On the indexes of beauty.
Keywords Dirichlet divisor function, index of beauty, elementary method.
[section] 1. Introduction and result
For any positive integer n, the famous Dirichlet divisor function d(n) is defined as the number of all distinct divisors of n. If [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] be the prime power factorization of n, then from the definition and properties of d(n) we can easily get
d(n) = ([[alpha].sub.1] + 1) x ([[alpha].sub.2] + 1) ... ([[alpha].sub.k] + 1). (1)
About the deeply properties of d(n), many people had studied it, and obtained a series results, see references ,  and . In reference , Murthy introduced an index of beauty involving function d(n) as follows: For any positive integer m, if there exists a positive integer n such that
m = n/d(n),
then m is called an index of beauty. At the same time, he also proposed the following conjecture:
Conjecture. Every positive integer is an index of beauty.
Maohua Le  gave a counter-example, and proved that 64 is not an index of beauty. We think that the conclusion in  can be generalization. This paper as a note of , we shall use the elementary method to prove the following general conclusion:
Theorem. There exists a set A including infinite positive integers such that each of n [member of] A is not an index of beauty.
[section] 2. Proof of the theorem
In this section, we shall prove our Theorem directly. For any prime p [greater than or equal to] 5, if we taking m = [p.sup.p-1], then we can prove that m is not an index of beauty. In fact, if m is an index of beauty, then there exists a positive integer n such that n = [p.sup.p-1] x d(n). From this identity we can deduce that [p.sup.p-1] | n. Let n = [p.sup.[alpha]] x b and (p, b) = 1, then from n = [p.sup.p-1] x d(n) and (1) we have the identity
[p.sup.[alpha]-p+1] x b = ([alpha] + 1) x d(b). (2)
From (2) we may immediately deduce that [alpha] [greater than or equal to] p-1. If [alpha] = p - 1, then (2) become b = p x d(b), this contradiction with (b, p) = 1. If [alpha] = p, then (2) become p x b = (p + 1) d(b), or
b = (1 + 1/p) x d(b). (3)
It is clear that (3) is not possible if prime p [greater than or equal to] 5. In fact if b = 1, 2, 3, 4, 5, 6, 7, then it is easily to check that (3) does not hold. If b [greater than or equal to] 8, then from (1) and the properties of d(b) we can deduce that b [greater than or equal to] 3/2 x d(b) > (1 + 1/p) x d(b). So (3) does not also hold. So if [alpha] = p, then (2) is not possible. If [alpha] [greater than or equal to] p + 1, let [alpha] - p = k [greater than or equal to] 1, the (2) become
[p.sup.k+1] x b = (p + k + 1) x d(b). (4)
Note that b [greater than or equal to] d(b) for all positive integer b, so formula (4) is not possible, since [p.sup.k+1] > p + k + 1 for all prime p [greater than or equal to] 5 and k [greater than or equal to] 1. Since there are infinite prime p [greater than or equal to] 5, and each [p.sup.p-1] is not an index of beauty, so there exists a set A including infinite positive integers such that each of n [member of] A is not an index of beauty. This completes the proof of Theorem.
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Department of Applied Mathematics, Xi'an University of Technology, Xi'an, P.R.China
(1) This work is supported by the Shaann Provincial Education Department Foundation 08JK398.
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|Date:||Sep 1, 2008|
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