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Prime symmetric divisor functions Woodford, Roger 2005

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Prime Symmetric Divisor Functions by Roger Woodford B . S c , Univers i ty of M a n i t o b a , 2003 A T H E S I S S U B M I T T E D I N P A R T I A L F U L F I L L M E N T O F T H E R E Q U I R E M E N T S F O R T H E D E G R E E O F Maste r of Science i n T H E F A C U L T Y O F G R A D U A T E S T U D I E S (Mathemat ics) The University of British Columbia Augus t 2005 © Roger Woodford , 2005 Abstract In this thesis, we s tudy a new class of divisor-related functions: the pr ime symmetr ic functions. T h e elementary pr ime symmetr ic functions are denned on the nonnega-tive integers. T h e y take the values of the elementary symmetr ic functions appl ied to the multi-set of pr ime factors w i t h repet i t ion of an integer n. These functions are first defined i n [20]. In this thesis, we refine some of the definitions presented there, and revisit some of the key results regarding such functions. T h e pr ime symmetr ic functions are polynomials over Q i n the elementary pr ime symmetr ic functions, w i t h values i n Z . W e look at basic properties of prime symmetr ic functions. We s tudy the effect of i te ra t ion of these functions, and look at the question: when does i tera t ing produce cycles? We consider the inverse question of when and i n how many ways a number n can be expressed as f(m) for certain predetermined pr ime symmetr ic functions / . A s well , we look at asymptot ic approximat ions for certain classes of pr ime symmetr ic functions. i i Contents A b s t r a c t ii Contents ii i Acknowledgements v Dedicat ion v i 1 Div i sor Functions 1 1.1 Perfect Numbers , Var ia t ions on Perfect Numbers , and A l i q u o t Se-quences 2 2 P r i m e Symmetr i c Div i sor Functions 5 2.1 E lementa ry P r i m e Symmet r ic Div i sor Funct ions 6 2.2 Bas i c Proper t ies of the Elementary P r i m e Symmet r i c Funct ions . . . 9 3 T h e Second P r i m e Symmetr ic Funct ion 13 3.1 fi(n)=3 14 3.2 n(n) = 4 14 3.3 n(n) = 5 15 3.4 A n Inverse P r o b l e m 16 4 Higher E lementary P r i m e Symmetr ic Functions , 23 4.1 T h e H u n t for s 3 -cycles 26 i i i 5 T h e E x p o n e n t P r i m e Symmetr ic Div i sor Functions 32 5.1 T h e Average Order of 35 6 Further P r i m e Symmetr ic Functions 41 A p p e n d i x A M a p l e A lgor i thms 46 B ib l iography 49 iv Acknowledgements I wou ld like to acknowledge the encouragement and support of m y supervisors D r . Izabel la L a b a , and D r . Denis Sjerve, as well as the N a t u r a l Sciences and Engineer ing Research C o u n c i l for their generous funding. R O G E R W O O D F O R D The University of British Columbia August 2005 For She who is better t han rubies, who dwells w i t h prudence, and finds out knowledge of w i t ty inventions, B y w h o m kings reign and princes decree justice, W h o loves t hem that love her, and is found by those who seek her early, Whose fruit is better than gold; and whose revenue than choice silver, W h o was set up from everlasting, from the beginning, or ever the ear th was, Whose delights were i n the sons of men. v i Chapter 1 Divisor Functions T h e divisor function a : N —* N defined by d\n has been thoroughly s tudied for centuries. T h e first most na tura l general izat ion of the divisor funct ion are the divisor functions aa defined by <rQ(n) = X > , d\n Where a > 0 is a real number. W i t h this definition, OYJ counts the number of posit ive divisors of n. Sometimes o~o is denoted by d, and a\ is s imply a. Other relevant divisor-related functions are to, wh ich counts the number of dis t inct primes d i v i d i n g a number n , and Q, which counts the number of primes d i v i d i n g n, w i t h mul t ip l ic i ty . T h e functions aa are mul t ip l ica t ive , that is if m and n are relat ively pr ime, then <ja(mn) = aa(m)crQ(n). Derivat ions for the asymptot ic formulae of the average orders for aa can be found i n . [1]. In case a = 0, we have that for a l l x > l , d{n) = s l o g x + (2C - l)x + 0(y/x), n<x I where C is Eu le r ' s constant. A l s o , J2°(n) = J2x2 + 0(x\og x). n<x Furthermore , for a > 0, and a ^ 1, we have ^ M » ) = ^ a + 1 + o ( / ) , n<x where (3 = m a x { l , a } , and ( is the Riemann-ze ta function. I n this exposi t ion, we w i l l define and s tudy a new class of divisor functions: the pr ime symmetr ic divisor functions. These functions need not be mul t ip l ica t ive . T h e notions of perfect numbers, and aliquot sequences, and some of their gener-al izat ions based on the usual mul t ip l ica t ive divisor functions, or related functions, w i l l be extended to the prime symmetr ic divisor functions. 1.1 Perfect Numbers, Variations on Perfect Numbers, and Aliquot Sequences Several good texts detai l ing the basic theory of perfect numbers exist, see for instance [2], [6], [9], and [17]. In addi t ion, many variations on perfect numbers have been denned and studied. Examples can be found i n [4], [7], [8], [10], [11], [14], [15] and [16]. L e t <x*(n) = a(n)—n, that is cr*(n) sums al l the proper divisors of n . W e say that n is perfect i f a*(n) = n, excessive (or abundant) i f <7*(n) > n , and defective (or deficient) i f a*(n) < n. T h e first few perfect numbers are 6, 28, 496, 8128, 33550336. It has been known since E u c l i d ' s day that if p, and q = 2 P — 1 are pr ime, then the number n = 2 P _ 1 ( 2 P — 1) is perfect. T h i s can be readily checked, since its proper divisors correspond to the set { 1 , 2 , . . . , 2 P _ 1 , q, 2q,..., 2p~2q}, and hence 2 their sum is: a*(n) = 1 + 2 + • • • + 2p~l + q + 2q+--- + 2p~2q = 2 P - 1 + q{2p~l - 1) = (2p - l)2p~l = n. Euler proved conversely, that any even perfect number must be of this form. The proof is the same as that found in [12]. Theorem 1 An even number n is perfect if and only if n = 2p~lq, where p, and q — 2p — \ are both prime. Proof. We need only prove the only if part. Suppose n is an even perfect number. Write n = 2k~1m, where 2 f e~1||n, and so k > 2, and m is odd. Since o(n) = 2n, we have:' 2n = 2km = o{2k-1m) = (2k - l)er(m). Hence, 2k — 1 divides m, so we may write m = (2fc — Using this value of m, we get 2ke = o((2k -1)£). If £ > 1, then 1, £, and (2k — 1)£ are distinct divisors of m, and so 2k£ = o((2k - 1)£) >l+£+(2k-1)£ = 2k£ + 1, a contradiction. Thus £ = 1 and so 2k = a(2k - 1) = 1 + (2fe - 1) + ]T d, d\(2k-l) 1 < d < 2k - 1 so clearly, 2k — 1 must be prime. Furthermore, it is trivial to see that if k is composite, then 2k — 1 factors, thus q is also prime. • It is still not known if there are infinitely many even perfect numbers, or equivalently, if there are infinitely many Mersenne primes (primes of the form 2 P —1). It is also unknown whether or not there exists an odd perfect number. 