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Topic: The Versatility of Numbers--Part I
Replies: 1   Last Post: Jun 14, 1996 11:21 AM

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William Schneeberger

Posts: 4
Registered: 12/12/04
Re: The Versatility of Numbers--Part I
Posted: Jun 14, 1996 11:21 AM
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In article <4pktv0$s3h@newsbf02.news.aol.com> wlauritzen@aol.com (WLauritzen) writes:
> 2) very versatile number (dominant antiprime): a
>number that has a greater or equal number of factors to any number
>between zero and double itself. Here are the first very versatiles:
>1 (1), 2 (2), 6 (4), 12 (6), 60 (12), 360 (24), 2520 (48), are there
>more? (In mathematical language, n is very versatile if f(n) >= f(x)
>where f(n) and f(x) are the number of factors of n and x, for all x
>such that 0 < x < 2n.)


No, there are no more.

Suppose you have a number n=2^(n2-1)*3^(n3-1)*5^(n5-1)*7^(n7-1)*... .
Then the number of factors of n is equal to n2*n3*n5*n7*n11*... ; all
but finitely many factors will be 1, so this product converges.

If n is versatile, we clearly must have n2>=n3>=n5>=n7>=... .

If n is very versatile and divisible by 2, then #(3n/2)<=#(n), so
(n2-1)(n3+1)<=n2*n3, so n2<=n3+1. Of course this last inequality
also holds if n is not divisible by 2.

If n is very versatile and divisible by 3, then #(4n/3)<=#(n), so
(n2+2)(n3-1)<=n2*n3, so 2*n3<=n2+2, which also holds if n is not
divisible by 3. From the above inequality, we see 2*n3<=n2+2<=n3+3,
so n3<=3 and n2<=4.

Also if n is very versatile and divisible by 17, then #(32n/17)<=#(n),
so (n2+5)(n17-1)<=n2*n17, so 5*n17<=n2+5<=9. So n17=1 so 17 in fact
does not divide n.

So now we know n2<=4, n3<=3, n5<=3, n7<=3, n11<=3, n13<=3, and for
p>13, np=1. So now it's a finite check.
--
Will Schneeberger
william@math.Princeton.EDU
http://www.math.princeton.edu/~william







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