sums of distinct divisors

Sums of Distinct DivisorsAuthor(s): B. M. StewartSource: American Journal of Mathematics, Vol. 76, No. 4 (Oct., 1954), pp. 779-785Published by: The Johns Hopkins University PressStable URL: http://www.jstor.org/stable/2372651 .

Accessed: 09/12/2014 21:04

Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .http://www.jstor.org/page/info/about/policies/terms.jsp

.JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range ofcontent in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new formsof scholarship. For more information about JSTOR, please contact [email protected].

.

The Johns Hopkins University Press is collaborating with JSTOR to digitize, preserve and extend access toAmerican Journal of Mathematics.

http://www.jstor.org

This content downloaded from 128.235.251.160 on Tue, 9 Dec 2014 21:04:45 PMAll use subject to JSTOR Terms and Conditions

http://www.jstor.org/action/showPublisher?publisherCode=jhup

http://www.jstor.org/stable/2372651?origin=JSTOR-pdf

http://www.jstor.org/page/info/about/policies/terms.jsp


SUMS OF DISTINCT DIVISORS.*

By B. M. STEWART.

1. Introduction. For a given positive integer .Il, let cx(M) indicate thle nutimber of integers n which can be written in the form

(I) n Ed

wh ere the d are distinct positive divisors of M. Obviously a (M) ? ((M), the sumn of all the divisors of M.

For a given set S of integers M, define S* to be the subset of S con- taining those integers for which a (M) -a (MI) is minimal.

Thus for the set I of all positive integers, the set 1* contains all integers, such as M1 2k, for which a(M) - ax(M) = 0. However, for the set 0 of all odd integers 1l1> 5, the set O contains all integers, such as M = 945, for which a(M) - ax(M) = 2, for certainly both 2 and a(M1) - 2 defy representation in the form (1).

In this paper we give complete structure theorems for l* and Q* and indicate their application to problems of the Egyptian fraction type. We also show that the function a(M)l/a(Ml) is everywhere dense on the interval O to 1.

2. Structure theorems for 1. It is clear that i1 = 2k belongs to 1*, because any n in the range 1 < n < a(M) = 2k+1- 1 when written in the binary notation is in the form (1).

THEOREM 1. If M belongs to 1* and p is a prime with (p, M) = 1, then a necessary and sufficient condition that M' pkM, k > 1, belong to 1* is that p a (M) + 1.

Proof. The smallest divisor of Ml' not a divisor of M is p. If p > a (M) + 1, then n = o(M) + 1 defies representation with respect to M'-

Conversely, suppose p ? (M) + 1. We show by induction that pk ? < (p-lM) + 1. This is obvious for kc = 1. Using the induction hypothesis on k, we have

* Received May 23, 1953, revised January 28, 1954. Presented to the American Matematical Society, April 25, 1953.

779



780 B. M. STEWART.

pk+1 ? pu(p'-1M) + p ? p(pk-lM) + o(M) + 1 +(pM) -- 1 We assume as an induction hypothesis that pk-lM belongs to I*. To show that f11' - pkM belongs to I*, we consider the intervals ranging from Qpk

to Qpk + o(pk-lM) for Q = 0, 1, u la(M). It follows that no integer n in the range 1 _ n -_ a (M') is omitted from all these intervals. For on the one hand, because pk ? Ur(pk-lM) + 1, we have

(Q + l) pk Qpk + a(pk-IM) + 1,

hence the intervals are overlapping or contiguous. On the other hand, the intervals include 1 and pko(M) + (plk-lM) - a(M'). Thus each n can be written

n Qpk + R, O Q a(M), Q 0 R ? (pk-lM).

Since M and pk-lM are assumed in I*, we can write Q = E d, where the d are distinct divisors of M, and we can write R =

' D, where the D are distinct divisors of pk-lM. Then

n E d'+ ED

where the d' - pkd are distinct from the D because of involving the factor pk and both the d' and the D are divisors of M'. Hence M' belongs to I*.

