# Why does the number of divisors of a superior highly composite number is always a highly composite number up to 720720 ? (the only exception is 120)

I've calculated the number of divisors of every superior highly composite number up to $10^{27}$:

The number of divisors of a superior highly composite number is always a highly composite number up to 720720.

The only exception is 120 which has 16 divisors.

After 720720, the number of divisors of superior highly composite numbers seem to be numbers with a lot of divisors but which are not highly composite: they have a prime factorization having a lot 2, and a few 3 and 5 and sometimes a 7, but no larger prime number.

Why does this arise?

And why is 120 the only exception?

• Can you please explain what is your definition for highly composite, and why $16$ is not highly composite? – barak manos Jan 5 '15 at 8:21
• I don't understand,WIKi says that it can be shown that all SHC are also highly composite !?? – Gary B Jan 23 '15 at 15:15

I can think of one extreme case, numbers much larger than you are using. Given a real number $$\delta > 0,$$ the exponent of some prime $$p$$ in the superior highly composite number associated with $$\delta$$ is $$k = \left\lfloor \frac{1}{p^\delta - 1} \right\rfloor.$$ For small primes these exponents are roughly proportional to $$1 / \log p,$$ so they decrease fairly quickly at first, in any case they never increase, and no prime is skipped over. These numbers, and all highly composite numbers, are the product of primorials.
For example, taking $$\delta = 0.085$$ gives us the S.H.C. number $$N_{0.085} = 2^{16} \cdot 3^{10} \cdot 5^6 \cdot \mbox{more}.$$ The number of divisors is $$d(N_{0.085}) = 17 \cdot 11 \cdot 7 \cdot \mbox{more}.$$ This number is divisible by $$17$$ but not by $$13,$$ hence it cannot be a highly composite number.
Taking $$\delta = 0.05$$ gives us the S.H.C. number $$N_{0.05} = 2^{28} \cdot 3^{17} \cdot 5^{11} \cdot \mbox{more}.$$ The number of divisors is $$d(N_{0.05}) = 29 \cdot 18 \cdot 12 \cdot \mbox{more}.$$ This number is divisible by $$29$$ but not by any of $$23,19,17,13,$$ hence it cannot be a highly composite number.
We can invert things: given a prime $$p$$ and a target exponent $$k,$$ the largest $$\delta$$ that will assign the exponent $$k$$ to the prime $$p$$ is $$\delta = \frac{\log \left( \frac{k+1}{k} \right)}{\log p}$$
So, you see, it is possible to generate S.H.C. numbers on your own, and prescribe one of the exponents as you see fit. Note that the exact value, which will cause numerical problems, is not necessary; my preference is to take $$\delta$$ a rational number with $$\frac{\log \left( \frac{k+1}{k} \right)}{\log p} > \delta > \frac{\log \left( \frac{k+2}{k+1} \right)}{\log p},$$ which still gives the correct $$k.$$ I wanted the exponent of $$2$$ to be a prime minus $$1,$$ but the exponent of $$3$$ to be smaller than the previous prime minus $$1;$$ you can see how I did that now.
Edit: factoring; the ratio of consecutive S.H.C. numbers is a single, very small, prime. If you have factored one of these, you can factor the next one by taking the ratio, and just correcting the appropriate exponent by $$1.$$ Often, the exponent increased is for a new prime, exponent bumped from $$0$$ to $$1.$$ Anyway, you can keep doing this to factor all of them that anyone has listed, such as http://oeis.org/A002201 and text list here http://oeis.org/A002201/b002201.txt Oh, not by the way, the increased exponent found by ratio tells you what $$\delta$$ can be for the larger S.H.C. number.