On the Averages of the Divisors of a Number

On the Averages of the Divisors of a Number
Author(s): Oystein Ore
Source: The American Mathematical Monthly, Vol. 55, No. 10 (Dec., 1948), pp. 615-619
Published by: Mathematical Association of America
Stable URL: http://www.jstor.org/stable/2305616 .
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ON THE AVERAGES OF THE DIVISORS
OF A NUMBER
OYSTEIN ORE, Yale University
Let n be some integerand let
n=
al
pi
ar
...
pr
be its decompositioninto primefactors.As usual we shall denote the numberof
divisorsof n by v(n) and the sum of the divisorsby a(n) so that
V(n) = (al
1) ...
1)(a2 +
+
(ar +
1),
and
paj+1_
a,rr_1-
IPi -1
Pi1
dIn
Pr
pr
-1
The various means of the divisorsmay be computedeasily fromthe formulas
that follow.For the arithmetic
mean one has
(1)
A (n)
u(n)
=
___
v(n)
and forthe geometric
mean the result is
v(n)
(2)
G(n)
=
H d -/n.
d/n
Finally the harmonicmean is definedby
1
1
H1(n)
V(n)
1
d/n
d
and since
nE-=
d/n
1
d
n
2E-=
d/n
d
Ed
d/n
it followsthat
(3)
H(n) = nv(n)
0(n)
The combinationof the three formulas(1), (2), and (3) gives
H(n)A (n) = n = G(n)2;
in other words, the geometricmean of the divisorsis the geometricmean of the
arithmeticand harmonicmeans of thedivisors.
615
616
ON THE AVERAGES OF THE DIVISORS OF A NUMBER
[December,
One may now ask in which cases these means may be integers.For the geometricmean this is trivial: The geometricmean of thedivisorsis an integeronly
for square numbers.
For the other two means the problem is by no means simple. We may observe firstthat both the arithmeticand harmonicmeans have the multiplicative
property
A(a.b) = A(a)*A(b),
H(a* b)
=
H(a) *H(b)
provideda and b are relativelyprime.Thus one may only look forthe primitive
integralmeans, that is, such numbersthat theyare not the productof relatively
prime factors,each of which has an integralarithmeticor harmonic mean.
Let us considerthe arithmeticmean in a fewspecial cases. For an odd prime
p one has
A(p)
=
P+_
2
and this is always integral,while p =2 is an exception since A (2) = 3/2 is not
integral. From the multiplicative property we conclude that every number
odd primeshas an integralarithmeticmean of
which is the product of different
divisorswhile in the case of different
primefactorsincluding2 the mean is only
integralifone of the primesis of the form4k -1.
One may consider also when the powers of a prime may have an integral
arithmeticmean of divisors.Since
A(pa)
pa+l
_=-
(p-
-1
1)(a+1)
this is a problemclosely connectedwith the solution of the congruence
X=
e
1 (mod k).
We shall not go into details about this problem. In certain cases the arithmetic
mean is integral;forinstance,
A (58) = 39,
A(55) = 651.
It can be shown that no power of 2 can have an integralarithmeticmean.
The correspondingproblemsforthe harmonicmean seem more interesting.
In certain simple cases it can be established easily that the harmonicmean of
the divisorscannot be integral.We mentionfirst:
For thepowerof a primetheharmonicmean is notintegral.
If namely n =
then
pa
pH)(a
+ 1)
pa+pa-l+
. . .+p+1
Here pa is relativelyprime to the denominatorand since a+1
is a numberless
1948]
ON THE AVERAGES OF THE DIVISORS OF A NUMBER
617
than the denominatorthe harmonicmean cannot be integral.
Anotherobservationis:
Whenn z6 is theproductof different
primefactorstheharmonicmean cannot
be integral.
We write
n = PlP2 ... Pr
where the primesare arranged in increasingorder,and one finds
11(n)
Pr
PlP2
=
*2r.
(Pi + 1) . . . (pr+ 1)
Let us assume firstthat n is odd, and let us write
1l(n)=
Pr
Pi+'
Pr
p+ 1
Pi1+
2
2
Here the denominatorcontains at least r prime factors, and since Pr is not
among them the quotient cannot be integral.Next let Pi=2 so that
2(+1
H(n)=
4Pp2 *
p2 * * * Pr
3 (P2+ 1) ... (Pr+ 1)
P2 +
2
Pr
Pr +
2
and this expressioncan only be integralif P2 = 3 so that
H(n)=
11(n)
~2ps*..
__
-
_
P3+
2
Pr
__
Pr+1
.
