Comments on: EDP9 — a change of focus http://gowers.wordpress.com/2010/02/24/edp9-a-change-of-focus/ Mathematics related discussions Wed, 15 May 2013 08:08:31 +0000 hourly 1 http://wordpress.com/ By: Klas Markström http://gowers.wordpress.com/2010/02/24/edp9-a-change-of-focus/#comment-6524 Sat, 06 Mar 2010 08:46:01 +0000 http://gowers.wordpress.com/?p=1542#comment-6524 That’s an interesting sequence. I would not have expected the different p:s to give such different lengths.

However the fact that one can find sequences as long a the ones you have found for some primes, suggest that any proof showing that the maximum length is finite for any p will have to be rather indirect, not giving a good bound for the maximum length.

]]> By: Alec Edgington http://gowers.wordpress.com/2010/02/24/edp9-a-change-of-focus/#comment-6515 Fri, 05 Mar 2010 22:20:01 +0000 http://gowers.wordpress.com/?p=1542#comment-6515 I’ve now gone a bit further with this, and found:

min p flipped  max n (n+1 prime)
5              886
7              946
11             330
13             >= 1380
17             408
19             >= 2646
23             22
29             >= 546
31             >= 9906
37             >= 27508
41             >= 690
43             >= 24420
47             46
53             52
59             >= 1326
61             >= 15072

It seems that flipping a prime of the form 3k+1 is much better than flipping a prime of the form 3k-1. (Perhaps this isn’t too surprising, as it means we don’t immediately require f(3k+4) to change sign to keep the discrepancy at 2, but I’m not sure that fully explains it.)

]]> By: Alec Edgington http://gowers.wordpress.com/2010/02/24/edp9-a-change-of-focus/#comment-6508 Fri, 05 Mar 2010 18:55:35 +0000 http://gowers.wordpress.com/?p=1542#comment-6508 Well, I’ll have to eat my words. It finished almost immediately, having got only as far as 330:

+-0+-0+-0++0--0+-0+-0--0++0+-0+-0+-0+-0+-0-+0+-0+-0-+0--0+-0+-0++0+-0+-0--0++0+-0
+-0++0--0--0+-0+-0+-0++0-+0-+0+-0+-0+-0+-0+-0+-0--0+-0+-0+-0+-0++0+-0+-0--0+-0+-0
+-0++0+-0--0++0+-0-+0+-0--0+-0+-0++0+-0+-0+-0-+0-+0-+0++0-+0--0+-0+-0+-0+-0+-0+-0
+-0--0++0--0+-0++0+-0--0+-0+-0++0+-0+-0+-0+-0+-0--0+-0++0--0+-0++0+-0+-0++0--0+-0
--0+-0

(Primes sent to -1: 2, 5, 13, 17, 23, 29, 41, 43, 47, 59, 71, 73, 83, 89, 101, 109, 113, 131, 137, 149, 173, 179, 181, 191, 211, 223, 227, 233, 239, 257, 263, 269, 281, 293, 311.)

I had assumed that the maximum length would be a monotonic function of the first prime flipped, but evidently not…

]]> By: Alec Edgington http://gowers.wordpress.com/2010/02/24/edp9-a-change-of-focus/#comment-6507 Fri, 05 Mar 2010 18:49:17 +0000 http://gowers.wordpress.com/?p=1542#comment-6507 In this case (flipping the value at 7) the maximal length of sequence (or rather, the maximal length that is of the form p-1 — which is what I should have said above too) is 946:

