I think computing the partial sums is slower than just plain
counting the distinct elements of the table... :-(.
 
----- Original Message -----
From: Michael Kleber
To: math-fun
Sent: Saturday, July 05, 2003 10:46 PM
Subject: Re: [math-fun] Annoyed

> A027424 annoys me.

I agree with Rich that maybe the first differences seem
easier to attack, but this might be deceptive... this is
asking how many of {n,2n,3n,...,n^2} are not ij for
i,j<n, and doing this based on the factorization of n
means recognizing things like "No, 34*16 isn't new
for n=34, because it already appeared as 32*17."
Maybe for each n you can look at its prime factorization
and calculate one list of likely factor swaps to look out
for?  Ugh.

--Michael Kleber
   kleber@brandeis.edu