a(n)/sigma(n) <= a(n+1)/sigma(n+1)
Edwin Clark
eclark at math.usf.edu
Wed May 28 19:21:38 CEST 2003
>
> 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).
It's not possible to find such a permutation of the natural numbers:
Suppose a(1) = 1 and a(n) = 2 where n>=2. This cannot
be since a(1)/sigma(1) = 1 and a(n)/sigma(n) <= 2/(n+1) <=2/3.
Suppose a(1) = k where k > 1 and let a(n) = 1 where n >=2. This is also
not possible since a(1)/sigma(1) = k and a(n)/sigma(n) <=1/(n+1)<=1/3.
Has this question been raised for other sequences? For which sequences
b(n) is it possible to find a permutation a of the positive integers
satisfying
a(n)/b(n) <= a(n+1)/b(n+1)?
Edwin
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