+-0+-0--0++0-+0+-0+-0-+0++0--0+-0++0--0+-0-+0-+0+-0-+0-+0+-0+-0++0--0--0++0+-0--0
+-0++0--0++0--0+-0++0++0-+0-+0--0--0++0+-0+-0+-0-+0-+0+-0-+0+-0+-0-+0+-0+-0-+0+-0
+-0+-0+-0-+0-+0+-0+-0-+0-+0+-0+-0+-0+-0-+0+-0-+0-+0-+0-+0-+0-+0+-0-+0++0--0-+0+-0
+-0--0+-0+-0+-0++0-+0-+0--0+-0++0-+0--0+-0++0+-0+-0++0--0+-0+-0+-0+-0++0-+0--0+-0
--0+-0+-0++0+-0++0-+0-+0-+0--0+-0+-0+-0+-0++0--0++0-+0--0+-0+-0+-0--0++0+-0+-0-+0
--0+-0++0++0-+0-+0--0-+0--0++0-+0++0--0++0-+0--0+-0-+0-+0+-0--0++0+-0++0--0+-0+-0
--0+-0++0++0--0++0--0++0--0+-0+-0--0++0--0++0-+0-+0++0--0++0--0--0++0--0++0+-0+-0
+-0+-0-+0--0++0+-0+-0+-0-+0-+0+-0+-0+-0+-0+-0+-0+-0--0++0+-0-+0++0--0-+0+-0+-0+-0
-+0+-0--0++0+-0++0--0-+0+-0++0--0--0++0--0++0-+0++0--0-+0+-0--0++0+-0+-0--0+-0+-0
++0+-0--0++0++0--0--0++0--0++0+-0+-0++0--0+-0+-0+-0-+0++0--0-+0--0++0--0++0++0--0
-+0+-0--0++0+-0-+0+-0--0++0-+0++0--0-+0+-0++0--0-+0++0--0--0++0-+0+-0--0++0++0--0
 -+0++0--0-+0+-0+-0+-0-+0--0++0-+0--0+-0++0++0--0-+0--0+

(Primes sent to -1 in this example: 2, 5, 7, 13, 17, 29, 37, 41, 43, 59, 67, 71, 79, 83, 89, 109, 113, 137, 139, 157, 167, 179, 191, 197, 211, 223, 227, 229, 251, 257, 269, 277, 281, 293, 311, 349, 353, 359, 379, 383, 389, 401, 421, 431, 443, 457, 467, 479, 487, 491, 499, 521, 541, 547, 557, 563, 569, 577, 587, 599, 617, 619, 641, 647, 653, 683, 701, 709, 719, 761, 769, 773, 787, 809, 811, 857, 859, 881, 911, 929, 937.)

I’ll kick off a search with 11 flipped, but won’t hold my breath for it to finish!

]]> By: Alec Edgington http://gowers.wordpress.com/2010/02/24/edp9-a-change-of-focus/#comment-6422 Wed, 03 Mar 2010 08:54:46 +0000 http://gowers.wordpress.com/?p=1542#comment-6422 OK, I’ll try that when I get back to my computer this evening.

]]> By: Klas Markström http://gowers.wordpress.com/2010/02/24/edp9-a-change-of-focus/#comment-6417 Wed, 03 Mar 2010 08:03:30 +0000 http://gowers.wordpress.com/?p=1542#comment-6417 Alec, could you try that with the value at 7 flipped instead? It would be interesting to see how much further one gets by following the discrepancy 1 function out to different primes.

]]> By: Alec Edgington http://gowers.wordpress.com/2010/02/24/edp9-a-change-of-focus/#comment-6413 Tue, 02 Mar 2010 21:37:03 +0000 http://gowers.wordpress.com/?p=1542#comment-6413 That search finally terminated — so that is the longest such sequence we can get that sends 5 to +1.

]]> By: Alec Edgington http://gowers.wordpress.com/2010/02/24/edp9-a-change-of-focus/#comment-6411 Tue, 02 Mar 2010 20:16:24 +0000 http://gowers.wordpress.com/?p=1542#comment-6411 Here’s a plot of f(n)/n for 1 \leq n \leq 10\,000:
http://www.obtext.com/erdos/diagonals_gradient.png

]]> By: gowers http://gowers.wordpress.com/2010/02/24/edp9-a-change-of-focus/#comment-6408 Tue, 02 Mar 2010 16:45:23 +0000 http://gowers.wordpress.com/?p=1542#comment-6408 At some point I may ask whether somebody can write a piece of code to work out this binary operation that I am working out laboriously by hand. But for now, here are a few more values, which perhaps give us tiny further clues.

7*17=49, at which point I think it is a pretty safe conjecture that if p*q=2(p+q)+1, then p*r=2(p+r)+1 for all r\geq q.

11*19=67, so we’re not there yet with 11.

11*23=69=2(11+23)+1 so now we have got there.

It looks as though a necessary and sufficient condition for p*q=2(p+q)+1 is that q\geq 2p. What’s more, it looks as though p*q is constant before that. So the right formula should involve maxes and things.

