Counting divisors

11 Pages·2014·0.089 MB·English

by  Graham J. O. Jameson

Counting divisors
G.J.O. Jameson, Math. Gazette 98 (2014)
The divisor function (cid:28)(n)
The \divisors" of n are the positive integers (including 1 and n itself) that divide into
n. We will denote by (cid:28)(n) the number of these divisors. (This notation owes its origin to
German \Teiler"; English speakers often use d(n), but in my view, the letter d is in far too
much demand to be tied up in this way!).
Here we outline a number of results about (cid:28)(n), culminating in Dirichlet’s splendid
theorem on the cumulative sum of its values. For any readers with the appetite for it, we
then go on to show how the methods extend to give analogous results about the number of
ways of expressing a number as a product of three factors.
Every divisor j comes with a natural partner, its \codivisor" n=j. Another way to
describe (cid:28)(n) is: the number of ordered pairs (j;k) of positive integers with jk = n (just
think of j as de(cid:12)ning the divisors in turn). Really (cid:28) is (cid:28) in a sequence of functions (cid:28) ; we
2 k
return to this later.
First, a pleasant fact that follows at once from this pairing idea:
PROPOSITION 1. (cid:28)(n) is odd if and only if n is a square.
Proof. If n is not a square, then all the divisors can be listed in pairs (j;k), with
j < pn and jk = n, so the number of them is even. If n is a square, say n = m2, then the
divisors consist of these pairs together with m, so the number is odd. (cid:3)
It is quite easy to give an expression for (cid:28)(n) in terms of the prime factorisation of n.
Note (cid:12)rst that if p is prime, then (cid:28)(p) = 2 and (cid:28)(pa) = a+1, since the divisors of pa are
1;p;:::;pa. For a general number n, we have the following expression for (cid:28)(n):
PROPOSITION 2. Suppose that n > 1, with prime factorization n = m paj. Then
j=1 j
(cid:28)(n) = m (a +1).
j=1 j Q
Q
Proof. Becauseofuniqueprimefactorisation,thedivisorsarethenumberspb1pb2 :::pbm,
1 2 m
where 0 b a for each j. For each j, there are a +1 possible values for b , and each
j j j j
(cid:20) (cid:20)
combination gives a di(cid:11)erent divisor. (The divisor 1 is obtained by choosing b = 0 for each
j
j, and the divisor n by choosing b = a ; it would be strange to exclude these two.) (cid:3)
j j
For example, since 60 = 22 3 5, we have (cid:28)(60) = 3 2 2 = 12.
(cid:2) (cid:2) (cid:2) (cid:2)
1
As a further illustration, we will describe, in terms of the possible patterns of prime
factors, the numbers for which (cid:28)(n) = 8, and (cid:12)nd the smallest such number. The factor-
izations of 8 are 8, 4 2 and 2 2 2. So (cid:28)(n) = 8 if n is one of the forms p7, p3q, pqr
(cid:2) (cid:2) (cid:2)
(where p, q, r are distinct primes). The smallest of each type is: 27 = 128, 23 3 = 24,
(cid:2)
2 3 5 = 30. So the smallest is 24. (The reader might like to repeat this exercise with a
(cid:2) (cid:2)
di(cid:11)erent value for (cid:28)(n), for example 12.)
Of course, Proposition 1 follows easily from Proposition 2: if n is a square, then each
a is even, so m (a +1) is odd.
j j=1 j
Q
A way inwhich the divisor function seems toappear outofthe blue is asfollows. Recall
that the Riemann zeta function is de(cid:12)ned by (cid:16)(s) = 1 1=ns for real s > 1 (or indeed for
n=1
complex s with Re s > 1). So
P
1 1
1 1
(cid:16)(s)2 = :
js ks
! !
j=1 k=1
X X
In this product, consider the terms that equate to 1=ns for a (cid:12)xed n. There is such a term
for each ordered pair (j;k) with jk = n, so 1=ns occurs (cid:28)(n) times. Hence
1
(cid:28)(n)
(cid:16)(s)2 = :
ns
n=1
X
(Again, it is clear that 1 and n must be counted as divisors.)
This is actually a special case of the following. Given arithmetic functions a(n), b(n),
the convolution a b is de(cid:12)ned by
(cid:3)
n
(a b)(n) = a(j)b = a(j)b(k):
(cid:3) j
jjn (cid:18) (cid:19) jk=n
X X
Clearly, (cid:28) = u u, where u is the \unit function" de(cid:12)ned by u(n) = 1 for all n. Convolution
(cid:3)
de(cid:12)nes the coe(cid:14)cients when two Dirichlet series are multiplied: if we write 1 a(n)n(cid:0)s =
n=1
F (s),andsimilarlyF (s),thenF (s)F (s) = F (s). Convolutionsareveryusefulinnumber
a b a b a(cid:3)b P
theory, but they will not be used in this article.
