OK, here's the computation
Jan. 8th, 2010 06:56 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
sniffnoySo: d(n)<10(2-3log_3(2)) iff n is of one of the following forms: 2^a*3^k, a≤9; 5*2^a*3^k, a≤4; 7*2^a*3^k, a≤3; 19*2^a*3^k, a≤2; 13*3^k; 55*3^k; or (3^m+1)3^k.
With this, and our previous stuff (not all of which has appeared here, mind you :P ) it's not too hard to show that f(2^20*3^k)=40+3k.
Why do I get the distinct suspicion that it will, in fact, turn out that A_(k-ε)(x)=Θ((log x)^k) for k an integer, as Josh suggested? Or at least when k=2? And that these won't get too much more complicated until I pass k=2 so I should be able to push up to about 29 or 30 or so?
-Harry
With this, and our previous stuff (not all of which has appeared here, mind you :P ) it's not too hard to show that f(2^20*3^k)=40+3k.
Why do I get the distinct suspicion that it will, in fact, turn out that A_(k-ε)(x)=Θ((log x)^k) for k an integer, as Josh suggested? Or at least when k=2? And that these won't get too much more complicated until I pass k=2 so I should be able to push up to about 29 or 30 or so?
-Harry
