a(n)/sigma(n) <= a(n+1)/sigma(n+1)

[Dean Hickerson]

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