COROLLARY 1. M belongs to I* if and only if M is of the form M = 2% aO 0, or of the form

M =2pap1ai** . p..e, a > 1l aj _ 1, j 1, 2, * *,k

where the primes pj satisfy the following conditions:

2 < pi < * * < pk, pl _? .(2a) + 1, and

pj+l ?C (2aplai . . . pja) + 1, j=1, 2, * * *, k 1.

Proof. It is easy to see that 1 is the only odd number belonging to 1*. We have noted that M = 2a belongs to 1*. The corollary then follows from Theorem 1 by induction on 7k.

3. Practical numbers. A. K. Srillivasan [1] has defined a practical number 11 (i. e., a number of advantage for use in weights, measures, or coinage), requiring that 31 have property (1) for all n with 1 ? n ? M. But he failed to find a complete structure theorem for these numbers.

It is obvious that every number belonging to I* is a practical number. WVe show below that the converse is true. It follows that Corollary 1 answers



SUMS OF DISTINCT DIVISORS. 781

Srinivasan's question about the complete structure theorem for practical numbers.

To establish that every practical number belongs to 1*, we simply review the proof of Theorem 1, noting that if 31i is assumed practical, the criterion of Theorem 1 applies to decide whether M' -pkI is practical. For if p > r(M) + 1, then not only does n = o(M) + 1 defy representation in the form (1) relative to M', but also n < M'. Thus the condition of Theorem 1 is necessary for M' to be practical. Since the condition is sufficient to make M' belong to 1*, it is sufficient to make M' practical.

4. Values of at(M)/a(M). We need the following lemma.

LEMMA 1. If p is a prime such that p > a(T), then

a (pT) =a (T) { (T) + 2}.

Proof. Every sum of divisors of pT must have the form n = n1p + n2 where n, and n2 are either 0 or a sum of divisors of T. Furthermore, since the maximum value of n2 is u (T) and p > r(T), no two of these n can be equal unless they have the same n1 and n2. Since there are a (T) + 1 choices for each of n1 and n2, except that the choice 0, 0 must be omitted, we have

ac(pT) {ac(T) +1}2- 1 (T){a(T) +2}.

We define s(M) a (M)/la(M).

THEOREM 2. The values of s (M) are everywhere dense on the interval O to 1.

Proof. For given real numbers x and y such that 0 < x < y ? 1, we must produce an integer K such that x < s(K) < y. We try to find K = pM where M is in I* and p > a (M), for then by Lemma 1

s(K) = s(pM) = o(M){or(M) + 2}/(p + 1)u(M) = {u(M) + 2}/(p + 1).

If we set u X 1/y, u(1 + E) = 1/x, our problem is to find M and p so that

u{u(M) + 2} < p + 1 < U{r(M) + 2} (1 + e).

On the one hand we know from Theorem 1 that there exist arbitrarily large numbers M belonging to I*. On the other hand we have the Cahen-Stieltjes theorem that for any E> 0 and for sufficiently large v = u{, (M) + 2} there must exist at least one prime p in the interval v - 1 < p < (v - 1) (1 + e), hence such that v < p + 1 < v(1 + E). Since u ? 1, such a prime p has the required property p > oa(M), which completes the proof.

4



782 B. M. STEWART.

It seems difficult to decide whether s(M) will take on a specified rational value. Even s (M) = 1/3 requires some search before we find a solution such as M --12 - 89. For s (M) = 4/5 the simplest solution we know is MI 28 73 * 47269.

5. Structure theorem for O*. A necessary condition that M belongs to 0' is that M be a multiple of 15, for unless M has the factors 3 and 5, the numbers n = 3 and n = 5 will be exceptions to (1).

Arrange the - divisors dj of M in natural order

d =1 < d2 3 < d3a5 < . * < dT 1Q1M

and let oaj denote the sum of the first i of the dj.