3 T h e equivalent definit ion for a perfect number, i.e. n is perfect if a(n) = 2n motivates one generalization, that of multiperfect numbers. A number n is said to be &:-multiperfect i f <r(n) = kn. For instance, <x(120) = 360, so 120 is 3-multiperfect. A s already stated, there are numerous variations of the not ion of perfect number. T h e sequence {c*^(n)}^L0 is cal led the aliquot sequence of n , that is the aliquot sequence of n is obtained from n by i terat ing the sum of proper divisors funct ion a*. The aliquot sequence of a number n can fluctuate up and down. T h e unsolved Catalan-Dickson problem is to determine whether for every n , the sequence {t7*(fc) (n) }^LQ is eventually periodic. Instances are known when the shortest per iod is longer than 1. For instance: 12496 14288 15472 14536 14264 12496. T h e case when the per iod is of length 2 has a special name. If o-*^2\n) = n ^ cr*(n), then the pair {n,o*(n)} is cal led amicable. One such amicable pair is {220,284}. 4 Chapter 2 Prime Symmetric Divisor Functions T h e pr ime symmetr ic divisor functions are defined i n terms of the elementary sym-metr ic divisor functions. In this paper, we w i l l denote the set of nonnegative integers N u { 0 } b y N o . Definit ion 1 Let k € N o . Define : N o —> N o as follows: if n = 0, then Sfc(0) = 0, for all k. Forn > 0, if k = 0, Sfc(n) = 1. Ifk>0, andn = p\ • • -pr, where r = Q(n) is the number of prime factors (with multiplicity) of n, then where the sum is taken over all products of k prime factors from the multi-set { p i , . . . ,pr}- We say Sk is the kth elementary prime symmetric function. Note that i f k > r , then the s u m is empty, so Sfc(n) = 0. Definit ion 2 A function f : N o —> Z is called a prime symmetric function if it can be expressed as a polynomial over Q in the elementary prime symmetric functions S 0 , S i , . . . . 5 N o w we w i l l extend the not ion of perfection, aliquot sequences, etc. to more general classes of functions. Definition 3 Let S C Z , and suppose f : S —» Z . If n e S satisfies f(n) = n, then we say that n is f-perfect. If f(n) > n, then we say that n is f-excessive. If f(n) < n, then we say that n is f-defective. If f(n) > n, then we say that n is f-special. If the sequence { / ^ ( n ) } ? | 0 is well defined, then it is called the f-sequence of n. If the sequence {/^ Hn)}f=o * s well-defined, we say it is an f -sequence of length e. A finite sequence {no,... ,n(} is an f-cycle of length t if the following con-ditions are satisfied: 1. t> 1, 2. n o , . . . , nt-i are distinct and n( = no, and 3. f(ni) = n i + 1 , fori = 0,1,...,£-1. 2.1 Elementary Prime Symmetric Divisor Functions Observe that i f J7(n) < k, we have s^(n) = 0. There is an alternate way of defining the functions Sfc. G i v e n n = pi • • -pr e N, set R Sn(x)=Y[(l+Pix). i=l T h e n Sfc(n) is the coefficient of xk i n Sn(x). T h e empty product is taken to be 1. Example 1 s 0 (12) = 1, S!(12) = 2 + 2 + 3 = 7, s 2 (12) = 2 - 2 + 2 - 3 + 2 - 3 = 16, s 3 (12) = 12, and s 4 (12) = 0. If f2(n) = k, then we t r i v i a l l y have that n is s^-perfect. T h i s is rather uninteresting, so we define a related version of Sfc-perfection to avoid this. 6 D e f i n i t i o n 4 Let n € N satisfy Q(n) > k. If Sfc(n) = n, then we say that n is s*k-perfect. If s^(n) > n, then we say that n is s^-special. E x a m p l e 2 If p is prime, then pp is an s*_i-perfect number, since sP-i{f)=(j)P_^J?-l=Pp. In fact, this example has a form of converse, first proved i n [20]: T h e o r e m 2 The prime power pa is s*k-perfect if and only if a = p and k = p — 1. Proof. We have seen that this is sufficient, now suppose k < a, and Sk(pa) = pa • T h e n For 1 < k < a — 1, (£) is divis ible by two dis t inct pr ime factors, hence we must have k = 1 or a — 1. N o w 4 = 2 2 , is the only sj-perfect number, and corresponds to the case where k = 1 = a — 1. Hence we may assume k — a — 1, wh ich from (2.1) implies that ct — p and k = p — 1. T h i s proves the theorem. • T h e next na tura l question to ask is: when is paq@ an s^-perfect number for some fc? T h i s turns out to be substantial ly more difficult t han the previous question. However, we w i l l come up w i t h some necessary condit ions and also look at some special cases. T h e o r e m 3 Suppose that p and q are distinct primes and that a,(3>0. If paq@ is s*k-perfect, then a^=k, and (3 ^  k. Proof. Suppose a = k. T h e n Sk{paq13) — paq13 is equivalent to T h i s is impossible as on ly the right side of this equat ion is divis ible by q. W e get a s imilar contradic t ion if /? = k. • 7 F r o m Compute r searches, there are two numbers of the form paqP known to be s£-per fec t for some k. 48 = 2 4 • 3 is s^-perfect, and 46875 = 3 • 5 6 is Sg-perfect. Note that these are bo th of the form paq. It is this form that we w i l l specialize our next theorem to. Theorem 4 Let p and q be distinct primes. (i) Suppose that p > q. If p > k + 1, and q > (a + l)/(a — k + 1), then paq is not s*k-perfect. (ii) Suppose that p < q. If p > k + 1, then paq is not s^-perfect. Proof, (i) In the first instance, suppose that paq is s^-perfect. W e have sk(Pag) = (f)Pk + (fc a_ ^Pk~\ = Pag. T h i s implies that or a — kl \a — k + 1 ' L e t t i n g £ = a — k + 1, we have Since p > q, we have that _ 1 P + ( i 1 = PQ-a \ (a\ ( a \ Ia\ (ot + 1 If we show that the assumption i n (i) implies that a + 1 ^ (2.2) £ then we are done. B u t ' a + l \ (a + l \ ( a \ fa - £ + 3 \ (a - £ + 2 £-1 \ 2 \ 1 There are £ terms i n this product . T h e first t e rm < q by assumption. Since a a - l ^ a - £ + 2 ^ e - i - e - 2 - ~ 1 by assumption, (2.2) holds and we have proved (i). (ii) T h e proof of (ii) is s imilar . In this case sk{paq) < ( a ^ 1 ) « ' where £ is as before. Hence it suffices to prove that Since p> a — £ + 2 = k + I, this indeed holds, so (ii) is proved. • 2.2 Basic Properties of the Elementary Prime Symmet-ric Functions T h e following propos i t ion from [20] is an immediate consequence of the definit ion. P r o p o s i t i o n 5 If n = 1 • • • p^T, then ii + ••• + ir = k ii, . . . , ir > 0 P r o p o s i t i o n 6 k •'• v sk{mn) = ^sk-i{m)si{n).- ~ Proo/ . If TO = 1, or n = 1, the result is immediate , as it is i f k = 0. If k > 0, TO = p i . . . p r , and n = q\... qs, let S, = {Pi,---,Pr,<2 ,i,-..,?s}. 9 T h e n s f c (mn) = ]P n---rk. {ri,...,rk}CS W e collect the terms of this sum having k — i factors from ra, and i factors from n. T h e sum of these is equal to sk-i(m)si(n). S u m m i n g as i ranges from 0 to A: gives the desired result. • Corollary 7 Let n, k £ N, and let p and q be primes, with p < q, and suppose Q(pn) > k. If pn — sk(pn) > 0, then pn — sk(pn) < qn — sk(qn). Proof. Since fl(n) > k, we have that Sfc(n) > 0. Therefore, since pn - sk(pn) > 0, it follows that n > Sfc_i(n) + sk(n)/p > Sfc_i(n). qn - sk(qn) =qn- qsk-i(n) - sk{n) > pn-psk-i{n) - Sfe(n) = pn- sk(pn). • In searching for s^-cycles and s£-per fec t numbers, it is essential to know when sk(n) > n. W e search by f ixing fi(n), and systematical ly checking a l l products of Q(n) primes. T h e corol lary tells us that i f sk(pn) < pn, then for any q > p, qn is also Sfc-defective. Lemma 8 Let k, n e N. Then there exists anr > k such that if Q(n) > r, then n is sk-defective. Let r(k) denote the least such r. Then ^ ~r-k • Hence, r(fc) = m i n { r : ( < 2r~k } . i k 1 10 Proof. There is an r > k such that the function 't satisfies f(t) < g(t) for a l l t>r, where g(t) = since / is a po lynomia l , and g is an exponential function. N o w suppose t >r, and let pi,... ,pt be t primes. T h e n < 2* fe, wh ich implies fc) G . t \ _ 1 where the sum is taken over a l l i i , . . . , it-k such that 1 < i\ < • • • < it-k < -^ T h i s implies YlPh • • •Vik <Pi---Pt, where the s u m is taken over a l l i\,... ,ik such that 1 < i\ < • • • < ik < t. T h i s i n t u r n implies that Sfc(pi • • -Pt) < P i • • -Pi-N o w we prove the second statement. T h e inequal i ty holds for a l l k > 1, and so r(k) > 2k. T h i s i n m i n d , let r(k) be as c la imed i n the statement of the theorem. We argue inductively. L e t t > r, and suppose that T h e n 11 Since t > 2k, we have t < 2(t — k), and so Hence t\ _ t ( t - l ) - - - ( t - k + 1) 2(t-l)(t-2)---(t-k) _ 2 f t - l k) ~ k\ K k\ 1 k and the proof is complete by induct ion . • Remark 1 An instructive way of viewing Lemma 8 is the following equality of sets: <2r-k}={r(k),r(k) + l,...