0sinl/32"
+10,
/'(ö)=2Cisin-l/32«öcos2"öSinl/32"
+ 2^isin-2/3
at a max, we have
+1ö
272+ lö cos 222+ 10 sin2/3272ö
tan 2"+ 6 = - tan 2"0.
The only solutions
of this are
2"0= 77/3, 277/3, 477/3, 57r/3, 0, 77. A quick check shows then that
maxe|sin2/3 2n0sin1/32"+10| = V3£. Thus
fc-1
Jyß\k-1 = | V3 C2ln3/ln4)
n d-22") <2
max
22=0
and the lemma is established.
II.
a
Suppose
that we had a general
••• and we wanted to determine
sequence
the relative
number of ones in its binary expansion.
of integers
frequency
a y «2, ••• ,
that a. has an even
Let A(n) be the counting
function
for the sequence.
If the following
number of ones.
integral
Choose
is 0(^(22)) then
k so that
r
2
M of the a.'s
11* nU-z2
) /
«»-J|.|..-=ST—
(,
dz.
a;<N
We will establish
Theorem.
that the square
Let \a.\
free numbers
be an arithmetic
have an even
< N < 2 + ,
divide
evenly.
progression.
Then
CN-S~ log/V.
Proof. |pov)| < /H=1 ¡nnfe=0i- z2"\ \\i<Nz-a'\
In-§I we showed that on the unit circle
nd-
„2"i
I 22=0
dB.
the max of
<C|Vlog3/log4
logJ_<
log 4
So |PQV)|<CN-*-f\2.z-"i\d6.
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g_
P(N) <
74
D. J. NEWMANANDM. SLATER
Now if fß(i is an arithmetic
n terms
progression,
with difference
a, containing
< N,
/
-J"
11-
J-n
a;<N
Set
Q= a =/_n T=T dd * C(Jo*nde+L f) * <*«*
- lo*«■
Differentiating,
/|Sa<N2-a^l9
of course
we find the expression
minimized for k = ifri so
< C log N and for \a\ an a.p. |P(/V)| < CN-8""logW. This,
shows that P{N) = o{A{N)).
We might remark that the above argument
is quite general.
We have
P{N)
< N-8~L1{AN). For equal frequency we need P{N) = o{A{N)) so lHan)
o{A{N)/N-8~) is sufficient.
Returning
to the question
above that the L
of arithmetic
progressions
of many arithmetic
with respect
to the digits in the binary expansions.
N
progressions
we mean the set of integers
principle
on the collection
progressions
If our set
we see from the
norm of an a.p. is only log N. Thus a set which is a
combination
a.p.'s,
would have to be equidistributed
By a combination
of
formed by using the inclusion-exclusion
of a.p.'s.
Under these
conditions
we can combine
and our set will be equidistributed.
ÍC¿!, C. ¿ C,
i 4 7» is made up of a.p.'s
in this way, then
.2
z~*\+...<N''
f\T.*c\<A\L,"<\,-A\Z
We note in passing
and has positive
square-free
=
that if a set is a complement
logN,
of one of the above sets
density, then»it also has the equidistribution
property.
The
numbers are such a set.
We first form the set of numbers containing
combine multiples
of 2
and 3
then subtract
a square factor as follows:
off multiples
of 6
etc., or
if Sj is the set of multiples of d < N we form - %x< . ßj u{d)S¿.
Now while each integer
we have too many arithmetic
We observe
progressions,
a square factor occurs exactly
namely
however that our set of progressions
sets of progressions:
one set containing
which our theorem applies
o{N) integers
containing
once,
A/"', to use our estimate.
can be divided
fewer than Af
into two
progressions
and the other set of progressions
to
which contains
all together.
