[seqfan] Re: A144925: non-trivial divisors of composite numbers

[Alois Heinz]

franktaw at netscape.net schrieb:
> %p A144925 numcompz(n):=taylor(sum(1/(z^i*(z^i-1)),i,2,n),z, inf,n)$
This is not valid Maple code ...
> But more seriously, it doesn't appear to be correct.  I don't really 
> know either Maxima or Maple, so I don't know for sure what it does 
> there; but what it appears to do is nonsense.  I can't make any sense 
> out of any interpretation of it I can think of, either.
>
> Can anyone make sense of this?  If  not, I'm going to delete it.
from a Maple help page:

 series (expr, eqn, n)

 expr - expression     
 eqn  - equation (such as x = a) or name (such as x)
 n    - (optional) non-negative integer

 The series function computes a truncated series expansion of expr,
 with respect to the variable x, about the point a, up to order n.
 If a is infinity then an asymptotic expansion is given.

Here is Maple code for 110 terms:

s:= series (sum (1/(z^i*(z^i-1)), i=2..80), z=infinity, 150):
a:= n-> numer (op (n, sort (s,z))):
seq (a(n), n=1..110);

output is:

1,2,2,1,2,4,2,2,3,4,4,2,2,6,1,2,2,4,6,4,2,2,2,7,2,2,6,
6,4,4,2,8,1,4,2,4,6,2,6,2,2,10,2,4,5,2,6,4,2,6,10,2,4,
4,2,6,8,3,2,10,2,2,2,6,10,2,4,2,2,2,10,4,4,7,6,6,6,2,
10,6,2,8,6,2,4,4,2,2,14,1,2,2,4,2,10,6,2,6,10,2,2,6,6,
6,10,2,2,2,13,2

the series statement gives:

1/(z^4)+2/z^6+2/z^8+1/(z^9)+2/z^10+ ...
>
> I also wonder about the link to Huen's home page.  I don't see anything 
> relevant there.
I found a paper that has more information:
http://web.singnet.com.sg/~huens/paper13.htm

Alois