}. T h e first few values of r(k) are given i n the following table: k r(k) 1 3 2 6 3 10 4 14 5 19 6 23 7 27 8 31 9 36 10 40 T h e properties of s i-perfect ion etc., corresponding to the first elementary pr ime symmetr ic function s i are easily characterized. T h e si-perfect numbers are the primes and 4 is the only s|-perfect number. A l l other numbers are si-defective. C lea r ly there are no si -cycles . We now investigate these properties i n the second pr ime symmetr ic function. 12 Chapter 3 The Second Prime Symmetric Function L e t n > 1 be an integer. B y a fami ly Ek(n,r) of s^-special numbers we mean a set Ek(n, r) = {npi • • -pr\pi, • • • ,pr are primes} such that i f m £ Ek(n,r), then m is s£-spec ia l . The fami ly ^ ( 4 , 1 ) of numbers of the form 4p, where p is pr ime, is one such set, since the elements satisfy S2(4p) = 4p + 4 > 4p. Ek(n,0) merely denotes the singleton set of the number n satisfying Sfc(n) > n, and fl(n) > k. In t roducing the definit ion Ek(n,r), is product ive for classifying a l l sjji-special numbers for a fixed k. W e w i l l see later that w i t h finitely many exceptions, a l l such numbers are i n fact s^-excessive, and that the families Ek+i{n,r + 1) can be determined completely from the s^-special numbers. A s wel l , a l l but finitely many s^-special numbers belong to such families. To f ind a l l s^-perfect numbers and a l l S2-cycles we need to f ind a l l numbers n such that 2 < f i (n) < 6, w i t h S2(n) > n, since r(2) = 6. To do this we use the a lgor i thm mentioned after corol lary 7, the approach taken i n [20]. 13 3.1 fi(n)=3 s 2 ( 2 - 2 - p ) = 4p + 4 > 4 p , s 2 (2 • 3 • p) = 5p + 6 < 6p, when p > 6. T h i s shows that there are no other infinite families of sij-special numbers satisfying fi(n) = 3. Be low we find a l l s^-special numbers not belonging to this family. 5 2 ( 2 3 3) = 21 > 18, s 2 (2 3 5) = 31 > 30, s 2 (2 3 7) = 41 < 42, S2 (3 3 3) = 27, s 2 (3 3 5) = 39 < 45. T h u s 27 is the only s^i-perfect number satisfying fi(n) = 3. I terat ing on the above s 2 -excessive numbers shows none belong to s 2 -cycles . For example 3.2 fi(n) = 4 s 2 ( 2 • 2 • 2 • p) = 6p + 12 < 8p, when p > 6. T h u s there are no infinite families of Srj-special numbers w i t h Q(n) = 4. I terating on 8p for p = 2, 3, 5, shows that none belong to an s 2 -cyc le . Check ing other cases: s 2 ( 2 2 3 3) = 37 > 36, s 2 (2 2 3 5) = 5 1 < 60, s 2 ( 2 3 3 3) = 45 < 54. 14 Hence there are no sj-perfect numbers satisfying il(n) — 4. I terat ing on the above S2-excessive numbers shows that none belong to an S2-cycle. 3.3 Q(n) = 5 s 2 ( 2 • 2 • 2 • 2 • p) = 8p + 24 < 16p, when p > 3. T h u s there are no infinite families of s^-special numbers w i t h J7(n) = 5. I tera t ing on 16p for p = 2, 3, shows that 48 is i n fact s^-perfecti and 32, wh ich is S2-excessive, does not belong to an S2-cycle. Check ing other cases: s 2 (2 • 2 • 2 • 3 • 3) = 57 < 72, Hence 48 is the on ly s^-perfect number satisfying fi(n) = 5. We have proved the fol lowing theorem. T h e o r e m 9 27 and 48 are the only s^-perfect numbers. T h e o r e m 10 There are no si-cycles. Proof. A n S2-cycle must have a least element that is S2-excessive. It is easily verified (as it was above i n the case n = 18), that a l l of the finitely many s 2 -excessive numbers not of the form 4p do not belong to S2-cycles. Thus any such element must belong to the fami ly of numbers of the form 4p. We w i l l show that i n a l l but a few t r i v i a l cases S2(s 2(4p)) < 4p, g iv ing a contradict ion. N o w , s 2 (4p) = 8 ( ( p + l ) / 2 ) . We may assume that p is odd, and set m = ( p + 1 ) / 2 . T h u s we w i l l have a cont radic t ion if the fol lowing holds: s 2 (8ra) < 8 m - 4. T h i s is equivalent to 12 + 6 s i ( m ) + s2{m) < 8 m - 4 , (3.1) 15 which is equivalent to 16 A 1 V - 1 + 6 > : + > < Pl---Ps ^Pl---Pi---Ps l<£Ji<sVl---Pi---Pj---Ps where m = p\ • • • ps. Here p\ • • • pi • • • ps is defined to be p i • • • ps/pi, and p\ • • • p% • • • pj is defined to be p\ • • -ps/piPj. Since pi > 2, this expression is impl ied by: 16 _6s_ s(s-l) 1 wh ich holds for a l l s > 4. If s = 1, then m = p is pr ime, and so condi t ion (3.1) becomes: . " , 12 + 6p < 8p - 4 , . wh ich holds for a l l p > 8. It is easily verified for p = 2, 3, 5 and 7, that 8p does not belong to an S2-cycle. , For s = 2, we can wri te m = pg. T h e only values of m for wh ich (3.1) fails are determined by the pr ime pairs (p,q) = (2,2) , (2,3) . In bo th cases, 8 m does not belong to an S2-cycle. F i n a l l y for s = 3, i f in = pqr, only for the t r iple (p, q, r) = (2, 2,2) does m fai l to satisfy (3.1). A g a i n , i n this case, 8 m does not belong to an S2-cycle. • R e m a r k 2 The longest increasing si-sequence is 8 Jl> 12 J i » 16 J** 24 ^ 30 31. 3.4 A n Inverse Problem Definit ion 5 Lei fe e No- We define : N —> No 6t/ ^(n) = IK 1 [{n}] | 16 E x a m p l e 3 r i ( l ) = 0, but for all n > 2, r\(n) > 1. In fact, lim^oo r\(n) = oo. To see this, simply set n = si{2aib) = 2a + 36, and observe that the number of pairs (a, b) satisfying this equation can be made arbitrarily large for all n sufficiently large. W e now prove a weaker result for ri, as found i n [20]. T h e o r e m 11 There exists an N e N such that for all m> N, r2(m) > 1. Proof. It suffices to show that for m sufficiently large, m = S 2 ( 2 a 3 b 5 c 7 d ) , for some a, b, c, and d > 0. In general, S 2 ( 2 « 3 W , = 4 ( « ) + 9 Q ) + 2 5 ( « ) + 4 9 g ) - O ( 0 - G ) ( ? + - ( 9 ( f = i [(2a + 36 + 5c + 7d)2 - (4a + 96 + 25c + 49d)] So, given m, we need only find solutions to the equations: 2a + 36 + 5c + 7d = R, 4a + 96 + 25c + 49d = R2 -2m, w i t h nonnegative integers a, b, c, d, and R EN. These equations are equivalent to: 2a 10c - 28d = 3R - R2 + 2m,. (3.2) 36 + 15c + 35d = R2 - 2R - 2 m , (3.3) Since a and 6 must be nonnegative integers, we have the following necessary and sufficient condit ions for a solut ion to (3.2) and (3.3): 1. 2 m = R2 + R + d (mod 3), 17 2. R2 - 3 R - 10c - 28d < 2m, 3. 2m < R2 - 2R - 15c - 35d. No te that equat ion (3.2) is always satisfied modulo 2. C o n d i t i o n 1 results from t ak ing equat ion (3.3) modulo 3, and condit ions 2 and 3 are derived from the fact that a, b > 0. Consider the interval IR(c, d) = [R2 - 3 R - 10c - 28d, R2 - 2R - 15c - 35d]. For fixed d, let cR(d) be the least c such that £(IR(c,d)) < 15, where £ ( / ) denotes the length of an interval / . We use the nota t ion L(I) and R(I) to denote the left and right endpoints of an interval I, respectively. Since R(IR(C, d)) = i ? ( / # ( c + l , <i)) + 15, when they exist, we have that cR(d) [j IR(c, d) = [R2 - 3 R - 10cR(d) - 28d, R 2 - 2 R - 35d]. c = 0 Denote the above interval by Z]i(d). B y definit ion of ca{d), £(IR{cR(d), d))=R- 5cR(d) - Id < 15, so - 1 0 c f i ( d ) < -2R + 30 + 14d, and cR(d) is the least such c. Consider the interval Hd=o ^ R ( ^ ) - C lea r ly R(f]^-Q 1R{d)) R2 — 2R — 70. W e now wish to find an upper bound for L(Cfd=r)lR{d)). F r o m the above inequality, we have that L(IR(d)) = R 2 - 3 R - 10cfl(d) - 28d < R2 - 5 R - 14d + 30. 