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Re: The Versatility of NumbersPart I
Posted:
Jun 14, 1996 11:21 AM


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^(n21)*3^(n31)*5^(n51)*7^(n71)*... . 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 (n21)(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)(n31)<=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)(n171)<=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



