The number crunch
Jan. 15th, 2010 12:19 am![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
sniffnoySo for the past few days I've been crunching numbers to compute all numbers n with d(n)<k(2-3log_3(2)), for k=10, 11, 12... well, so far I've made it to 13. [From hereon I'm just going to refer to 2-3log_3(2) as α_1.] Each iteration has become more and more tedious and I'm not sure I'm actually going to want to go all the way to 19 or 20 like I planned. I've compiled a giant table, in order, of all possible values (both approximate and exact) of d(n) less than whatever I have so far; what n yield those values; and what D(n) is for those n. I'm still not far enough out to see any "strange" behavior, disappointingly.
The process is as follows: Say we want numbers with defect less than kα_1. Well, we look at pairs (i,j) with i+j=k+1, and look at products of pairs (n, m) with d(n)<iα_1 and d(m)<jα_1. Check the defect to see if it actuallly works. Then, check n+1 for all n with d(n)<(k-1)α_1. (This doesn't work in general, but for k this low, it does.) This is harder than it sounds because of course these sets are not finite, nor are they just finite sets times powers of 3, which is why I haven't quite figured out how to automate it yet. (Well, I may have a way, but it'll almost certainly break at 19 or so. I may want to actually try it.)
Well, the table should allow us to reduce our bounds on Akα(x), because rather than having to rely on our induction past k=1, we can actually determine the correct rate of growth explicitly when k is small enough that we're within the table. Picking α=5-3log_3(5) (which I will in the future refer to as β_2), because that's been so good here in the past, we can get explicitly for k=2. What does this actually get us? I have no idea, because I've been doing basically nothing but computing these sets! I haven't been using them to prove f(2^a3^b)=2a+3b either... just computing them.
Actually, here's something neat I thought of: It's not in general true that f(m3^k)=f(m)+3k. But it is true that for every m, there exists a K such that if we let M=m3^K, then f(M3^k)=f(M)+3k for all k. Proof? For any n≠1, D(3n)≤D(n) with equality iff f(3n)=f(n)+3. (If m=1, well, we already know it's true in that case.) But also for any n, D(n)≥0, and D(n) is always an integer, so the sequence D(m), D(3m), D(9m), ..., must stabilize, yielding the above.
Can you possibly get a bound on K? I have no idea and I'm not going to think about it.
-Harry
The process is as follows: Say we want numbers with defect less than kα_1. Well, we look at pairs (i,j) with i+j=k+1, and look at products of pairs (n, m) with d(n)<iα_1 and d(m)<jα_1. Check the defect to see if it actuallly works. Then, check n+1 for all n with d(n)<(k-1)α_1. (This doesn't work in general, but for k this low, it does.) This is harder than it sounds because of course these sets are not finite, nor are they just finite sets times powers of 3, which is why I haven't quite figured out how to automate it yet. (Well, I may have a way, but it'll almost certainly break at 19 or so. I may want to actually try it.)
Well, the table should allow us to reduce our bounds on Akα(x), because rather than having to rely on our induction past k=1, we can actually determine the correct rate of growth explicitly when k is small enough that we're within the table. Picking α=5-3log_3(5) (which I will in the future refer to as β_2), because that's been so good here in the past, we can get explicitly for k=2. What does this actually get us? I have no idea, because I've been doing basically nothing but computing these sets! I haven't been using them to prove f(2^a3^b)=2a+3b either... just computing them.
Actually, here's something neat I thought of: It's not in general true that f(m3^k)=f(m)+3k. But it is true that for every m, there exists a K such that if we let M=m3^K, then f(M3^k)=f(M)+3k for all k. Proof? For any n≠1, D(3n)≤D(n) with equality iff f(3n)=f(n)+3. (If m=1, well, we already know it's true in that case.) But also for any n, D(n)≥0, and D(n) is always an integer, so the sequence D(m), D(3m), D(9m), ..., must stabilize, yielding the above.
Can you possibly get a bound on K? I have no idea and I'm not going to think about it.
-Harry