We note that this approach
formula for evaluating
is valid because
the variance
in distribution,
instead
of using the integral
we can simply use as an
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75
BINARYDIGIT DISTRIBUTION
upper bound the number of integers
in a set.
progressions
S¿ as follows:
progressions
plus the remaining integers
With this in mind we divide the
S¿, d < k, and S ,, k < d < yN.
i<d<\ß
\k<
Thus the total estimate
iß J/
This gives
k
which total
\<N/k + yß.
for the variance
this is minimized for k = N' /yjlog N.
is N'8k log N + N/k + y/Ñ
Thus we find that the variance
distribution
is < /V9 \A°g N.
This establishes
the equidistribution
square-free
numbers provided
that numbers containing
and
in
for
a square factor have
density < 1.
In order to establish
-£
this, .we must evaluate
p(d)dW2K-N
Z
i^ +o(VÑ)
sU)
and since
the sum
d>yJN d
£(2) = 77/6 the number of numbers
< N containing
is (n2 - 6)/V/t72 and this is of course a set of density
The set of square-free
integers
follows that the set of square-free
Theorem.
The probability
is the complement
integers
that a square-free
of one's in its binary expansion
set of square-free
integers
arbitrary
of the above set.
density
It
and also
integer has an even number
In the previous
section
out all multiples
more general
those integers
set, are the remaining
Of course
set.
by sieving
we pose the following
from the set of all integers
< 1.
is %.
III. Some general sieved sets.
In this section
has positive
a square factor
integers
of the perfect
question.
which are multiples
equidistributed
the answer to this is no if no restriction
we formed the
squares.
If we sieve out
of some
in the binary system?
is made on the sieving
For example we might sieve with the set of all odd primes.
In that case
the remaining set would be only powers of 2.
We will prove the following
Theorem.
Given a sequence
< oc then the set of all integers
equidistributed
of integers
sieved
of any of the a.'s
l/a.
is
in the binary system.
We will prove the theorem in two steps.
set has positive
progression
a , a., ••• such that 2~«
that are not multiples
estimates
lower density.
Step 1 will establish
that our
Step 2 will show that the arithmetic
of §11 apply.
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76
D. J.NEWMAN AND M. SLATER
Lemma 1. Let A{n) be the counting
We note that
of the sieved
Il°^0 (l — l/a¿) > 0 since the infinite
if and only if ^°10 ^-la- converges
that the density
function
and this was given.
of such a set actually
set.
product
converges
Erdös [2] has shown
exists.
Proof of Lemma 1. The lemma will follow if we can show that the
density
of a set sieved by the first
by the convergence
Nli>Ml/a
M of the a¿'s is > Il^Kl
of 1.a., sieving
< eN integers
by the remaining
by proper choice
The proof is by induction.
a.'s removes
for
at most
of M.
If we sieve with only aQ then DQ > (l - l/ßn)
and we need to show that Dk > (l - l/a¿)ofe_j
sieving
- l/a);
where Dk is the density
after
with k + 1 integers.
We note that there
exists
a number not being divisible
an exact probability
or density
by any of k integers
associated
with
namely
k-l
P.1-TÍ+
Z
i=0
where
I I denotes
'
—-(-D*-
iV<*-llflf
the least
ay!
i<Vßfe-l!
common multiple.
Thus we can phrase
the argu-
ment in terms of probability.
We want to establish
P{n not divisible
that for n < N,
by any aQ, ...
, afe)
> Í1 - — JP{n not divisible by any aQ, ... , ak_x)
or
— P{n not divisible
ak
by any aQ, ...
, ak_))
> P(n not divisible by aQ, ...
- P{n not divisible
, a, _j)
by flQ, ...
, a^)
> P{n divisible by a^ but not aQ.ak-?
> P{ma, not divisible
1
{a., a)
...
, «¿.j)
aQ
a
= ——P In not divisible
by ;--,
...,
is the g.c.d.
step follows
since
ak
where
/
by aQ, ...
\
The last
{aa, a.)