THEOREM 3. Necessary and sufficient conditions that M = 1ST belong to 0* are that for i> 3,

either (a) di+1 ? o -2, di+1, = - 4;

or (b) di+1, = - 4, d+2 0-( 2.

Proof. For numbers of 0, it is clear that 2 does not have the form (1). Because of the ordering of the dj, it follows that if di+, > -i - 2, then n (r,--2 does not have the form (1). If di+, - 4, then n c=r- 2 = di+1 + 2 again defies representation unless di+, = - 2. Therefore the conditions (a) or (b) are necessary when i ? 3 if M11 is to belong to O*.

Conversely, suppose (a) or (b) satisfied when i ? 3. As an induction hypothesis on i assume when 3 ? i < ? - 1 and 2 < x < -i 2 that either x = d using distinct d?< di or that x 2 (u -2=di + 2 di+,. This hypothesis is easily verified when i == 3.

Consider y where 2 < y < uj+1 -2. When y < ui -2 set y =x and use the induction hypothesis to write y E d where d ?< d+. When vi-2?< y < -+, 2 set y di+, + x and note from (a) or (b) that O? x < j -2. Except when x 2 or x= d+1, d + 2, it follows from the induction hypothesis that y = d with distinct d E dj+j

WThen x = di+, di + 2 .= - 2 we find that y = di+, + x = di + ai-, n. When x= 2 the conditions Tj -2 ? y = di+, + 2 and di+, ? - 2

imply that either y , - 1, or a - 2. The first two cases obviously allow y E= d with d ? d+,. The last case y di+, + 2 = - 2 because of (b) has the representation y d+,

Thus we have established for the case i + 1 the exact counterparts of the induction hypothesis on i. It follows that every n satisfying




2 < i < u(M) -2 has a representation (1); so do n = 1, ao(M) -1, and (1MI), hence Mll belongs to 0.

THEO1SEM 4. If MI belongs to 0* and p is an odd prime, then Ml' pM belongs to 0* if and only if

2p ?<a(M) -2, 2p v(M) -4.

Proof. Let the divisors and partial sums of M1' be indicated by d,' and o-', respectively, and note that the only divisors of M' which are not divisors of M are multiples of pa1+ where 11 paA, a > 0, (p, A) 1.

Supose that the first t divisors of M' coincide with those of 1l, so that d,- d/, i < t, 3 < t. Then the properties (a) or (b) of M as described in Theorem 3 carry over to M' for 3 ? i < t.

Next consider cases t ? r < k where dr ? dk' < dk+1' < dr+,. By Theorem 3 applied to M we have

dk+l' ? dr+1 Ur - 2 < Tr + pia+l -4 -4

heiice case (a) holds for 111' and k. Finally, consider cases where M ?- dk' < d k+1 < Mt. Then dk+l' =M'd

where d1 < p hence dj divides M as the notation indicates. When M/dj > 5, it follows from Theorem 3 applied to M for i ? 3 that

Ml/dj (3? ) - Ml/dl - M/d2 - - M/dj -2. Hence

dk+=' pMI/dj pa(M) - M'/d1 - Md2 - ] M'/dj - 2p

< u(A) 4- por(Ml) - llId - M'/d2 - - 3I'/dj -4 -Ok 4,

so that case (a) holds for 1ll' and k. There remain onlv the possibilities that dk+l' =- p, 3p, or 5p with iM < dk+l'. In case 3p < M < -p we have p < M/3 < K(M) -4, hence dk+' =5p < U17(M) + p + 3p-4 ==-- '-4 so that (a) holds for M' and k. In case p < M < 3p or M < p, the condition dk+l' -3p < Uk' - 2 = 0 (M) + p -2 is necessary for M' to belong to 0*. The contingency suggested by (b) is ruled out for it would require dk+l' 3p dk+2- 2 =- 5p - 2, hence case (a) must hold with 3p H4 a(21i) + p - 4. Thus the conditions of Theorem 4 are necessary. Conversely, when these conditions are satisfied, then

dk+l' 5p < (M) + p + 3p-4 o4 -4;

and if M1 < p, the condition dk+_' p < C (M) --4 -k' 4 holds. Therefore for all k ? 3 the conditionis of Theorem 3 lhold an1d 311'

belongs to 0*.