18 T h u s 2 L( p) lR(d)) = m a x { i ? 2 -3R- 10cR(d) - 28d\d = 0 ,1 ,2} < max{R2 - 5 R - Ud + 30|d = 0 ,1 ,2} = R2 - 5R + 30. Le t JR = [R2 -5R + 30, R 2 - 2 R - 70]. T h e n JR c f|S=o JR(D)- N O W L(JR+I) < R(JR), i f and only i f R2 - 3R + 26 < R2 - 2R - 70, wh ich holds for a l l R > 96. So i f 2 m > L(Jg^) = 8766, then there is an R > 96 such that 2 m € JR C fld=o-^ R(^)- Choose d G {0 ,1 ,2} such that condi t ion 1 is satisfied. Since 2 m s IR(d), there is a c > 0 such that 2 m € IR(C, d). For these values of R, c, and d, condit ions 2 and 3 are satisfied. In other words, there exists an n such that m = S2(n). T h i s completes the proof. • L e t us t r y and improve on this result. T h e o r e m 12 l i m m _ ( 0 0 r2(m) = oo Proof. Le t C O i=l where pi denotes the i th pr ime, Xi > 0, and xi = 0 for a l l but finitely many values of i. If O O 1=1 then s 2 ( n ) - i [ A 2 - ^ ] . Set S2(n) = m and then set = R, as before. T h i s gives two equations i n the unknowns x\,X2, - • • , and R, namely: 19 2 x i + 3a; 2 + • • • = R (3.4) 2 2 x i + 3 2 x 2 + ••• = R2 -2m. (3.5) E l i m i n a t i n g from the first and x\ from the second gives 2 x i - 1 0 x 3 {p2r- 3pr)xR = -R2 + 3R + 2m 3 x 2 + 1 5 x 3 + \-{p2r- 2pr)xR + • • • = R2 - 2R - 2m. If ( x 3 , X 4 , . . . , R), satisfy the conditions (1) 2m > R2 - 3R - 1 0 x 3 - 2 8 x 4 - •. • • - {pi - 3pr)xr , (2) 2m < R2 - 2R - 1 5 x 3 - 3 5 x 4 {pi - 2pr)xr , (3) 1 5 x 3 + 3 5 x 4 H + {pl~ 2pr)xr + --- = R2+R + m (mod 3), then they uniquely determine a solut ion ( x i , x 2 , . . . ,R) to the equations (3.4) and (3.5). Observe that given R and m , for any choice of X 3 , X 5 , x$,..., we can choose x 4 uniquely from the set {0 ,1 ,2} so that the congruence (3) holds. We w i l l do this i n the following way. L e t qi, ( j 2 , . . . be the odd primes congruent to 2 modulo 3, l is ted i n increasing order. T h e n for yi > 0, and r > 0, we have that 15^ /1 + 99y2 + • • • + (g? - 21r)yr = 0 (mod 3). Denote this sum as D = D{y\,..., yr), and let r C = ^ ( l i ~ 3 9i)2 / i i=0 20 L e t q be any o d d prime congruent to 2 modulo 3, k = q2 — 2>q, £ = q2 — 2q, and suppose a > 0. Consider intervals of the form IR(a) = [R2 -3R-28d-C- ak,R2 -2R- 35d — D — a£}. W e w i l l handle the case when R, m = 0 (mod 3). T h e other cases are v i r t ua l l y ident ical . In this case, choosing d = 0, we have that i f 2 m lies i n an interval IR{Q) = [R2 -3R-C - ak, R2 - 2R- D - a£], then m satisfies condit ions (1), (2), and (3), w i t h the primes qi, q. Note that d is p laying the role of X4 above. W e w i l l show that the number of such intervals containing 2 m tends to inf ini ty as m does. T h i s implies the theorem. If ml = 0 (mod 3) and i f 2 m ' lies i n the interval / f l - i ( 0 ) = [R2 - 5R + 4 - C, R2 - AR + 3 - D], then since R = 0 (mod 3), condi t ion (3) is satisfied w i t h respect to the primes occur r ing i n the terms of C and D, R — 1 instead of R, and m! instead of m. G i v e n R >> 0, C, D, k, £ and q, let aR > 0 be the largest number such that (using nota t ion as i n the previous theorem): (i) L(IR(aR)) < R(IR(aR)), and (ii) R(IR(aR + l))>L(IR(aR)). These condit ions ensure that U w A=0 is an interval , and of longest possible length. Since we always have R(IR(a + 1)) < R(IR(a)), then condi t ion (ii) implies condi t ion (i). C o n d i t i o n (ii) holds i f R2-2R-D-{aR + l)£> R2 -3R-C - aRk, which i n t u r n holds i f and only i f R > (D - C) + aR{£ -k) + e. 21 T h u s we have R (D-C) e-k Define JR=\J W. We wish to show that JRC] IR-I(0) is nonempty. T h a t is that L(JR) < R(IR-I(0)). T h i s is true i f and only i f L(IR(CIR)) < R(IR-I(Q)), i.e. i f and only i f R2 - 3R - C - aRk < R2 - 4R + 3 - D. Subs t i tu t ing i n the value for otR into the above gives Rz - m - C - R-i-(D-C) e-k k<R2-4R + 3-D. T h i s inequal i ty is impl i ed by the inequal i ty R 2 _ 3 R _ c + k-[R \ ( f C))k<R2-4R + 3-D, £ — k which is i n t u r n equivalent to the inequali ty R+(D-C)-3 + k< R-e-(D-C) e-k k. T h e coefficient of R on the left hand side of this inequal i ty is 1. O n the right hand side it is k/(e — k) — q — 3 > 1, and so for R large enough, this inequal i ty holds. Since q, and each qi could be taken to be arb i t ra ry o d d primes congruent to 2 (mod 3), we can vary the primes i n the terms of C and D. For each such choice, and each choice of q say w i t h q > qr, we obta in dist inct solutions for every rn such that 2 m lies i n JR, i n this case where R, m = 0 (mod 3). T h e remaining cases, i n which R and m need not be congruent to 0 (mod 3) are handled similar ly, by first selecting a different value of d G { 0 , 1 , 2 } , and showing that the corresponding interval JR reaches low enough to intersect the interval 7R_I(0). ' • W e end this section w i t h a conjecture. Conjecture 1 For every k G N, limn-^ oo r^(n) — oo. 22 Chapter 4 Higher Elementary Prime Symmetric Functions T h e o r e m 13 (1) Let n G N . If n is s*k-special then pn is Sk+\-excessive for every prime p. (2) Ifpn is sk+1-special for every prime p, then n is s\-special, and hence by (1) , pn is Sh+i-excessive for every prime p. Proof. (1) Suppose n is s^-special. T h e n since J7(n) > k, we have sk+i(n) > 0. So Sfc+i (pn) = psk (n) + s f c + i (n) >pn + sk+i(n) > pn. (2) If sk+i(pn) = psk(n) + sk+i(n) > pn, for every pr ime p, then sk(n) > n — sk+\(n)/p. L e t t i n g p —> oo, we have sk(n) >n. • C o r o l l a r y 14 For k € N , there are only finitely many s*k-perfect numbers. Proof. B y the previous theorem, any infinite family Ek(n, r ) , r > 0 contains only sk-excessive numbers. W e c l a i m that there are only finitely many s£-spec ia l numbers not belonging to any such family. Indeed if there were inf ini tely many, then by 23 L e m m a 8, there would be an r > k such that there are infini tely many such s£-special numbers n satisfying Q(n) = r. Wr i t e them as follows: s = {py--.p?\p?.-.p?\pw.--P(v,...