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Curious Inequality

by MathNerd
Tags: curious, inequality
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MathNerd
#1
Feb19-04, 10:38 AM
P: n/a
I know that this isnít 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 (doesnít 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 donít know if this has been previously discovered but giving its simplicity it wouldnít surprise me if it has.
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matt grime
#2
Feb19-04, 10:52 AM
Sci Advisor
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P: 9,396
Originally posted by MathNerd
I know that this isnít 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 (doesnít have to be distinct prime factors e.g. [tex]12 = 2^2 \ 3 [/tex] has got three prime factors (2,2,3)), Then

[tex] d(n) \leq \sum_{k=0}^{p} _{p} C_{k} [/tex]

I donít know if this has been previously discovered but giving its simplicity it wouldnít 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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