2
Since none of the r-2 factorsin the denominatorare equal to prwe conclude
that each of themmust be equal to one of the otherprimefactorsin the numerator. But the assumption
P3+
1
2
2
or Ps=3 was already excluded.
A result of some interestis the following:
A perfectnumberhas an integralharmonicmean of divisors.
Since a perfectnumberis definedby the propertythat
a(n) = 2n
it followsthat
H1(n)
nv(n)
r
-
2
v(n)
618
ON THE
AVERAGES
OF THE
DIVISORS
[December,
OF A NUMBER
When n is an even perfect number it has the form n=20 p so that v(n)
=(a+1) *2 and H(n)=a+l
is integral. When we have an odd perfectnumber it is known that one of the exponents in the prime factordecompositionis
odd so that v(n) is even also in this case.
By means of the tables of J. W. L. Glaisher giving the values of a(n) and
z(n) one can determinerathereasily the values of n below 10000 forwhichH(n)
is integral.One findsthe followingtable:
TABLE I
n
A (n)
H(n)
6=2*3
28=22.7
140 =22 . 5 . 7
270 =2 *33*5
2
3
5
6
28
45
496=24 -31
672 =26 3 *7
1638 =2 *32-7 13
2970-2 33-5*11
5
8
9
11
99
84
182
270
6200 =23. 52-31
8128=26.127
8190=2 *32 .5 *7 *13
10
7
15
620
11614.
546
3
94
It would be of interestto make a systematic examination of such numbers
up to highernumericallimits.This requiresan extensionof Glaisher's Numberdivisor tables, a project which would be desirable also in connectionwith other
numbertheoreticalinvestigations.
There are, however,various methods by means of which one can construct
new numbers with integral harmonic means fromgiven ones. Let us suppose
that n has this propertyso that
n. (n)
=
a. (n),
a = H(n).
We multiply both sides by some number k relatively prime to n and write
n1=k n. Then one finds
H(ni)=-
fl
ni. v(ni)
(n-
)
a -k.v(k)
a(k)
Thus if the right-handexpressionis an integerthe numbern1 will also have an
integralharmonic mean. This may again be examined by means of Glaisher's
tables. The values of a in the last column of the table given above lead successively to the followingnumbers:
1948]
POLYGONS
WITH
TWO EQUIANGULAR
619
POINTS
TABLE II
n
2 32-7 -13.17
2 32-5.7.13*29
23-3 -52-31
23.33.52.31
23-33-52 17. 31
23.52. 19.31
23-52-72 19. 31
22-52 72 134 19*31
23-3-52-29-31
23-52 19-31 .37
IH(n)
.17
29
15
27
51
19
49
91
29
37
n
23 33.-52-31 *53
26- 13. 127
26-32. 13. 127*
26-32-5 . 13 *127
26-32 . 13 . 17. 127
26-32-5 *13 -17*127
26-32-5 *13. 29 *127
26-32 . 13 . 53. 127
26-32-5-13-89-127
H(n)
53
13
27
45
51
85
87
53
89
Table I shows that there are eleven numbers below 10,000 with integral
harmonicmean forthe divisors.Among these are the fourperfectnumbersbelow this limit.One verifiessimply that an even perfectnumbercannot have an
integralarithmeticmean of the divisors. However, if these are excluded from
Table I the remainingnumbers have the propertythat also their arithmetic
means of divisorsare integral.One mightbe led to the conjecturethat thiswould
be a general propertyand the computation of Table II was executed with this
in mind. However, the number marked by an asterisk proved to be a counter
example. It is not perfectand its arithmeticmean of divisors is not integral.A
much more interestingconjecture, however,appears fromour numerical computations,as an extensionofthe famousconjectureforperfectnumbers,namely,
that a numberwithintegralharmonicmean of divisorsmustbe even.
POLYGONS WITH TWO EQUIANGULAR POINTS
ARTHUR WORMSER, Riverside,Illinois
1. Introduction.In 1816 A. L. Crelle [1] discovered that in every triangle
ABC there are two points G and G' such that AGAB= GBC=GCA =G'BA
= G'CB = G'A C =w. AfterH. Brocard [2] rediscoveredthis featurein 1875, he
and other mathematiciansfounda great numberof related facts [3 ]. Since then
the points G, G' and angle w have been called Brocard's points and Brocard's
angle. Interest in the subject remained at a high level for many years, and in
1886 R. Tucker [4] foundthat Moebius' harmonicquadrangle [5] has a similar
property.Althoughthis type of geometryof the trianglefascinatedquite a few,
all effortsfailed to constructor calculate polygons with more than four sides
having two equiangular points [6] as definedabove. In 1930 K. Hagge [7] pub-