} and assume that p\^ < Pi+V C lea r ly we must have that l i r n 3 _ 0 0 p ^ = oo, otherwise, we must have the same number occurr ing twice i n the sequence. If p^\ contains a bounded subsequence, then there must be an ident ical product p^p • • • p^_\\ occur r ing for infini tely many values of I. T h i s gives a contradict ion, since it implies that Ekipf0^ • • -Pr- i> 1) intersects S, where £n is one such value oil. T h u s l i r n 3 _ + 0 O p ^ 2 1 = oo. Inductively, by s imilar arguments, we conclude that l i m ^ o o P m ^ = oo, for m = 1,2,... ,r. B u t then J -oo pU) . . . pU) T h i s implies that the set S contains Sfc-defective elements, a contradic t ion. • T h e preceding results of this section were first proved i n [20], by the author. C o r o l l a r y 15 Let k e N . There are only finitely many d < 0 such that the equation n — Sfc(n) = ti has infinitely many solutions. Proof. T h e only such d are those for which — d is s£_ ] -pe r f ec t , of which there are only f ini tely many. • T h u s the infinite families of S3-excessive numbers are: £ 3 ( 4 , 2 ) , £ 3 ( 1 6 , 1 ) , £ 3 ( 1 8 , 1 ) , £ 3 ( 2 4 , 1 ) , £ 3 ( 2 7 , 1 ) , £3 (30 ,1 ) , £3 (32 ,1 ) , £ 3 ( 3 6 , 1 ) , £ 3 ( 4 0 , 1 ) , £ 3 ( 4 8 , 1 ) . T h a t is, those corresponding to the s^-special numbers £ 2 ( 4 , 1 ) U { 1 6 , 1 8 , 2 4 , 2 7 , 3 0 , 3 2 , 36 ,40 ,48} . B y exhaustive search (as was done w i t h k = 2), a l l other s^-special num-bers can be found. T h e y constitute the following set: {42p|p = 7 , 1 1 , . . . , 41} U {.56p|p = 7 ,11, . . . , 43} U {64p|p = 2 , 3 , . . . , 37}U {726,858 ,250 ,350 ,225 ,315 ,968 ,1144,300 ,420 ,162 ,270 ,378 ,243 ,400 ,560 , 216 ,360 ,504 ,324 ,288 ,480 ,672 ,432 ,256 ,384 ,640 ,576 ,512 ,768} . 24 S3 197660 = 4 5- 9883 S3 248636 = 4 61 • 1019 S3 315452 = 4 17 •4639 S3 352508 = 4 13 • 6779 S3 428636 = 4 13 • 8243 None of the elements i n the above sets are Scj-perfect, hence there are no s^-perfect numbers. T h e diversi ty of possible increasing S3-sequences makes it difficult to rule out the existence of S3-cycles as we d i d S2-cycles. T h i s is i l lus t ra ted i n the following example. E x a m p l e 4 If pi, q\ are odd primes, then s^(Apxq-\) = 4 ( p i g i + pi + qi). It is possible that p\q\ + pi+ q\ = P2Q2, where P2, 92 are again odd primes, and so on. Several such sequences exist the longest one with piq± < 50000, andpi,qi > 3 is: 184892 = 4 • 17 • 2719 195836= 4 • 173 • 283 237212 = 4 • 31 • 1913 244988 = 4 - 7 3 - 8 3 9 252956 = 4 • 11 • 5749 275996 = 4 • 7 • 9857 334076 = 4 • 47 • 1777 341372= 4 • 31 • 2753 379676 = 4 • 11 • 8629 414236= 4 • 29 • 3571 461660 = 4 • 5 • 41 • 563. It seems h igh ly unl ikely however that an increasing S3-sequence be infinite. T h i s is part of the following conjecture: Conjecture 2 Any increasing Sk-sequence is finite. A related conjecture is given by: Conjecture 3 For any n S N o , the Sk-sequence of n is ultimately periodic. T h e evidence for these conjectures lies p r inc ipa l ly i n the fact that for a fixed k, w i t h finitely many exceptions, a l l s£-excess ive numbers belong to one of the families Ek(n,r). T h e union of a l l such families has asymptot ic density 0 i n the integers. Be low is a complete list of a l l s^-perfect numbers for k = 1,2,3, and 4, found by exhaustive search. 25 k sjl-perfect numbers 1 4 2 27, 48 3 none 4 3125, 9315, 31280 4.1 The Hunt for S3-cycles A s just noted, ru l ing out the existence of s 3 -cycles is not a t r i v i a l matter , since there seems no apparent m a x i m u m length on increasing S3-sequences. Nonetheless, there are a number of things which can be done to get a better picture of what such a sequence can look like. U s i n g maple, computer searches were done to verify that many numbers do not belong to an s 3 -cyc le . These include: {42p\p = 7 , 1 1 , . . . , 41} U {56p|p = 7,11, . . . , 43} U {64p\p = 2 , 3 , . . . , 37}U {250,350 ,225 ,315 ,968 ,1144,300 ,420 ,162 ,270 ,378 ,243 ,400 ,560 ,216 ,360 ,504 ,324 , 288 ,480,672,432,256,384,640,576,512,768} , that is, a l l those Sg-special numbers not belonging to any of the infinite families. It was also shown for each of the following sets: #3(16,1), #3(18,1), #3(24,1), #3(27,1), #3(30,1), #3(32,1), # 3 (36 ,1 ) , # 3 (40 ,1 ) , #3(48,1), wh ich are a l l o f t h e form {np\p is pr ime}, for a l l p < 1000, that none of the numbers belong to an S3-cycle. F ina l l y , the same was shown for the set # 3 (4 ,2 ) = {4pq\p, q are prime} i n the fol lowing ranges: 2<p< 30,2 < q < 1000, and 2 < p < 100,2 < q < 100. T h u s an s 3 - cyc le must have a least element belonging to one of the 10 infinite sets E%(n,k). W e label these sets i n the following way: we say m is type 1 i f m e £3(4,2). S i m i l a r l y we identify elements i n the following sets: . • £ 7 3 ( 1 6 , l ) - t y p e 2 , ' • # 3 (18 ,1 ) - type 3, 26 •£3(24,1) - type 4, •# 3 (27 ,1 ) - type 5, •#3(30,1) - type 6, •# 3 (32 ,1 ) - type 7, • # 3 ( 3 6 , 1 ) - type 8, •#3(40 ,1) - type 9, • # s ( 4 8 , l ) - type 10. T o analyze what increasing sequences begin w i t h elements of types 1 to 10, we w i l l consider the effect of apply ing S3 to such a number, and consider what type the result ing number may be, i f any type. We w i l l use the nota t ion type a —> type b to indicate the following: it is not impossible for d iv i s ib i l i t y reasons for to have numbers m and S3 (m) of type a and b respectively. We have noted that type 1 —> type 1 is possible. Be low is a complete list of a l l the possibil i t ies. Note that we are assuming that the numbers do not lie i n the ranges ruled out by the computer searches. • • • • type 1 — • type 1, type 8, • type 2 — * type 9, • type 7 —> type 9. • type 10 — • type 9. T h i s list was arr ived at by s imply ru l ing out possibili t ies. For example, i f rn = 18p is type 3, then S3(18p) = 21p + 18 = 3(7p + 6). Since we may assume p > 1000, p is clearly not 2 or 3, so 3||s3(18p), and 2 \ S3(18p). T h u s it is impossible for s$(l%p) to be of any of types 1 through 10. T h e list arr ived at above includes a l l those cases that could not be ruled out i n like manner. N e x t we w i l l rule out the possibi l i ty that numbers of the types that can not be of another type upon appl icat ion of S3, are not least elements of any S3-cycles. T h e o r e m 16 / / n is type 3, 4, 5, 6, 8, or 9, then n is not the least element of an ss-cycle. 27 Proof. F i r s t suppose n = 18p is type 3. We may assume that p > 1000. A p p l y i n g S3 we get S3(18p) = 21p + 18 = 3(7p + 6). L e t m = 7p + 6. If m is pr ime, then the s 3-sequence terminates, as it does if m is the product of 2 primes. If m = 919293 is the product of 3 primes, then we have s3 (n) < n if and only i f 3(<7i<?2 + 9193 + 9293) + 919293 < 18 ^ 9 l 9 2 ^ 3 — i f and on ly i f 108 + 21(9192 + 9^3 + 9293) < I l 9 i 9 2 9 3 - (4.1) Since p > 1000, we have that 919293 > 7006. A l s o , clearly q^ ^  2 ,3 , or 7. Unde r these constraints, the inequali ty (4.1) holds, and so n cannot be the least element of an S3-cycle. Hence we may assume m = 91 • • -qr, where r > 4. T h e same holds for m > 7006, and qi ^ 2 ,3 , or 7. In this case, s 3 (n) < n i f and only i f 108 + 21 (9 i92 + h qr-iqr) + 7(9l9293 + r- 9 r _ 2 9 r - l 9 r ) < 1891 • • ' 9r-U p o n d i v i d i n g each side by m, and using the facts that m > 7006, and 9, > 5, we have that the latter inequal i ty is impl ied by the inequal i ty 108 21 fr\ 7 fr\ + r - ^ L + - T L < 1 8 . 