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