Summation of (cid:28)(n)
Individual values of (cid:28)(n) (cid:13)uctuate wildly. However, this is smoothed out when the
cumulative sums of these values are considered, and in fact it is possible to give a very
satisfactory estimation of such sums. For all real x > 0, write
T(x) = (cid:28)(n):
n(cid:20)x
X
2
We do not restrict x to integer values: the advantage of this will be seen in the applications
below. Thenotation means thatsummationisovertheintegers nsuch that1 n x.
n(cid:20)x (cid:20) (cid:20)
Of course, as a function of a real variable, T(x) is constant between integers and jumps by
P
(cid:28)(n) at the integer n.
There are two ways, both obvious, in which T(x) can be described as enumerating
ordered pairs of positive integers:
(A) the number of ordered pairs (j;n) with j n and n x;
j (cid:20)
(B) the number of ordered pairs (j;k) with jk x.
(cid:20)
Geometrically, ordered pairs (j;k) of integers are called \lattice points". Note that (B) can
be described as the number of lattice points in the (s;t)-plane lying below the hyperbola
st = x.
Denote by [x] the largest integer not greater than x, and write x for x [x], the
f g (cid:0)
fractional part of x. Clearly, 0 x < 1.
(cid:20) f g
From (A), by a neat example of \double counting", we have at once the following
expression for T(x):
PROPOSITION 3. We have
x
T(x) = : (1)
j
j(cid:20)x(cid:20) (cid:21)
X
Proof. Consider the pairs in (A). For a (cid:12)xed j (instead of (cid:12)xed n), the values of n
allowed are the multiples kj not greater than x, so that k x=j. The number of such k is
(cid:20)
clearly [x=j]. The stated expression follows. (cid:3)
This expression gives a way to evaluate T(x) without calculating individual values of
(cid:28)(n). However, a better alternative will be described shortly.
To derive a formula approximating T(x), we need an estimation of the harmonic sum
1
H(x) = :
n
n(cid:20)x
X
x
As the reader may be aware, comparison with the integral (1=t)dt = lnx shows that H(x)
1
is roughly lnx. For greater precision, and for later application, we reiterate here how such
R
estimations work in general. The basic underlying result for a decreasing function is:
3
LEMMA 1. Let f(t) be a decreasing, non-negative function for t 1, and let
(cid:21)
x
S(x) = f(n); I(x) = f(t)dt:
n(cid:20)x Z1
X
Then for all x 1,
(cid:21)
I(x) S(x) I(x)+f(1): (cid:3)
(cid:20) (cid:20)
When x is an integer, this result is obtained by combining the obvious inequalities
r
f(r) f(t)dt f(r 1)
(cid:20) (cid:20) (cid:0)
Zr(cid:0)1
for 2 r x. The version for non-integer x follows quite easily; e.g. see [Jam1, p. 19].
(cid:20) (cid:20)
Applied with f(t) = 1=t, Lemma 1 gives:
LEMMA 2. For all x > 1,
lnx H(x) lnx+1: (cid:3) (2)
(cid:20) (cid:20)
The following estimation of T(x) now drops into our lap.
THEOREM 1. For all x > 1,
xlnx x T(x) xlnx+x: (3)
(cid:0) (cid:20) (cid:20)
Proof. Since x 1 [x] x, (1) gives
(cid:0) (cid:20) (cid:20)
n
x x
1 T(x) ;
j (cid:0) (cid:20) (cid:20) j
j(cid:20)x(cid:18) (cid:19) j(cid:20)x
X X
which equates to
xH(x) [x] T(x) xH(x):
(cid:0) (cid:20) (cid:20)
(cid:3)
Inserting (2), we obtain (3).
We now show how to obtain a much more accurate estimate. Both the result and the
method were presented by Dirichlet in 1841. Peter Gustav Lejeune Dirichlet (1805{1859)
grew up in the German Rhineland, in a family of mixed German and French origins. He
taught for most of his life in Berlin, and married a sister of the composer Mendelssohn. He
made it his mission to make the awe-inspiring works of Gauss better known and understood,
buthealsomadeimportantcontributionsofhisowninseveraldi(cid:11)erentareasofmathematics.