7006 5 r - 2 V 2 / 5 r - 3 V 3 , T h i s indeed holds for r > 4. T h u s numbers of type 3 cannot be least elements of S3-cycles. T h e remain ing cases, types 4, 5, 6, 8, and 9 are handled i n v i r tua l ly ident ical fashion. • Remark 3 A consequence of the above proof, is that all but finitely many maximal increasing S3-sequences of length greater than 2 have as least element a type 1 num-ber, and have as second last element either a type 1 or a type 8 number, with all 28 intermediate numbers type 1. That is, the sequence must be of the form type 1 —> type 1 —> • • • —•> type 1 —> m , or of the form type 1 —> fr/pe i —> • • • —• fj/pe. i —> type <? —> ro, where m satisfies S3(m) < rn. W e have accumulated enough informat ion to prove the non-existence of the simplest case of s 3 -cycles . T h e o r e m 17 There are no s^-cycles of length 2. That is, if sf\n) = n, then s 3 (n ) = n. Proof. A n y such s 3 -cyc le must have a least element n of type 1, 2, 7, or 10. F i r s t assume that n = Apq is type 1. Because of the computer searches, we may assume that the pair (p, q) lies outside of the ranges 2 < p < 3 0 , 2 < q < 1008, and 2 < p < 100, 2 < q < 100, and that p < q. L e t ro = pq+p + q. In the allowable range, the m i n i m u m value for m occurs when p = 2, and q = 1009, which gives m = 3029. No te that 2 \ m. W e have 4 2 ) (4p9) = s 3 (4ro) = 4s i ( ro) + 4s 2 ( ro) + s 3 (ro) . If s 3 2 ) ( n ) = t n e n 4si ( ro) + 4s 2 ( ro) + s 3 ( m ) = 4pg. (4.2) We w i l l take cases depending on i1{m). In the following, qi always indicates a pr ime d i v i d i n g m . If m = q\ is pr ime, then we are done since then s 2 (4ro) = 4 m . If ro — q\q%, then (4.2) implies qi + 92 + 9 i 9 2 = pq. Subs t i tu t ing qio2 = p o + p + o , and canceling gives qi+q2+P+q = 0, an impossibi l i ty . 29 If m = gig2<?3; then (4.2) implies that 4 s i ( m ) + 4s2(m) + m = 4pq. T h i s is impossible, since 2 { ra. In case m = q\ • • -04, we w i l l show that (4.2) is i n fact a str ict inequality. D i v i d i n g (4.2) through by m gives 4 ( _ L . + ... + - ! _ ) + 4 ( - L + ... + - L W I + ... + I ) . ^ a _ . V919293 929394/ \q\qi qsq^J \ 9 i q±) pq + p + q W e can min imize the right hand side of this equation. In the range i n question, 4pq/'(pq+p+q) is min imized w h e n p = 2,q = 1009, and this gives us that 4pq/(pq + p + q) > 2.6649. It is an easy check that if > 3, and m = g i • • -94 > 3029, then we always have 1 1 \ . / 1 1 \ / 1 1 + ••• + + — + ••• + — + ••• + - + 4 + • 910293 025304/ \q1q2 9394/ \ 9 l < 2.6649 < — . pq + p + q A s imilar approach can be used when il(m) = 5 . If r > 6, then 1 1 \ , ( 1 1 \ / 1 1 1 1 4 - 4 1 1 + -+ ••• + + 4 + ••• + + — + ••• + — 919293 qr-2qr-iqrJ \q1q2 qr-iqrJ \qi 9r / r \ 4 / r \ 4 fr\ 1 1 y 3 r - ! V2 /3 r - 2 V 3/ 3''" 3 < 2.6649 < — — , pq + p + q and we are done the type 1 case. N e x t suppose that the least element n = 16p is type 2. We may assume that p > 1009. N o w 4 2 ) ( 1 6 P ) = S3(8(3p + 4)). L e t m = 3p + 4, so 16p = 16(m - 4 ) / 3 . T h e n 2 ,3 \ m, and m > 3031, So s 3 2 ) ( 1 6 P ) = 8 + 12s i (m) + 6s2(m) + s3(m).~ T h e supposi t ion that s 3 '(n) = n is equivalent to (m — 4 N 8 + 12s i (m) + 6 s 2 ( m ) + s 3 ( m ) = 16 30 which is equivalent to 88 + 36si(m) + 18s2(m) + 3s3(m) = 16m, (4.3) We will do cases depending on the value of il(m). In the following, each g, is assumed to be prime. If m = gi is prime, then (4.3) becomes 88 + 36gi = 16gi, which is impossible. If m = g i g 2 , then (4.3) is also impossible for m in the allowable range. If m — 919293; then (4.3) is equivalent to 88 + 36(gi + q2 + g3) + 18(gig2 + 9i93 + 0293) = 13gig293-Dividing through by m, and using the facts that m > 3031, and qi > 5, we have the inequality 8 R + 3 6 r - i - + J _ + J L > ) + 1 8 f I + I + I , < 1 3 3031 V°i°2 9i93 9293/ \ 9 i 92 93, This contradicts (4.3). A similar contradiction occurs when fl(m) > 3. The proofs of the cases when n is type 7 'or 10 follow the same lines as the proof for case 2. • 31 Chapter 5 The Exponent Prime Symmetric Divisor Functions Anothe r na tura l class of pr ime symmetr ic divisor functions are the functions whose values on n G N are the sum of the kth powers of the primes d i v i d i n g n, w i t h repet i t ion. D e f i n i t i o n 6 Let k G No- Define the function : No —> No by e/c(0) = 0. For n = pi • • -pr G N , where the pi are the primes dividing n, with repetition, set r efc(n) = ]TP*-i=l We call the function the kth exponent prime symmetric divisor function. R e c a l l we were able to classify when a pr ime power pa was s^-perfect for some k. W e now present an analogous result per ta in ing to e/c-perfection. T h e o r e m 18 Let n=pa be a prime power. Then there exists a k G N such that n is ek-perfect if and only if k = p® — and a = p® for some j3 > 0. Proof. ek{pa) = pa i f and only i f apk = pa. Hence we may wri te a = pP for some (5 > 0. T h i s implies that p8^ = p^, and so k = p3 - /?. Conversely, it is t r i v i a l to see that pf^ is always e p^_^-perfect. • 32 R e m a r k 4 The set {pi3 - (3\p is prime and ft > 0} has density 0 in the integers. So, for "most" values of k, there is no prime power pa that is ek-perfect. W i t h the functions ek, we can also characterize which numbers are e^-perfect for some k for products of two pr ime powers. ' . T h e o r e m 19 Let n = paqP', where p and q are primes, and a, (5 > 0. Then n is not ek-perfect for any k. Proof. W i t h o u t loss of generality, p < q. Furthermore, since e i = s i , and the statement has been proved for s\, we may assume that k > 2. If ek(paq^) = Paq^, then apk + (3qk=paq0. (5.1) F i r s t assume that a,j3 > k. T h e n since Pqk=pk(pa-Y-a), we have that pk divides (3. S imi l a r ly qk divides a. Wr i t e (3 = bpk, and a = aqk. Clea r ly a, b > 0, and equat ion (5.1) becomes (a + b)pkqk =paikq bP". W e w i l l show that we instead have a strict inequali ty {a + b)pkqk < p a q q b p k . (5.2) Since a,b > 0, p > 2, and q > 3, it is easily seen that bp^1 < bpk — k, and aqk-i < agk _ ^ a n c i n e n c e t n e inequali ty (5.2) is impl i ed by o + 6 < p a « * " l 9 ' * * ~ 1 . T h i s is i n t u r n impl i ed by a + b < ab(pq)k~l 33 N o w pq > 6, and k > 2, so the latter inequali ty holds i f a + b < 6ab. T h i s is indeed true for a, b > 0. T h e second case we consider is when a, (3 < k. T h e n i n s imilar fashion to the previous case, we can wri te (3 = bpa, and a = aq@, where a, b > 0. T h i s however gives an immediate contradict ion, since then a = aqP = aqbp" > a. N e x t suppose that a < k, and (3 > k. T h e n for some a > 0, we have that T h i s is a contradic t ion. A similar contradic t ion results i n the remaining case, thus, R e c a l l that we considered the functions r^, wh ich counted the number of representations of n as Sfc(m) for some m. We now define their analogue for the exponent pr ime symmetr ic functions. Definit ion 7 Let : N —> N Q be defined by It is much easier to work w i t h the functions than w i t h their analogues rfe to the elementary pr ime symmetr ic functions: Theorem 20 For k £ N , we have that limn-^ oo £fc(n) = oo. Proof. Le t t £ {0, l , . . . , 2 f c - 1}, and consider the sequence {n2k + £}^=1. L e t q be any pr ime congruent to 1 modulo 2 f c . Numbers of the form 2nq(, satisfy ek{2nql) = n2k + £qk, and £qk = £ (mod 2k). Since there are infini tely many such o, we have that lim^oo ek{n2k + £) = oo. Since this holds for any £ e { 0 , 1 , . . . , 2k — 1}, we have the stated result. • F r o m the theory of part i t ions (see for instance [1]), we can construct gener-a t ing functions useful i n calculat ing £fc(n). Indeed, we have a = aqk > aqa > a. the theorem is proved. • £k(n) = \ek (n)|. 