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 MathNerd Feb19-04 10:38 AM

Curious Inequality

I know that this isn’t very practical but I discovered the following curious inequality when I was playing around with $$d(n)$$ where $$d(n)$$ gives the number of divisors of $$n \ \epsilon \ N$$. If $$n$$ has $$p$$ prime factors (doesn’t have to be distinct prime factors e.g. $$12 = 2^2 \ 3$$ has got three prime factors (2,2,3)), Then

$$p + 1 \leq d(n) \leq \sum_{k=0}^{p} _{p} C_{k}$$

I don’t know if this has been previously discovered but giving its simplicity it wouldn’t surprise me if it has.

 matt grime Feb19-04 10:52 AM

Re: Curious Inequality

Quote:
 Originally posted by MathNerd I know that this isn’t very practical but I discovered the following curious inequality when I was playing around with $$d(n)$$ where $$d(n)$$ gives the number of divisors of $$n \ \epsilon \ N$$. If $$n$$ has $$p$$ prime factors (doesn’t have to be distinct prime factors e.g. $$12 = 2^2 \ 3$$ has got three prime factors (2,2,3)), Then $$d(n) \leq \sum_{k=0}^{p} _{p} C_{k}$$ 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.

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