a curiosity

Mon, 2010-02-15 23:29

Just calculated that
where τ(n) is the number of divisors of n (e.g., τ(6)=4 since 1, 2, 3, and 6 are the divisors of 6).

This is the smallest such number.

Since τ(n)≥2 for n>1,
and τ(n)=2 only if n is prime,
and among four consecutive integers there are at most two primes,
any number with the property that 819000 has must have at least 36 divisors.
In fact, only one integer among four consecutive integers can have exactly three divisors, so (2)(2)(3)(4)=48 divisors are required.

Now, is there an efficient way to generate only numbers which have at least 48 divisors? I suppose looping over the number of prime divisors, then over the exponents in the prime factorization might be reasonable. Complicated little piece of code that would be.