784 B. M. STEWART.

COROLLARY 4. If the prime p divides At and MI belongs to O*, them 1' pM belongs to 0*.

Proof. We have noted that M = 15T hence when p > 5, we have a(M) > 15p > 2p + 4; and if p = 3, or 5, we have a(M11) > 15 > 2p + 4; so the conditions of Theorem 4 hold.

We cannot give any simple catalog for O* like that for I* in Corollary 1, because there are infinitely many different types of "base" numbers in 0* instead of the single type 2a noted for I*. However, a few examples will help fix ideas.

Since ill - 1ST has a3 = 9, it follows from Theorem 3 that if M is in 0*, then d4 T 7. This implies u, = 16 and hence that d5 = 9, 11, or 13. It happens that all of 32 .5.7, 3*5*7*11, and 3*5a7*13 fail to belong to 0*, but only because dT > a, - 2. It is then easy to check, for example, that ]Ii 32.5.*7p is in Q* for any odd prime p ? 103. In particular, when p = 3, we find 211 = 945, the smallest integer in Q*. For use in the sequel, we apply Theorem 4 to see that M - 34 5 * 7 is in 0* for a ? 3.

Another special case of interest is that if Mk is the product of the first k odd primes, then Mk is in Q* when ki_? 5.

The author is indebted to the referee for noting and insisting on possi- bility (b) in Theorem 3, despite the fact that the smaller numbers in 0* consistently s7atisfy (a). By considering the arguments of Theorems 3 and 4, the author has constructed the following example: 11i 32.5 I2.72. 229pq where prime p = 2,641,663 and prime q = 7,924,991, such that M is in 0* and has d56 3p -455 4andd57= q =o 55-2, so that case (b) occurs for i 55.

6. Egyptian fraction problems. Let a positive rational number with numerator 1 be called a unit fraction. Then the usual Egyptian fraction problem is to show that any given positive rational number x may be written as a sum of a finite number of distinct unit fractions. A variation of this problem proposed by E. P. Starke [2] is to show that if x has an odd denominator, then x may be written as a sum of a finite number of distinct unit fractions with odd denominators.

Whether u1, 1/n or u,n 1/(2n + 1), the sequence {St}, where St u-- 1u + u * * *- + ut, is divergent, hence it is easy to replace the problem of representing an improper fraction x by that of representing a proper fraction A/B = x - St where A/B < ut,, is so small that unit fractions used in its representation cannot duplicate those used in the partial sum St.




For the usual Egyptian fraction problem easy constructions are known, but an amusing variant is provided by our knowledge of numbers belonging to 1*. By Corollary 1 we can find C, one choice being C - 2a for a sufficiently large a, so that BC 1M will belong to J*. Since AC < M, we may write AC = E d, using distinct divisors d of M. If we set dd' = M, then

A/B = AC/BC = i d/M = E 1/d' solves the problem.

For Starke's problem, our knowledge of numbers belonging to 0O pro- vides the same sort of solution. We can find C, one choice being C - 3a'(945) for a sufficiently large a, so that BC = M, by Theorem 4, will belong to 0* and so that 2 < AC < M, and then proceed as explained above.

MICHIGAN STATE COLLEGE.

REFERENCES.

[1] A. K. Srinivasan, "Practical numbers," Current Science (1948), pp. 179-180. [2] E. P. Starke, "Advanced problem 4512," American Mathematical Monthly, vol. 59

(1952), p. 640.



sums of distinct divisors

Documents