4
The key step is to replace (1) by the following expression for T(x), which is known as
\Dirichlet’s hyperbola identity":
PROPOSITION 4. For all x > 1,
x
T(x) = 2 [x1=2]2: (4)
j (cid:0)
j(cid:20)Xx1=2(cid:20) (cid:21)
Proof. In the expression (B), let N , N be the number of pairs (j;k) with j x1=2
1 2
(cid:20)
and k x1=2 respectively (in the diagram, these are the points in A B and A C). By
(cid:20) [ [
symmetry, N = N . For (cid:12)xed j x1=2, the number of k such that jk x is [x=j]. Hence
1 2
(cid:20) (cid:20)
N = [x=j].
1 j(cid:20)x1=2
P x
k
B
jk = x
x1/2
A C
1 x1/2 x
j
In the sum N + N , the points satisfying both j x1=2 and k x1=2 are counted
1 2
(cid:20) (cid:20)
twice (in the diagram, these are the points in A). Hence T(x) = 2N N , where N is the
1 0 0
(cid:0)
number of such points. Clearly, N = [x1=2]2, so (4) follows. (cid:3)
0
Note that the summation in (4) is over the shorter range j x1=2 instead of j x. We
(cid:20) (cid:20)
illustrate this by using (4) to calculate T(100). We tabulate the values of [100=j] as follows:
j 1 2 3 4 5 6 7 8 9 10
[100=j] 100 50 33 25 20 16 14 12 11 10
Addition gives 10 [100=j] = 291. So T(100) = 2 291 102 = 482. Using (1), we would
j=1 (cid:2) (cid:0)
have needed to consider [100=j] for all j 100.
P
(cid:20)
At the same time, we deploy a better approximation than (2) for H(x): such an
approximation is lnx + (cid:13), where (cid:13) is Euler’s constant. The exact statement, in the form
that we require, is as follows; a detailed proof (for those who wish) can be seen in [Jam1, p.
24{25].
5
LEMMA 3. For all x 1
(cid:21)
H(x) = lnx+(cid:13) +q(x); (5)
where q(x) 1=x. (cid:3)
j j (cid:20)
With (4) and (5), the path to Dirichlet’s theorem is clear. It says:
THEOREM 2. For all x 1,
(cid:21)
T(x) = xlnx+(2(cid:13) 1)x+(cid:1)(x); (6)
(cid:0)
where (cid:1)(x) 4x1=2. In other words, T(x) is approximated by xlnx+(2(cid:13) 1)x, with the
j j (cid:20) (cid:0)
error no greater than 4x1=2.
Proof. Let N , N be as in Proposition 4. Then N = [x1=2]2. Since x1=2 1 [x1=2]
0 1 0
(cid:0) (cid:20) (cid:20)
x1=2, we have N = x q (x), where 0 q (x) 2x1=2.
0 1 1
(cid:0) (cid:20) (cid:20)
Recalling that [x] = x x and 0 x < 1, we have
(cid:0)f g (cid:20) f g
x x
N = = xH(x1=2) q (x);
1 2
j (cid:0) j (cid:0)
j(cid:20)Xx1=2(cid:18) (cid:26) (cid:27)(cid:19)
where 0 q (x) x1=2. By (5),
2
(cid:20) (cid:20)
H(x1=2) = 1 lnx+(cid:13) +q(x1=2);
2
where q(x1=2) 1=x1=2, so
j j (cid:20)
xH(x1=2) = x(1 lnx+(cid:13))+q (x);
2 3
where q (x) x1=2. Put together, we obtain
3
j j (cid:20)
T(x) = 2N N = xlnx+(2(cid:13) 1)x+q (x) 2q (x)+2q (x):
1 0 1 2 3
(cid:0) (cid:0) (cid:0)
Clearly, q (x) 2q (x) 2x1=2, and hence (cid:1)(x) 4x1=2. (cid:3)
1 2
j (cid:0) j (cid:20) j j (cid:20)
Another striking example of the importance of Euler’s constant!
Recall that the notation O[g(x)] denotes a quantity f(x) that satis(cid:12)es f(x) Kg(x)
j j (cid:20)
for some constant K throughout the range of de(cid:12)nition. In this notation, (6) can be stated
as follows:
T(x) = xlnx+(2(cid:13) 1)x+O(x1=2):
(cid:0)
6
The theorem can be interpreted as saying that (cid:28)(n) averages out as if it were lnn+2(cid:13),
because the method of integral estimation gives
(lnn+2(cid:13)) = xlnx+(2(cid:13) 1)x+O(lnx):
(cid:0)
n(cid:20)x
X
The following table compares some actual values of T(x) with the estimate
xlnx+(2(cid:13) 1)x.
