a(n)/sigma(n) <= a(n+1)/sigma(n+1)
Dean Hickerson
dean at math.ucdavis.edu
Thu May 29 01:04:36 CEST 2003
Amarnath Murthy asked:
> Does the following sequence make sence?
> rearrangement of natural numbers in an ascending order such that
> a(n)/sigma(n) <= a(n+1)/sigma(n+1).
>
> sigma(6) = 12, sigma(12)= 28.
> 12/28 < 6/12
> hence 12 comes before 6 in the sequence
Edwin Clark asked the question as asked, but it appears from the example
that the definition should have been:
rearrangement of positive integers such that
a(n)/sigma(a(n)) <= a(n+1)/sigma(a(n+1)).
But again there's no such sequence. If a(n)=1 for some n>=1, then a(n+1)>1
so sigma(a(n+1)) > a(n+1). Hence
1 = a(n)/sigma(a(n)) <= a(n+1)/sigma(a(n+1)) < 1.
Dean Hickerson
dean at math.ucdavis.edu
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