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Exploring HCNs (Posted on 2012-10-26) Difficulty: 3 of 5
A highly composite number (HCN) is a positive integer having more divisors than any smaller positive integer (sequence A002182 in OEIS).

Please prove the following:

1. There is an infinite number of highly composite numbers.
2. For any highly composite number (n= p1c1*p2c2* p3c3*...pkck) the k given prime numbers pi must be precisely the first k prime numbers ( i.e. 2, 3, 5,7,...).
3. The sequence of exponents ck must be non-increasing.
4. Only in two special cases (which?) the last exponent ck is greater than 1.
Rem: Although number 1 does not exactly comply with my definition it is considered an HC number.

No Solution Yet Submitted by Ady TZIDON    
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Some Thoughts First three and part of 4. | Comment 1 of 5

1. Since there are an infinite number of primes, there are an infinite number of numbers of factors. At any given number only a finite number of these have been used, so there will always be a next number that will eventually be achieved, and therefore for the first time.

2. If any prime were to be skipped, such as in 2^3 * 5^2 * 7, substitution of 2 and 5 for 5 and 7 in this case, would result in the same number of factors, but a lower number, and therefore an earlier occurrence, so the later occurrence would not be the first.

3. In order to be the lowest number with the given number of factors, the primes used would have to be the lowest available for the given power. For example:

   2^2 * 3^3 * 5^6
  
would have a lower value and still have the same number of factors if changed to:

   2^6 * 3^3 * 5^2
  
as the number of factors depends solely on the c(k), or rather, on the set of c(j){j=1 to k}: (c(1)+1)*(c(2)+1)*(c(3)+1)* ... *(c(k-1)+1)*(c(k)+1). (Subtract 1 or 2 from this if not counting 1 and/or n itself as factors.)

4. This requires some exploration.  The above proofs all depended on showing that if a given condition was not met, then there'd be a lower number with the same number of divisors. But since the definition does not state that the sequence is that of the smallest number with a given number of divisors, but rather the smallest number with at least that number of divisors. So, for example, 576 is the smallest number with exactly 21 divisors, as it is 2^6 * 3^2, while 360, a smaller number, has more divisors: 24, as it is 2^3 * 3^2 * 5.

Examining A002182, it would seem that only the numbers 4 and 36 have c(k)>1, with 4 = 2^2 and 36 = 2^2 * 3^2. I don't have a proof that these are the only two.


  Posted by Charlie on 2012-10-26 12:58:26
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