Sums-Of-Powers over Minimum Divisors

[Leroy Quet]

I said:  

> Let s(r,m) =
> 
> ---
> \      r
>  >    k
> /
> ---
> k|m
> 1<= k <= sqrt(m)
> 
> (which is, in linear-mode)
> 
> sum{k|m,1<= k<= sqrt(m)}   k^r.
> 
> So, we have s(r,m) is 
> the sum of the r-powers taken over the lower half of the positive
> divisors of m.
> 
> For example, s(1,m) is:
> 
http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A066839
>  
> 
>  If r is > 0 (r = any *positive* real), then:
> 
> 
> limit{m -> oo} 
>            m
>           ---
>   1       \
> -------    >   s(2r,k)       =
> m^(r+1)   /
>           ---
>           k=1
> 
> 
>     1
> ----------    (?)
> 2 r (r+1)
> 
> 
> Linear-mode:
> 
> limit{m->oo} 
>       (1/m^(r+1)) sum{k=1 to m} s(2r,k)  = 
> 
> 1/(2 r (r+1))     (?)
> 
> 
> 
> (I am err-prone today, so I hope I thwarted fate...)
> 
> Example: If I am right, the sum of the first m terms of the EIS's
> A066839 divided by m^(3/2) approaches 2/3.



By the way,

If s(r,m) is such that, for q = integer >= 2,

---
\      r
 >    k
/
---
k|m
1<= k <= m^(1/q)

(which is, in linear-mode)

sum{k|m,1<= k<= m^(1/q)}   k^r ;

then (?):

limit{m -> oo} 
           m
          ---
  1       \
-------    >   s(qr,k)       =
m^(r+1)   /
          ---
          k=1


    1
----------    
q r (r+1)


Linear-mode:

limit{m->oo} 
      (1/m^(r+1)) sum{k=1 to m} s(qr,k)  = 

1/(q r (r+1)) 


(unless I made a mistake)

thanks,
Leroy Quet