UNITARY HARMONIC NUMBERS. CHARLES R. WALL Trident Technical College, Charleston, SC (Submitted October 1981) 1. INTRODUCTION

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1 <> <> CHARLES R. WALL Tridet Techical College, Charlesto, SC 911 (Submitted October 1981) 1. INTRODUCTION Ore [] ivestigated the harmoic mea H() of the divisors of ad showed that H() x () I'o(), where, as usual, x() ad o() deote, respectively, the umber ad sum of the divisors of. A iteger is said to be harmoic if H() is a iteger. For example, 6 ad are harmoic, sice H(6) ad #() 5. Ore proved that ay perfect umber (eve or odd) is harmoic, ad that o prime power is harmoic. Pomerace [] proved that ay harmoic umber of the form p a q b 9 with p ad q prime, must be a eve perfect umber. Ore also cojectured that there is o odd > 1 which is harmoic, ad Garcia [1] verified Ore T s cojecture for < ; however, sice 0re T s cojecture implies that there are o odd perfect umbers, ay proof must be quite deep. A divisor d of a iteger is a uitary divisor if g.c.d. (d 9 /d) 1, i which case we write d\\. Let T*() ad o*() be, respectively, the umber ad sum of the uitary divisors of. If has oo() distict prime factors, it is easy to show that T*( ) U)() ad o*() O (1 + p e ), p e \\ both fuctios beig multiplicative. Let be the harmoic mea of the uitary divisors of. It follows that - T*()/o*() II ^ V e \\ 1 + V 6 We say that is uitary harmoic if is a iteger. I this paper we outlie the proofs of two results: 18 [Feb.

2 Theorem 1 There are uitary harmoic umbers with 0)() < (see Table 1). Theorem There are uitary harmoic umbers < 6. These umbers, which iclude all but oe of those i Theorem 1, are give i Table. TABLE 1 w(w) o)() , , , , , ,95 55, i ,58 185,96 661, lf , ,80 6,9 5 - ^ ,65 0,1, V 1,560 5 * TABLE 1 1 9, , , , , , (cotiued) 198] 19

3 TABLE (cotiued) \ ,6,00 1,500 0,950 6,9 51,08 55,15 6,60 66,58 81,900 8,60 95,550 1,60 16,800 1, ^- 1 lf ,96,0 5,00 0,50,60 6,90 565,8 598,500 66,5 661,500 16,65 90,98 859, ,900 99,880 tf tf ' lf ** The complete proofs of Theorems 1 ad are quite tedious, requirig may cases ad subcases. However, the techiques are quite simple, ad are adequately illustrated by the cases discussed here.. TECHNIQUES FOR THEOREM 1 If p ad q are (ot ecessarily distict) primes ad p a < q b, the it is easy to show that H*(p a ) > H*(q b ). This fact ca be used, oce u)() ad are specified, to fid a upper boud for the smallest prime power uitary divisor of \ for each choice, the process is repeated to fid choices for the ext smallest prime power uitary divisor, ad the process cotiues util all but oe of the prime power uitary divisors is foud; the largest prime power ca the be solved for directly, without a search. Of course, this procedure is iterrupted ay time it becomes obvious that the as yet ukow portio of must have more prime divisors tha allowed by the prespecified size of o)() With 0() ad give, the problem is to fid with /o*() #*()/T*(rc) beig a prespecified fractio, which i tur requires that ay odd prime 0 [Feb.

4 that divides o*() must also divide. Also., sice T*() is a power of, ay odd prime that divides must also divide. Several of the cases are shorteed by usig results of Subbarao ad Warre [] for the special case o*() I (i.e., for beig uitary perfect). We preset here the proof for the case a)(), 15, oe of the loger ad subtler cases of Theorem 1. Throughout, let pqrs with p < q < v < s ad p, q, r, ad s powers of distict primes (though ot ecessarily prime). Note that because /o*() 15/16, 5\pqrs. Also, if has a prime power uitary divisor which is cogruet to (mod ), the must be eve. If p > 59, the /o*() > 15/16, so p < 5. p 5: q < 61, so q 59, which requires that 5 PS, a cotradictio. p 9: q < 6. But q 61 implies 5 * l rs, ad q 59 requires * * 5 rs; both of these are impossible. If q 5, the r < 9, but there are o powers of or 5 betwee 5 ad 9. p : q < 6 ad * 5 qps, so ^ 6, from which follows the impossibility 5 1 z»s. p : * 5* ll\qrs, a cotradictio. p 1: q < 1 ad * 5 e gi>s. The oly possibility is ^ 9, which requires v < ad * 5 rs. This i tur forces r 81, which implies s 15. Thus we have a uitary harmoic umber, sice #*( 5 1) 15. p : q < 9 ad» 5 19 grs 5 a cotradictio. p : <y < 8 ad 5 * ll qrs, so q 81. But the 5 11 l ps, a cotradictio. p 1: q < 89 ad *» 5 grs. There are three upalatable choices: q 81 requires that» 5» 1 rs, ad ^ 6 implies - 5«1 rs, while q forces 5 ll ps. p < q < 9, ad rs is divisible by at least three primes uless q is 89, 81, 59, or 9. If q 89, the v < ad there are o powers of or 5 betwee 89 ad. If q 81, the v < 9 ad 5 l ps, a cotradictio. If q 59, the p < 16 ad the oly possible cases are r 15, which implies s, ad v 81, which forces 5 1 s. If q - 9, the v < 19, so either r 15, which does ot leave the required 5 i the umerator of /o*(), or r 81 5 which forces 5» 1 \s» Thus p 9 is impossible. 198] 1