(cid:0)
x 100 1;000 10;000 100;000 1;000;000
actual 482 7;069 93;668 1;166;750 13;970;034
estimate 476 7;062 93;648 1;166;736 13;969;942
A table like this only gives a very partial picture, because the di(cid:11)erence (cid:1)(x) ob-
viously has irregular (cid:13)uctuations corresponding to those of (cid:28)(n), and can be positive or
negative. To illustrate this, note that since 100;000 = 2555, we have (cid:28)(100;000) = 36, so
that S (99;999) = 1;166;714, while the estimate is 1;166;723. So (cid:1)(99;999) 9 (to the
(cid:28)
(cid:25) (cid:0)
nearest integer), while (cid:1)(100;000) 14.
(cid:25)
With this said, (cid:1)(x) is visibly smaller than the stated estimate 4x1=2 for the values
listed. The problem of determining the true order ofmagnitude of(cid:1)(x) is called the \Dirich-
let divisor problem". It has been the subject of a great deal of study. The point of interest is
not the factor 4 ((which can in fact quite easily be reduced to 1) but the power of x. Denote
by (cid:18) the in(cid:12)mum of numbers (cid:18) such that (cid:1)(x) is O(x(cid:18)). It was already shown by Voronoi in
0
1903 that (cid:18) 1. A proof of this can be seen in [Ten, sect. 1.6.4]; it uses delicate estimates
0 (cid:20) 3
of \exponential sums" of the form eif(n). With very considerable e(cid:11)ort, the estimate
n(cid:20)x
has been reduced slightly in a long succession of small steps, rather like the 10,000 metres
P
record in athletics! The current record, held by M. N. Huxley, is (cid:18) 131 0:31490. On
0 (cid:20) 416 (cid:25)
the other hand, it was shown by Hardy and Landau in 1915 that (cid:18) is at least 1, and a bold
0 4
conjecture is that this is the true value.
Combining Proposition 3 and Theorem 2, we can derive a rather striking application
to sums of fractional parts (with the divisor function nowhere in sight):
PROPOSITION 5. We have
x
= (1 (cid:13))x+(cid:1) (x);
1
j (cid:0)
j(cid:20)x(cid:26) (cid:27)
X
where (cid:1) (x) 4x1=2 +1.
1
j j (cid:20)
7
Proof. Denote the sum by V(x). By (1) and (5),
x x
T(x) = = xH(x) V(x)
j (cid:0) j (cid:0)
j(cid:20)x(cid:18) (cid:26) (cid:27)(cid:19)
X
= x(lnx+(cid:13))+xq(x) V(x);
(cid:0)
where xq(x) 1. Now equating this to (6), we have
j j (cid:20)
V(x) = (1 (cid:13))x (cid:1)(x)+xq(x);
(cid:0) (cid:0)
(cid:3)
and the statement follows.
So the average of these fractional parts approximates to 1 (cid:13); one might have expected
(cid:0)
it to approximate to 1.
2
Expressions as products of three factors
We now widen our investigation to consider the number of ways of expressing n as
a product of three factors. More exactly, de(cid:12)ne (cid:28) (n) to be the number of ordered triples
3
(i;j;k) with ijk = n. We show how the previous methods and results extend quite naturally
to this case.
For a prime p, we have (cid:28) (p) = 3: the triples are (p;1;1), (1;p;1) and (1;1;p).
3
Again, we will consider the cumulative sums: let T (x) = (cid:28) (n). Clearly, this is
3 n(cid:20)x 3
the number of (positive) triples (i;j;k) with ijk x.
P
(cid:20)
We can relate (cid:28) and T to (cid:28) (= (cid:28) ) and T (= T ) as follows.
3 3 2 2
PROPOSITION 6. We have
(cid:28) (n) = (cid:28)(k); (7)
3
kjn
X
x
T (x) = T : (8)
3
k
Xk(cid:20)x (cid:16) (cid:17)
Proof. For a (cid:12)xed divisor k of n, the number of triples with ijk = n is the number of
pairs (i;j) with ij = n=k, that is, (cid:28)(n=k). When k runs through the divisors of n, so does
n=k. Hence (7).