OK, here is my revised formula: p*q=\max\{6p+1,2(p+q)+1\}. Phew!

To test the earlier conjecture, let me try 7*19. It works out to be 53, which is indeed what it should be.

And to test the new conjecture, let me try 13*19. The prediction is 79. What I actually get is … 79. OK, I think I believe this formula now, but it will need quite a bit more slog to get a formula for f(p^aq^br^c)/p^aq^br^c from here. I’ve spent several hours on this: the calculations are certainly routine enough to do on a computer, so I wonder if anyone could take over at this point. For instance, it would be good to give a formula for f(3^kpq). Does it change behaviour according to whether q<3p or q>3p?

]]> By: gowers http://gowers.wordpress.com/2010/02/24/edp9-a-change-of-focus/#comment-6407 Tue, 02 Mar 2010 16:21:36 +0000 http://gowers.wordpress.com/?p=1542#comment-6407 Correction: the formula does not give the right value for 7*13 either, so my understanding is worse than I thought. However, in a way that might be good news, since I found the idea of a formula that fails for consecutive primes a bit bizarre.

]]> By: gowers http://gowers.wordpress.com/2010/02/24/edp9-a-change-of-focus/#comment-6406 Tue, 02 Mar 2010 16:16:40 +0000 http://gowers.wordpress.com/?p=1542#comment-6406 After further investigation, I no longer think the value at 30 is anomalous, but I have discovered (with much more effort than I thought would be needed, because the pattern is clearly subtler than I thought it would be) the following formula for f(2^kpq)/2^kpq. We know from earlier that f(pq)/pq=1-1/p-1/2pq, so that gives us the value when k=0. If I call that value a_{pq}, then the formula, which is linear in k as we expect, is

\displaystyle a_{pq}+(1+\frac{p*q}{4pq})k,

where * is a binary operation that I do not fully understand. Let me tabulate the values that I have calculated so far:

3*5=19, 3*7=21, 3*11=29,3*13=33,

5*7=31,5*11=33,5*13=37,

7*11=43, 7*13=43,

11*13=67.

I had to calculate a lot of values before I realized that there is a nice formula for this binary operation except if p and q are consecutive primes. The formula is simply p*q=2(p+q)+1. But when p and q are consecutive we seem to get a little “kick” and the value is larger. In fact, by the time we get to 11*13 the kick is not all that little.

Anyhow, this observation partially explains the mysterious anomaly at 30, but really it replaces it by a bigger mystery: why should the values be sensitive to consecutiveness of primes?

Of course, at this stage I haven’t looked at all that many numbers, so I hope that I’ll be able to ask a more precise question in due course. So far, however, I don’t feel all that close to a formula for f(p^aq^br^c) …

]]> By: Moses Charikar http://gowers.wordpress.com/2010/02/24/edp9-a-change-of-focus/#comment-6405 Tue, 02 Mar 2010 15:55:19 +0000 http://gowers.wordpress.com/?p=1542#comment-6405 I looked over the code and as far as I can tell, it is correct. I now understand why the “bug” above did not really change anything.

]]> By: Moses Charikar http://gowers.wordpress.com/2010/02/24/edp9-a-change-of-focus/#comment-6404 Tue, 02 Mar 2010 15:36:21 +0000 http://gowers.wordpress.com/?p=1542#comment-6404 Uh oh … I spotted a tiny problem in the code which could potentially affect the calculation for rows and columns greater than n/2. Except that for some mysterious reason it doesn’t. i.e. the output is exactly the same after fixing the “bug”. Basically, I was not setting the constant term in the Taylor expansion correctly for diagonal entries > n/2. It should have been 1 and was being set to 0 instead. This could potentially affect the calculation of the coefficient of \alpha for all entries in columns > n/2, but apparently it does not. I’m checking to see if there are any other issues.

]]> By: Moses Charikar http://gowers.wordpress.com/2010/02/24/edp9-a-change-of-focus/#comment-6403 Tue, 02 Mar 2010 15:08:16 +0000 http://gowers.wordpress.com/?p=1542#comment-6403 I’ve added an even bigger file with diagonal entries upto 10,000 in this directory in case you are looking for more data points to verify guesses for formulae. The directory also contains the C code (cholesky.c) used to generate the lower triangular matrix of coefficients and another piece of code (cholesky2.c) to generate the diagonal entries only. I really hope the code is not buggy !