5 p : 5 < q < ad 5 l\qrs 9 so the oly possible values for q are 6 ad 9. If q 6, the 5 1 rs. If q 9, the 15 < r < 51 ad 5 rs, SO r - 18, whece 5 s, a cotradictio. p 5: 9 < q < 11 ad rs is divisible by three or more primes except whe q is,, 89, 81, 6, or 5. If q, the r < 15, while q - implies r < 18, ad q 89 forces r < 19; i each case, * 1 rs, a cotradictio. If q 81, the r < 15 ad 1* l rs, which is impossible. If q 6, the r < 11 ad * 1Ire; thus r 169, which forces 118, or r - 81, i which case 1 '11s. If q - 5, the r < 0 ad * 1 rs, so r is, 169, or 81; each of these possibilities forces s to be divisible by two distict primes. p : 5 < q < 1 ad 5\qrs. The possible values for q are 18, 15, 81, ad 6, but each of these forces rs to be divisible by three or more primes, a cotradictio i ay evet. p 19: 5 < q < ad $\qrs. Thus, q is 18, 15, or 81. Each of these possibilities is ruled out sice rs caot be divisible by three primes. p 1: 15 < q < 0 ad * b\qrs. To be withi the iterval, q caot be a power of 5, ad q forces r < 611 ad 5 6l rs, a cotradictio. Therefore, q is a prime power betwee 15 ad 0, cogruet to 1 (mod ), ad such that q + 1 has o odd prime factor other tha f s, 5 f s, ad at most oe 1. There are but two possibilities: q 69 ad q- 19. If q 69, the v < 5 ad ' 5 Irs, a cotradictio. If q 19, the 16 < r < 8 ad - 5 rs, so r 18, whece 5-5 s, a cotradictio. p 16: 55 < q < 65 ad 5 1\qrs 9 so q is 9,65, or 89, each of which would require that rs be divisible by three distict primes. Fially, if q < 16, the /o*() < 15/16.. TECHNIQUES FOR THEOREM Suppose that is uitary harmoic, i.e., that x*()/a*(w) is a iteger. Suppose also that < 6 ad that a. Sice T*() is a power of, ay odd prime that divides o* () must also divide. For a > 0, o*( a ) 1 + a, so a implies a (l + a ), ad hece a <. Except for a 0, the suppositio that a requires that be divisible by the largest prime dividig 1 + a, ad the restrictio that < 6 ca be used to determie how may times this prime divides. This gives rise to ewly kow uitary divisors of 9 ad therefore (usually) ewly kow odd primes dividig o*() ad hece. The procedure is repeated util all the possibilities are exhausted. [Feb-