34 T h e proof is by analogy w i t h [1] pp.308-310. W e can relate the functions sk and using the N e w t o n - G i r a r d formulas (see [18]). These formulas i m p l y that k (-l)kksk + Yl(-l)i+keiSk-i = 0. i=l T h e function ek can be expressed s t r ic t ly i n terms of the elementary pr ime symmetr ic functions s\, . . . , Sfc as a determinant (see also [19]): 1 0 0 . . . o 2 s 2 Sl 1 0 . . . o 3 s 3 S2 Sl 1 . . . o efc = ( - l ) f c 4 s 4 S3 S2 S l . . . o ksk Sfc-l Sfc-2 Sfe- 3 • • • S i Simi lar ly , we can express sk i n terms of efc as follows: Sfc e i §2. 2 3 4 2 §2. 3 £2 4 £1 3 £2 4 4 e^-i e*'.-2 efc-3 efc-4 fe-1 fc-1 efc_i e f c_2 e f c_3 A- /c /c 5.1 The Average Order of G i v e n an ar i thmet ic function / , we denote by / the function defined by / » = / ( l ) + / ( 2 ) + . . . + / ( n ) . In this section we w i l l come up w i t h upper and lower bounds for the functions ek. D o i n g so is more straightforward than for the functions sk for the simple reason that 35 the functions ek satisfy ek(ab) = ek(a) + ek(b). In the following analysis, p always denotes a pr ime, so for instance indexing p < n means over a l l primes less t han or equal to n. Theorem 21 The function ek satisfies the following inequalities: ( - - l ) E ^ - M n ) E I 4 ) < 5 f c ( n ) < ^ - 1 ^ ^ T -p<n p<n p<n Proof. Le t n G N . W e recall de Pol ignac 's formula (see [13]): i f p is pr ime then the largest number a such that pa\n\ is W e have N o w , ejfe(n) = efc(l) + h ek(n) = ek(n\) / OO = E K E n j=i ^ • Llogp(n)J = E[pk E p<n \ i=l Llog p(n)j E z=i [log (n)J < V ^ - A^l pt 1=1 1 _ ^ 1 j L l o g " ( n ) J \ < n P- 1 n P- 1 n P- 1 n — 1 ' i \ Llog p (n)J > l - ( - ) p - 1 36 T h i s establishes the second inequality. To establish the first, observe that Llogp(n)J E n L i o g , » J , We c l a i m that T h i s is equivalent to „ / /l\Llogp(n)J\ — x l - ( - ) - L l o g P ( n ) J . l o g » - Llog p (n)J > U i ) j \ ( l°6p («)J l L e t a; = l o g p ( n ) , and XQ = x — |_x_|. T h e n the above inequal i ty becomes x - I x l > [ - 4 T - — ) , L J _ p - 1 V P W Px J which is equivalent to ( p - l ) x 0 + l >pxo. T h e curve y = p ' , 0 < £ < l lies below the secant line y = (p — l)t + 1 through the points (0 ,1) , and ( l , p ) . T h u s this last inequal i ty holds, by t ak ing t = XQ. Consequent ly Li°gP(«)J E i=i n — 1 log(n) ~ p - 1 log (p ) ' wh ich implies the first inequality. W e w i l l now give more explici t asymptot ic bounds for e~k(n)-• Lemma 22 LetkeN. Then p<n fe+1 (fc + l ) l o g ( n ) ' 37 where the sum is taken over all primes p < n. That is l i m n—>oo (fc + l ) l o g ( n ) E p < n P f c ^ 1 Proof. T h e sum can be expressed as a Riemann-Stiel t jes integral: rn /~^pk = / xkdir(x). Integrat ing by parts, we have that rn rn i xk dir(x) = nkir(n) - k / xk~1ix(x) dx. J3/2 JZ/2 Since log(x) by the pr ime number theorem, we have that nk+l /•« „k rn Jz/2 X dw(x) -kT — log(n) 73/2 log(x) dx. ,k+l log(n) k ^k+l (fc + l)log(n) _ i r x  k + 1 J3/2 ( log(x)) ; dx ,fc+i (fc + l ) l o g ( n ) ' T h i s completes the proof. L e m m a 23 Let k € N. Then p<n log(p) (fc + l)(log(n))2-• Proof. A s before, we begin by expressing the sum as a Riemann-Stiel t jes integral: Y — p<n ft log(p) - f J3/2 log(x) dir(x). Integrating by parts we have: f A r t e ) = ^ - f f — - ^ ) n(x) dx k,2 l og (» ) ( j log(n) 73/2 l l o g ( x ) ( l og (x ) )2 j ffW 3/2 38 A p p e a l i n g to the pr ime number theorem as i n L e m m a 22, we have that r *k d7r{x)-nk*w !: r xk~^dx i r^iMdx 73/2 log(a;) log(n) J3/2 log(x) J3/2 ( l og (z ) ) 2 f^c+l rn xk rn xk ~ ( i o i ( n T p ~ f c y3/2 w d x + Uo^Wdx nk+i .. (log(n))2 k nk+l 2 rn xk (fc + l)(log(n))2 + fc+T i 3 / 2 ( b g M (fc + l)(log(n))2' comple t ing the proof. • C o m b i n i n g the results from the previous two lemmas and the theorem, we have the following: T h e o r e m 24 Let keN. Then 1 ^ • ,. M n ) l o g ( n ) Sfc(n) l o g ( n ) l — r < h m i n f i—r-1— < l i m s u p — , , , < —. k(k +1) ~ nfc+! _ F nk+l ~ k Proof. Since and we have that Pk fc-l fe-2 . 1 P — 1 P — 1 E ^ ' n ^1 ( n - l ) V - ^ ^ fclog(n) P<71 Simi lar ly , l 0 ^ ) E d ? ,fe n f e + 1 p < n l o g ( p ) (fc + l ) l o g ( n ) 39 by the previous l emma. T h u s by the inequalities i n theorem 21, we have that < ek(n) nfc+i fc l og (n ) ' T h i s implies the theorem. • 40 Chapter 6 Further Prime Symmetric Functions In this section we w i l l s tudy the properties of some more compl ica ted pr ime sym-metr ic functions. W e begin w i t h a definit ion, and some pre l iminary results. Definit ion 8 Letpi,... ,pr be primes such thatpi < ... < pr, and letn — p\- • -pr. We say that n is first Sk-defective of order r if n is Sk-defective and if whenever primes q\,..., qr satisfying q\ < ... < qr, are such that q\ • • • qr is Sk-defective and Qi < Pi for each i, then qi = Pi for each i. E x a m p l e 5 42 = 2 • 3 • 7 is first s^-defective of order 3. For a given k, and fixed r , there are only finitely many first s^-defective numbers of order r . L e t r(k) be defined as i n L e m m a 8. If r < k, or if r > r(k) then 2 r is the only first s^-defective number of order r . If r — k, there are no such numbers, since no such number is s^-defective. If n = pi • • • pr is Sfc-defective, then there exists an n ' = p[ • • • p'r wh ich is first Sfc-defective of order r , and such that p\ < pi for each i. We make use of the fact that there are only f ini tely many such numbers for each r i n the following lemma. 41 L e m m a 25 Let k G N , and let M > 0. Then there exists an N e N swc/i £fta£ i / n> N and n is sk-defective, then n — Sfc(n) > M. Proof. Le t n = p\ • • • pr. T h e inequal i ty n > Sfe(n) + M holds i f and only if T h i s is imp l i ed by V - 1 M 1 > 2^ + —. l<H<...<i r_ f c<rP i l n A " \ 1 M ,„ N There is an f? such that r > R implies that the inequal i ty (6.1) holds. O n the other hand, i f r < k, then we need only ensure N > M. T h u s we may assume k <r < R, for wh ich there are only finitely many possible values of r. F i x r i n this range. For the given r , there is a finite set of first Sfc-defective numbers of order r . L e t n' = p[ • • • p'r, where p'-, < ... < p'r, be one such number. T h e n since Sfc(n') < n', we have that i > E i l<U<...<ir_fc<r a. N o w i f n — pi • • • pr, where p\ < ... < pr and Pi > p\ for i = 1 , . . . , r , then 1 > a >- E — l<ii<...<ir_fc<r T h i s implies that Ph • • • Pir-k I M M > + — < c H . l<h<...<ir-k<r 1 T~k If we choose n large enough such that M l - a n~ < ~ ~ 2 ~ ' 42 then we have that n — Sk{n) < M as desired. There are only finitely many r for wh ich we must apply this analysis. For each such r, there are only finitely many first Sfc-defective numbers of order r , and as already stated, for any s^-defective number n = pi • • • pr, there is a first Sfc-defective number n' = p[ - • • p'r of order r such that Pi < Pi- T h i s proves the lemma. • T h e fol lowing Coro l l a ry is a companion to Coro l l a ry 15, and is an immediate consequence of the preceding L e m m a . C o r o l l a r y 26 Let k € N, and let d > 0. The equation n — Sk{n) — d has at most finitely many solutions. N o w let us look at some new prime symmetr ic functions. For 6 > 0, let / = Si + b. In [3], the case when 6 = 1 was studied It was shown that repeated i tera t ion of / eventually terminates i n the cycle 7 , 8 , 7 , 8 , e x c e p t when n < 7, i n wh ich case the sequence terminates i n 1 or 6. We w i l l show that for any 6 > 0, the funct ion / w i l l u l t imate ly terminate under repeated i tera t ion i