PROBLEM BELOW (Note, for this prob., you will need to know tau is the numbers of divisors function.
A positive integer n, n > = 1, is said to be highly composite if tau(m) < tau(n) for all pos. integers m < n. The first 6 highly composite integers are 1, 2, 4, 6, 12, 24, 36, 48.
1.) Show if n is highly composite int., and also m is a pos. int. with tau(m) > tau(n), then there will exist a highly composite int. k such that n < k <= m.
HINT: You don't need to use formula of tau(n); only use the fact that you're looking at the sequence of integers: tau(1), tau(2), tau(3), ..., tau(n), ..., tau(m).
2.) From #1, conclude that there exists an infinite number of highly composite integers,
3.) Prove if n is a highly composite int., then n = 2^(a_1)*3^(a_2)*5^(a_3)*...*p_(k)^(a_k), where p_k is the k-th prime and a_1 >= a_2 >= a_3 >= ... >= a_k >= 0.
HINT: prove by contrapositive. Two cases to consider --> 1st case: assume n is missing a prime which says there exists an i where p_i and P_(i + 2) are factors of n, but p_(i + 1) is not a factor of n.
2nd case: assume that the a's are not in non- increasing order, which says that there exists an i such that a_i <= a_(i + 1)
Note the converse is not true.
n = 72 = 2^3*3^2 doesn't skip primes and the exponents are decreasing. Yet, tau(6) = tau(72) = 12, and thus 72 is not highly composite.