6 No particular difficulty arises with this procedure, except whe oe rus out of primes with which to work, ad the the procedure breaks dow completely. I such a case we write Nk, where N\\ ad k is ukow. I light of Theorem 1, we may require (o() >, which imposes a lower boud o o)(/c) ; ad < imposes a upper boud o k ad hece o oj(k). There are also divisibility restrictios o k ad a*(k) from N ad a*(n). See the 5 ad a s cases i the discussio below. The odd (a - 0) case of Theorem is somewhat easier to hadle tha the others sice p b \\ implies p b 1 (mod ) i order to avoid havig too may? s i the deomiator of. We preset here the a 1 (i.e., ) case of Theorem : Immediate size cotradictios result if 1 \ or If b \\ for 6 < b ^ 11. If, the 6l, so either 61 or 6l, i which case l ; both possibilities make > 6. If If, the 1. If 1 or 1, the > 6, so l. The, ad > 6 if s \ or. If 1fe, the 1 < & < 1, ( e, fe) 1 ad o*(k) 18, so k is 5 or 1. Thus we have located two uitary harmoic umbers: H*( 5 1) 15, #*( 1 1) 1. If, the. Size cotradictios easily result if 6 or 5 \\ or k \\ or. If, the 5, ad > 6 if 5 ^ ^. If 5, the. 5 sice < 6, but a)(). Therefore, 5, so 1 ad hece 11» ad aother uitary harmoic umber is foud: #*( 5 1) 18. If, the - fe. It follows that tf*(w) 9H*(k)/. But H*(k) does ot have a eve umerator after reductio, so is ot a iteger. If, the 5. Size cotradictios occur if 5 \ or 5 6 \\ or 5^ ll, while there are too may f s i the deomiator of if 5 5 \\ or 5. Therefore, 5 j or 5 - If 5 j, the 1, ad > 6 if I h \ or 1 j or 1. Thus, 1, so \, but > 6 if 5 ad if there are too may 5 f s i the deomiator of, so. Therefore, w fe, where fe <, ( 5 1, k) 1 ad a*(fc) 0. This locates aother uitary harmoic umber: 198]

7 #*( 5 1) 15. If 5, the 5 k with ( - - 5, k) 1, fc < 11,111 ad (a*(fc), -5) 1, so k is composed of prime powers from the set {, 1, 1,,, 61, 6,, 9,, 11,... }. Sice a)() > 5, co(fc) >. However, ca(fe) < sice 1 1 > 11,111. If co(fc), the the smallest prime dividig k is, sice 1 1 > 11,111. Also, fe or else 19 fc, which is impossible if fc < 11,111. Thus, the oly possibility with b)(k) is k 1 * 1 1, which forces E*(ri) to be oitegral. If 0)(/c), the write 5 p q. Now, p <, sice - 11 > 11,111 ad 0*(q)\l6p 9 so the oly possibility is p ad q 1, ad aother uitary harmoic umber is foud: #*(- 5 1) 1. If, the - k with k < 166,666, (, fc) 1, (a*(fe), ) 1 ad oo(fc) >. But oo(fe) <, sice > 166,666. If co(fe), the smallest possible ext prime power is, sice > 166,666. But if -, the #*()'has at least oe excess i its deomiator. Therefore, (o(fe), so let fe pqr with p < q < r. Now, p < 9, sice > 166,666. We have the followig possibilities: p forces ll. But lljf, so > - 11 > 6. P implies 19. But 19f, so > 19 > 6. p 1 leaves extra f s i the deomiator of. p 5 requires 1, but 1 f. If 1, the > 6, ad the same is true if 1, because the 15 \. The 1, so 1 \ ad 1, so > 5 1^1 > 6. p - 19 forces 5 w, but 5. But > 6 if 5 6 or 5, ad there are extra 's i the deomiator of if 5 or 5. Therefore, 5, so 1 ad 1, but > 6 if 1, ad hece 1, whece 1j ad > 6. [Feb.

8 p 1 requires l\ $ ad l\. If 5 or l h \ or, the > 6. Thus,, so 5. The * * 5 1 k with fe <. The oly value of k that checks out is k 1: #*( * 5 1) 1. p leaves extra? s i the deomiator of. Sice, \ ad the >\ subcase is elimiated. Thus, the l\ of the theorem is proved. case. LARGE INTEGRAL VALUES OF It is ot at all hard to costruct with a large iteger. For example, oe may start with the fifth uitary perfect umber [5], 18 5^ * * , ad have However, substitutig for various blocks of uitary divisors yields the related umber 18 If 5 if lf l *19 1«* , for which ,15. The author cojectures that there are ifiitely may uitary harmoic umbers, icludig ifiitely may odd oes, but that there are oly fiitely may uitary harmoic umbers with a)() fixed. REFERENCES 1. M. Garcia. "O Numbers with Itegral Harmoic Mea." Amevica Math. Mothly 61 (195): Ore. "O the Averages of the Divisors of a Number." Amevica Math. Mothly 55 (198): C. Pomerace. "O a Problem of Ore: Harmoic Numbers." Private commuicatio.. M. V. Subbarao & L. J. Warre. "Uitary Perfect Numbers." Caadia Math. Bull. 9 (1966): C. R. Wall. "The Fifth Uitary Perfect Number." Caadia Math. Bull. 18 (195): ^o#ot 198] 5