n an / - c y c l e , or an /-perfect number. Fur thermore, we w i l l show that there are on ly finitely many such / - cyc les . For convenience, we w i l l consider an /-perfect number to be an / - c y c l e of length 1. T h e o r e m 27 Let 6 > 0 and let f — s\ + b. If n e N, then the f-sequence of n terminates in an f-cycle. Furthermore, there are only finitely many f-cycles. Proof. W e first need to show that the /-sequence of n must terminate i n an / - c y c l e . If not, then l i n i j _ > 0 0 f^(n) = oo. Le t p be the least pr ime not d iv id ing 6. If r > p, then elements of the sequence r, r + 6, r + 2 6 , . . . , r. + (p — 1)6 can not a l l be pr ime since one must be divis ible by p but s t r ic t ly greater t han p. B y L e m m a 25, there is an N such that i f n > N is not pr ime, then n > f(n) + 2pb. W h e n we iterate by / , a l l s t r ic t ly increasing subsequences of consecutive terms beginning above N are of the form q, q + 6 , . . . , q + ib, where q, q + 6 , . . . , q + 43 (i — 1)6 are pr ime, q + ib is not prime, and i < p — 1. In this case, we have that (g) = / ( g + i t ) < g + i& - 2p6 < q - pb. If g > max{7V + pb, / ( l ) + p6, / ( 2 ) + pb,..., f(N) + pb}, then fW(q) can never ascend above g + p 6 for any j. Thus the /-sequence of n, must terminate i n an / - c y c l e , since i n order for l i m ^ o o f^(n) = oo, it must contain such primes q. Showing there are only finitely many such cycles is done s imi la r ly to the first part of the proof. If there were infini tely many, we could choose a rb i t ra r i ly large least elements of such cycles. Clear ly , we can choose N large enough so that (1), i f n > iV is the least element of an / - cyc l e , then n is pr ime, and (2), i f m > N is not pr ime, then TO > / ( T O ) + pb. Le t n > N be the least element of an / - c y c l e . T h e n n is pr ime, and there is a least i < p such that n + ib is not prime. T h e n /( J + 1)(n) = f(n + ib) <n + ib — pb<n. T h i s contradicts that n is the least element of an / - c y c l e , and the Theorem is proved. • W e w i l l i l lustrate this Theorem w i t h the case when 6 = 2. Example 6 Let f = si + 2. Then for n € N , the f-sequence of n terminates in the f-cycle {8}. Indeed it is easily verified that this is so for n = 1 ,2 , . . . , 35. It is also readily seen that for n larger, if n is not prime, then f(n) + 4 < n. If there is another f-cycle, it must have a least element q > 35 which is prime. However, one of q + 2, and q + 4 can not be prime. In either case, if we apply f, we get a contradiction to q being the least element. Below is a table of various values of 6, and corresponding complete lists of / -cyc les , where / = s\ + b. 44 b / -cyc les 1 {1},{6},{7,8} 2 {8} 3 {9},{10} 4 {11,15,12}, {13,17,21,14} 5 {12},{13,18},{14} 6 {14,15} 7 {15} 8 {16} 9 {17,26,24,18}, {22} 10 {18}, {19,29,39,26,25,20} 45 Appendix A Maple Algorithms Computer searches and much of the experimentation for this exposition was done us-ing Maple. In this section I include some of the algorithms for computing some of the prime symmetric functions. First note that the wi th( l ina lg) : , with(numtheory) :, with(combinat) : and with(LinearAlgebra) : commands should be invoked at the beginning of the worksheet. The omega procedure with input n returns the value of ui{n), that is, the number of distinct prime factors of n. omega:=proc(n) om:=Dimension(Vector(ifactors(n)[2] ) ) /2: return(om): end proc: The pvect procedure with input n returns a vector of length fi(n) whose entries are the prime factors of n with repetition. pvect:=proc(n) c:=0: v:=array(l..1,1..bigomega(n)): 46 k:=omega(n): for i from 1 to k do for j from 1 to ifactors(n) [ 2 ][i] [ 2 ] do c:=c+l: v[l,c] :=ifactors(n) [2] [i] [1] : od:od: return(v): end proc: To calculate efc(n), use the following: e:=proc(k,n) v:=ifactors(n) [ 2 ] : r:=omega(n): s:=sum(v[j][2]*v[j] [ l ] ~ k , j = l - - r ) ; return(s); end proc: So, for example, if we wished to calculate e3(1444), we would type e(3,1444); Maple returns the correct value of 13734. To compute the function sk at n, use the following: s:=proc(k,n): b:=bigomega(n): v:=pvect(n): a:=choose(b,k); sk:=sum(product(v[l,a[i][j]],j=l..k),i=l..binomial(b,k)); return(sk); end proc: 47 For example, to compute ss(174960), we would type s ( 5 , 1 7 4 9 6 0 ) ; H i t t i n g re turn yields the correct output of 134942. 48 Bibliography [1] A p o s t o l , T . , Introduction to Analytic Number Theory, Springer, N e w Y o r k -Ber l in -Heide lberg-Hong K o n g - L o n d o n - M i l a n - P a r i s - T o k y o , 1976. [2] B u r t o n , D . M . , Elementary Number Theory, A l l a n and B a c o n , Inc. Bos ton-London-Sydney, 1976. [3] Cadogan , C . C . and Cal lendar , B . A . , A problem on positive integers N e w Zealand M a t h . M a g . 11 (1974), 87-91, 94. [4] C h a n d r a n , V . R . , On generalized unitary perfect numbers, M a t h . Student, 61 (1992), 54-56. [5] Cohen , G . L . and te Riele , H . J . J . , Iterating the sum of divisors function. Exper imen t . M a t h . , 5 (1996), 91-100. [6] D u n h a m , W . , Euler The Master of Us All, T h e M a t h e m a t i c a l Assoc ia t ion of A m e r i c a , 1999. [7] Hardy , B . E . and Subbarao, M . V . , On hyperperfect numbers, Congr . Numer . , 42 (1984), 183-198. [8] Hunsucker , J . L . and Pomerance, C , There are no odd super perfect numbers less than 7 x 1 0 2 4 , Indian J . M a t h . 17 (1975), 107-120. [9] Loweke, G . P . , The Lore of Prime Numbers, Vantage Press, N e w Y o r k -Wash ing ton-At lan ta -Los Angeles-Chicago, 1982. 49 [10] M c C r a n i e , J . S., A study of hyperperfect numbers, J . of Integer Seq., 3 (2000), 153-157. [11] M i n o l i , D . , Issues in nonlinear hyperperfect numbers, Ma themat i c s of C o m p u -ta t ion , 34 (1980), 639-645. [12] Nathanson , M . , Elementary Methods in Number Theory, Springer, N e w Y o r k -Ber l in -Heide lberg-Hong K o n g - L o n d o n - M i l a n - P a r i s - T o k y o , 2000. [13] N i v e n , I., Zuckerman, H . , Montgomery, H . , An Introduction to the Theory of Numbers, J o h n W i l e y and Sons, Inc., N e w York-Chiches ter -Br isbane-Toronto-Singapore, 1991. [14] Pomerance, C , Multiply perfect numbers, Mersenne primes and effective com-putability, M a t h . A n n . 226 (1977), 195-206. [15] Suryanarayana, D . , Superperfect Numbers, E l e m . M a t h . , 24 (1967), 16-17. [16] te Rie le , H . J . J . , Hyperperfect numbers with three different prime factors, M a t h -ematics of Compu ta t i on , 36 (1981), 297-298. [17] Tat tersa l l , J . J . , Elementary Number Theory in Nine Chapters, Cambr idge Unive r s i ty Press, 1999. [18] E r i c W . Weisstein . " N e w t o n - G i r a r d Formulas ." F r o m M a t h W o r l d - A Wol f r am W e b Resource. h t t p : / /ma thwor ld .wo l f r am.com/Newton-Gi ra rdFormulas .h tml [19] E r i c W . Weisstein et a l . " S y m m e t r i c Po lyno -m i a l . " F r o m M a t h W o r l d - A Wol f ram Web Resource, h t tp: / / m a t h w o r l d . w o l f r a m . c o m / S y m m e t r i c P o l y n o m i a l . h t m l [20] Woodford , R . , A Variation on Perfect Numbers, Integers: E lec t ron ic Jou rna l of Combina to r i a l N u m b e r Theory, 4 (2004), A l l . 50 

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