]]> By: gowers http://gowers.wordpress.com/2010/02/24/edp9-a-change-of-focus/#comment-6402 Tue, 02 Mar 2010 15:04:47 +0000 http://gowers.wordpress.com/?p=1542#comment-6402 Oh dear, for the first time I have run up against an anomaly in the data that may be hard to fit into a nice pattern. When p is any of 7,11,13,17 or 19 the value of f(6p)/6p is given by the formula 2-(2p-5)/12p, but when p=5 we get instead 2-(2p-7)/12p. But I’ll continue with this and try not to worry about f(30) too much for now. (But I would love to learn that it was wrong and should have been 57.5 instead of 58.5.)

]]> By: gowers http://gowers.wordpress.com/2010/02/24/edp9-a-change-of-focus/#comment-6401 Tue, 02 Mar 2010 14:52:51 +0000 http://gowers.wordpress.com/?p=1542#comment-6401 I’ve just wasted some time going about this a stupid way, which was to calculate f(pqr)/pqr for several values and try to spot a pattern. But I now see that that was unlikely to be easy, since the dependence on 1/p, 1/q and 1/r is trilinear. The values look quite strange too, but I have a new plan. That plan is to try to find a formula for f(6p^k)/6p^k when p is a prime greater than 3. That should give me some coefficients of a linear function in k. If I repeat the exercise for one or two other small products of two primes, then it should be easyish to spot a formula for the coefficients, and then I’ll have a formula for f(pqr^k), which will be a good start. In fact, it will be more than just a start, as from that and the multilinearity assumption I’ll be able to work out all the rest of the values at products of three prime powers.

]]> By: gowers http://gowers.wordpress.com/2010/02/24/edp9-a-change-of-focus/#comment-6400 Tue, 02 Mar 2010 14:05:32 +0000 http://gowers.wordpress.com/?p=1542#comment-6400 Just before I dive into the calculations, let me make the remark that the signs are that the formula will be an inhomogeneous multilinear one in the indices. More precisely, I am expecting a formula of the following kind for f(n)/n when n=p_1^{a_1}\dots p_r^{a_r} and the primes p_1,\dots,p_r are in increasing order:

\frac 12\sum_{A\subset\{1,\dots,k\}}\prod_{j\in A}(1+\frac{c_{A,j}}{p_j})a_j-1.

]]> By: gowers http://gowers.wordpress.com/2010/02/24/edp9-a-change-of-focus/#comment-6399 Tue, 02 Mar 2010 13:04:53 +0000 http://gowers.wordpress.com/?p=1542#comment-6399 Here’s a slightly more transparent way of writing the formula for f(p^aq^b)/p^aq^b:

\frac 12(1-\frac 1p)(1+\frac 1q)ab+\frac 12(1-\frac 1p)a+\frac 12(1-\frac 1q)b-\frac 12.

]]> By: gowers http://gowers.wordpress.com/2010/02/24/edp9-a-change-of-focus/#comment-6398 Tue, 02 Mar 2010 13:01:47 +0000 http://gowers.wordpress.com/?p=1542#comment-6398 I got the following formula for f(p^aq^b)/p^aq^b:

\frac 12(1-\frac 1p)a((1+\frac 1q)b+1)+\frac 12((1-\frac 1q)b-1).

I have tried it out on 225=3^2.5^2 and got the answer f(225)=577.5, which was correct. So I believe this formula too.

Just to repeat what I am doing here, I am assuming that the function f(n)/n takes geometric progressions to arithmetic progressions and deducing what I can whenever I know what two values are. In this way, knowing f for prime powers and for products of two primes was enough to give me f for all numbers of the form p^aq^b. Sorry — correction — I have assumed that only for GPs with prime common ratio. I haven’t looked at what happens when the common ratio is composite.

Come to think of it, if f(n)/n takes GPs with prime common ratio to APs, then \exp(f(n)/n) takes GPs with prime common ratio to GPs, which is a pretty strong multiplicativity property. That leads me to think that the eventual formula is going to be quite nice. I hope I’ll be able to guess it after working out f(pqr)/pqr, or perhaps I’ll have to do f(p^aq^br^c)/p^aq^br^c. But first I need some lunch.

