I know that this isn’t very practical but I discovered the following curious inequality when I was playing around with [tex]d(n)[/tex] where [tex]d(n)[/tex] gives the number of divisors of [tex]n \ \epsilon \ N[/tex]. If [tex]n[/tex] has [tex]p[/tex] prime factors (doesn’t have to be distinct prime factors e.g. [tex]12 = 2^2 \ 3 [/tex] has got three prime factors (2,2,3)), Then [tex] p + 1 \leq d(n) \leq \sum_{k=0}^{p} _{p} C_{k} [/tex] I don’t know if this has been previously discovered but giving its simplicity it wouldn’t surprise me if it has.
the sum you wrote down is just 2^p btw. and isn't that result rather obvious? I mean p distinct primes gives you 2^p divisors, so repeated primes naturally gives you fewer.