On Ore s harmonic numbers. (Tokyo University of Science, Japan)
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1 On Ore s harmonic numbers (Tokyo University of Science, Japan)
2 / Contents Definition of Ore s harmonic numbers Known facts Main result Application Demonstration of the computation
3 3/ Notation n: a positive integer The divisor function the sum of divisors of n the number of divisors of n is multiplicative, i.e.
4 4/ Notation H(n) := harmonic mean of divisors of n. Example. The function H is also multiplicative since
5 Definition of (Ore s) 5/ Harmonic Numbers A positive integer n is said to be harmonic if H(n) is integral. 1 is called a trivial harmonic number. 6 is the smallest nontrivial harmonic number. (Remark) The following rational number is also called (n-th) harmonic number.
6 6/ Ore s Theorem Ore proved that every perfect number is harmonic. Proof. Let n be a perfect number. Then It is easy to show that this is integral for a perfect number n.
7 7/ Ore s Conjecture Ore s conjecture: Every nontrivial harmonic number is even. (??) if true There does not exist an odd perfect number!
8 8/ Table of Harmonic numbers
9 9/ Known Facts Problem Which values does the harmonic mean take? Are there integers n satisfying H(n)=4,1,16,18,0,,? Theorem (Kanold, 1957) For any positive integer c, there exist only finitely many numbers n satisfying H(n)=c.
10 10/ Known Facts In 1997, Cohen listed all harmonic numbers n satisfying H(n) 13. There does not exist a harmonic number n with H(n)=4 or 1.
11 11/ Main Result An algorithm to find all harmonic numbers n satisfying H(n)=c for a given integer c. Use of a personal computer The list of all harmonic numbers n satisfying H(n) 100 (there are 1376 such numbers). Particular result If n is a nontrivial odd harmonic number, then H(n)>100.
12 1/ Nonexistence of Odd One There are no odd perfect numbers less than 10^300. (Brent, Cohen & te Riele, 1991) There are no odd nontrivial harmonic numbers less than 10^1. (Cohen & Sorli, 1998) There are exactly 435 harmonic numbers less than 10^1. (Sorli, Ph.D thesis)
13 13/ Application of the Main Result (my method) List all harmonic numbers n satisfying H(n) < List all harmonic numbers n satisfying H(n)^4 > n. The union of these lists contains all harmonic numbers less than 10^1, because n<10^1 H(n)<1000 or H(n)^4>n. By and 4 4.5, we obtain the fact: there are exactly 633 harmonic numbers less than 10^13.
14 14/ New Record of the nonexistence There are no nontrivial odd harmonic numbers less than 10^13. There are no nontrivial odd harmonic numbers less than 10^15. (Sorli, Ph.D thesis)
15 15/ Another Application (n):= the total number of primes dividing n If n is an odd perfect number, then (n) 37. (Iannucci and Sorli, 003) If n is a nontrivial odd harmonic number, then (n) 11. Proof.
16 16/ About Distinct Prime Factors (n):= the number of distinct primes dividing n If n is an odd perfect number, then (n) 8. (Chein, 1979, Hagis, 1980) If n is a nontrivial odd harmonic number, then (n) 3. (Pomerance, 1973)
17 17/ Open Problem Problem. If n=p^a q^b r^c is harmonic, is it even? Maybe, this problem can be solved by using the theory of cyclotomic polynomials. In the case of perfect numbers, it is known that there exist only finitely many odd perfect numbers n with a fixed (n) (in fact, n<^(4^ (n))), but I know no analogous fact in the case of harmonic numbers (or seeds).
18 18/ Harmonic Seeds more harmonic numbers H = 9 H = 37 H = 49 9 H = 9 /15 37 H = 37 / 19 7 H = 49 / H = H = 19 3 H = 3/ 19 H =19 / seed H = 10 H = 10
19 19/ Harmonic Seeds It was conjectured that the harmonic seed of a harmonic number is unique, but we found the following counterexample H = 55 5 H = 75/ 31 3 H = 3/ seed 5 7 H = 17 H = seed
20 0/ Basic Algorithm Let be the factorization of n. We call the pairs of exponents of n. Step 1. List possible pairs of exponents. For example, if H(n)=5, possible triple pairs are (1,1,1) and (,1,1). proof. If n has the pair (3,1,1), then a contradiction. Similarly, (,,1) is an impossible pair.
21 1/ Basic Algorithm Step. List possible prime factors. For example, suppose that H(n)=5, the pair of exponents of n is (,1,1), p is the smallest prime factor of n. Then On the other hand, Hence
22 / Improved Algorithm Proposition Let c be the numerator of H(n). In Step, we can cut many possibilities. We used UBASIC program which is useful to factor integers.
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