]]> By: Alec Edgington http://gowers.wordpress.com/2010/02/24/edp9-a-change-of-focus/#comment-6397 Tue, 02 Mar 2010 12:50:46 +0000 http://gowers.wordpress.com/?p=1542#comment-6397 All very interesting. Another thing to note about the \beta(i,j) is that the fractional part is \frac{1}{2} precisely when j \mid i and 4 \nmid j.

]]> By: gowers http://gowers.wordpress.com/2010/02/24/edp9-a-change-of-focus/#comment-6395 Tue, 02 Mar 2010 12:36:51 +0000 http://gowers.wordpress.com/?p=1542#comment-6395 For n=p^kq with p<q I get

f(n)=kp^{k-1}(p-1)(q+1/2)-p^k/2

or equivalently

f(n)/n=k(1-1/p)(1+1/2q)-1/2q.

I’ll add to this comment when I get a formula for p^aq^b, which should be easy now.

]]> By: Moses Charikar http://gowers.wordpress.com/2010/02/24/edp9-a-change-of-focus/#comment-6394 Tue, 02 Mar 2010 12:15:07 +0000 http://gowers.wordpress.com/?p=1542#comment-6394 Ok, I think we should have the diagonal values figured out soon, although understanding why these numbers arise will take more effort. Looking ahead, the hope was that large numbers are an indication that large b_i can be subtracted out from those entries without sacrificing positive semidefiniteness. At this point, I don’t have a clue about how these b_i should be picked. We can experiment with some guesses later.

Question for later:
The hope is that for any C, for suitably small \alpha, we can pick b_i such that \sum b_i > C/\alpha. At this point, I haven’t thought through why this should necessarily be the case. Is there a heuristic argument that we have enough mass on the diagonal to pull this through ?

I am going to have to sign off for a while and get back to this later in the day.

]]> By: gowers http://gowers.wordpress.com/2010/02/24/edp9-a-change-of-focus/#comment-6393 Tue, 02 Mar 2010 12:03:06 +0000 http://gowers.wordpress.com/?p=1542#comment-6393 For primes p,q with p<q the formula appears to be

f(pq)=q(p-1)-1/2. I inducted that from a few small instances, and then checked it on 407=11\times 37. My formula predicts 370-1/2=369.5 and that is exactly what I get.

In keeping with the general idea of looking at f(n)/n, let me also write this formula as f(pq)/pq=1-1/p-1/2pq.

A small calculation, which I shall now do, will allow me to deduce the formula for p^aq^b.

]]> By: gowers http://gowers.wordpress.com/2010/02/24/edp9-a-change-of-focus/#comment-6392 Tue, 02 Mar 2010 11:50:56 +0000 http://gowers.wordpress.com/?p=1542#comment-6392 It’s taken me longer than it should have, but I now see that we ought to be looking not at f(n) but at f(n)/n. Somehow, this is where the “real pattern” is.

When n=p^k we get f(n)/n=(1/2)((1-1/p)k-1), and when n=2^kp we get f(n)/n=(1/4)((2+1/p)k-2/p)=(1/2)(1+1/2p)k-1/p).

I think an abstract question is emerging. Suppose we know that, as seems to be the case, for any a and p we have a formula of the kind f(ap^k)=p^k(uk+v). With enough initial conditions, that should determine f. We already have a lot of initial conditions, since we know that f(1)=f(p)=-1/2 for every prime p. Those conditions are enough to determine f(p^k), but they don’t seem to be sufficient to do everything. For that I appear to need to know f at all square-free numbers, or something like that.

]]> By: gowers http://gowers.wordpress.com/2010/02/24/edp9-a-change-of-focus/#comment-6391 Tue, 02 Mar 2010 11:39:44 +0000 http://gowers.wordpress.com/?p=1542#comment-6391 Sorry — had to do something else.

But I’ve now checked the GP 3,6,12,… and we get the formula f(3.2^k)=2^{k-2}(7k-2).

Next up, the GP 5,10,20,40,… which gives the formula f(5.2^k)=2^{k-2}(11k-2).

So I’d hazard a guess that f(p.2^k)=2^{k-2}((2p+1)k-2). Let me do a random check by looking at f(88). If my guess is correct, this should be 2^{3-2}\times 23=46 and in fact it is … 134. Let me try that again. It should be 2^{3-2}(23\times 3-2)=2\times 67=134. So I believe that formula now.

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