For (8), take a (cid:12)xed k x. The number of triples with ijk x is the number of pairs
(cid:20) (cid:20)
(i;j) with ij x=k. By (B), this is T(x=k). Hence (8). (cid:3)
(cid:20)
8
In terms of convolutions, (7) says that (cid:28) = (cid:28) u = u u u. Corresponding to
3
(cid:3) (cid:3) (cid:3)
Proposition 2, we have:
PROPOSITION 7. Let n have prime factorization n = m paj. Then
j=1 j
m Q
(cid:28) (n) = 1(a +1)(a +2):
3 2 j j
j=1
Y
Proof: First consider pa for a prime p. By (7),
a a
(cid:28) (pa) = (cid:28)(pb) = (b+1) = 1(a+1)(a+2):
3 2
b=0 b=0
X X
When n is expressed as a product of three factors, these factors are of the form
m m m
pbj; pcj; pdj;
j j j
j=1 j=1 j=1
Y Y Y
in which b +c + d = a for each j. As just shown, for a (cid:12)xed j, the number of choices
j j j j
of (b ;c ;d ) is 1(a +1)(a +2). These choices combine to give distinct factorisations of n,
j j j 2 j j
hence (cid:28) (n) is as stated. (cid:3)
3
Given this expression, the reader may care to try showing that (cid:28) (n) is a multiple of 3
3
unless n is a cube, in which case it is congruent to 1 mod 3.
Using (8) and Theorem 2, we can derive a corresponding estimation of T (x). We now
3
need an estimate of [(lnn)=n]. This is delivered by the following variant of Lemma 1,
n(cid:20)x
which is proved by a slight extension of the same method (for details, see [Nath, p. 206{8]).
P
LEMMA 4. Suppose that f(t) is non-negative, increasing for 1 t x and decreasing
0
(cid:20) (cid:20)
for x x , with maximum value f(x ) = M. De(cid:12)ne S(x) and I(x) as in Lemma 1. Then
0 0
(cid:21)
I(x) M S(x) I(x)+M: (cid:3)
(cid:0) (cid:20) (cid:20)
Writing ln2x for (lnx)2, we deduce for our case:
LEMMA 5. We have
lnn
= 1 ln2x+r(x); (9)
n 2
n(cid:20)x
X
where r(x) e(cid:0)1.
j j (cid:20)
Proof. The function f(t) = (lnt)=t increases for 1 t e and decreases for t e,
(cid:20) (cid:20) (cid:21)
with f(e) = e(cid:0)1. Also, xf(t)dt = 1 ln2x. (cid:3)
1 2
R
9
Actually, [(lnn)=n] 1 ln2x converges to the \Stieltjes constant" (cid:13) 0:072816
n(cid:20)x (cid:0)2 1 (cid:25) (cid:0)
as x , but we don’t need this.
P
! 1
THEOREM 3. We have
T (x) = 1xln2x+(3(cid:13) 1)xlnx+O(x): (10)
3 2 (cid:0)
Proof. By (8) and (6),
x x x
T (x) = (lnx lnn)+(2(cid:13) 1) +(cid:1) :
3
n (cid:0) (cid:0) n n
Xn(cid:20)xh (cid:16) (cid:17)i
Write this as J J +J , where
1 2 3
(cid:0)
J = x(lnx+2(cid:13) 1)H(x);
1
(cid:0)
lnn
J = x ;
2
n
n(cid:20)x
X
x
J = (cid:1) :
3
n
Xn(cid:20)x (cid:16) (cid:17)
We deal with J and J (cid:12)rst. By (9),
2 3
J = 1xln2x+r (x);
2 2 2
where r (x) e(cid:0)1x. Also,
2
j j (cid:20)
x1=2
J 4 :
j 3j (cid:20) n1=2
n(cid:20)x
X
By Lemma 1, we see that (1=n1=2) 2x1=2. Hence J 8x.
n(cid:20)x (cid:20) j 3j (cid:20)
P
Now consider J . Substituting for H(x) by (5), we have
1
J = x(lnx+2(cid:13) 1)(lnx+(cid:13) +q(x))
1
(cid:0)
= x[ln2x+(3(cid:13) 1)lnx+c]+r (x);
1
(cid:0)
where c = (cid:13)(2(cid:13) 1) (so c < 1) and r (x) = xq(x)(lnx+2(cid:13) 1). Since q(x) 1=x and
(cid:0) 2 1 (cid:0) j j (cid:20)
2(cid:13) 1 < 1, we have r (x) lnx+1, hence r (x) x. Combining the three pieces, we
1 1
(cid:0) j j (cid:20) j j (cid:20)
obtain (10), with the term \O(x)" no greater than 10x. (cid:3)
With slightly less e(cid:11)ort, we could have used (2) and (3), instead of (5) and (6), to
derive the correspondingly less accurate estimate T (x) = 1xln2x+O(xlnx).
3 2
Of course, (cid:28) (n) and T (x) are de(cid:12)ned similarly for all k 3, enumerating k-tuples
k k
(cid:21)
instead of triples. In the same way, one has
x
T (x) = T ;
k k(cid:0)1
n
Xn(cid:20)x (cid:16) (cid:17)
10