—
{a k-l'
a^ma^
-*>)
if and only
if a\m{a.,
a.)fz —* a./{a.,
a,)\m
and also the number of numbers < N is a,tet'
i
i
i
te,
times as much as the number of numbers < N/ak and the probability
that a
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77
BINARYDIGIT DISTRIBUTION
number < N/a,
probability
is divisible
by a member of a certain
of a number < ¿V being divisible
sufficiently
large.
set is the same as the
by a member of the set for N
But our last inequality
P(n not divisible
by any an, ...
> P/72 not divisible
, «,)
a0
by
ak-
(aQ,ak)"
is true by inclusion.
Incidentally
Thus reversing
we may rewrite
the logic we arrive at the desired
the statement
k
i-Si+LA—»(-o
k j
2.7
'iak_vak)j
1
ana,
0 1 • • • a,k
>i-Z¿+
Z >a¿'
rAv--*-»'
i '
i.i;i<i
ay>
\an,
an inequality
which would seem quite difficult
Proof of theorem.
lower density.
tion property
Thus if we can show that its complement
with the first
may be formed in two parts
¿-integers
more than.once
remaining
set has positive
has the equidistribu-
sieving
adds
so we subtract
etc.
set.
This
we subtracted
we have
integers
First
puts
2
we
k progres-
have been placed
off the multiples
since
Altogether
Ai£. ,1/a.
estimate
as follows.
some integers
of \a., a., ak\
too many times
equidistribution
in our sieving
However,
Then add back multiples
numbers
,ak\
directly.
We have now shown that our sieved
sions into our complement.
complement
to establish
ax, ...
the theorem will follow.
The complement
sieve
lemma.
of Lemma 1 as
in the
of \a{, a.\.
off some
progressions.
to the complement.
The
Thus the
is
2kN-8logN+NZr
i>k '
and we want to show that this is o(N).
If we pick k = (2/11) log N we need only pick N so large that
£.>Al/a.
< { by convergence,
One application
integers
of the above theorem is of course that the square-free
are equidistributed.
A second
deficient
and the proof of the theorem is complete.
application
numbers
would be that both the abundant
have the equidistribution
is a number which is exceeded
number is abundant
any multiple
if it is abundant
numbers
An abundant
by the sum of its proper divisors.
is also abundant
d\A, Bd\BA and BA will be abundant.
abundant
property.
and the
number
If a
for if A is abundant
and
A number is said to be primitive
but not a multiple
of another
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abundant
number.
D. J. NEWMANAND M. SLATER
78
Erdös [2] has proven that the sum of the reciprocals
abundant
numbers
converges.
our theorem as the sieving
the sieved
buted.
Thus
set.
the primitive
Thus the abundants
set composed of deficient
of the primitive
abundants
and perfect
can serve
in
are equidistributed
and
numbers is also equidistri-
Since the perfect numbers are rare (Euler proved that all even
perfect numbers are of the form 22
PN ) we have equidistribution
- 2*-1
and all odds have the form
for the deficient numbers also.
A further example would be that the set of all numbers not divisible
any twin prime has the equidistribution
from Brun's result
property.
by
This follows at once,
that the sum of the reciprocals
of the twin primes con-
verges [3].
BIBLIOGRAPHY
1. L. Dickson, History of the theory of numbers. Vols. 1, 2, 3» Publ. no. 256,
Carnegie Inst., Washington, D.C., 1919, 1920, 1923; reprint, Steche«, New York.
2. P. Erdös,
(1934), 278-282.
3.
E. Landau,
On the density
Elementare
of abundant
Zahlentheorie,
numbers,
J. London Math. Soc. 9
Teubner,
Leipzig,
1927; English
transi., Chelsea, New York, 1958. MR 19, 1159.
4. D. J. Newman, On the number of binary digits
in a multiple
of three, Proc.
Amer. Math. S0c. 21 (1969), 719-721.
DEPARTMENTOF MATHEMATICS,BELFER GRADUATESCHOOLOF SCIENCE,
YESHIVAUNIVERSITY,NEWYORK, NEWYORK 10033
DEPARTMENTOF MATHEMATICS,CITY COLLEGE (CUNY), NEWYORK, NEWYORK
10031
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