• 1 Vote(s) - 5 Average
• 1
• 2
• 3
• 4
• 5
 My favorite integer sequence sheldonison Long Time Fellow Posts: 680 Threads: 24 Joined: Oct 2008 08/17/2014, 09:04 PM (This post was last modified: 08/17/2014, 09:08 PM by sheldonison.) (08/17/2014, 08:40 PM)tommy1729 Wrote: (08/17/2014, 05:54 PM)sheldonison Wrote: Then $\rho$ has zeros where f(n) crisscrosses the A(n) partition function. Here is a possible estimate of the zero count of rho: $\text{zerocount}(\rho(\ln(x))) \approx 2 \frac{d}{dx}\ln(f(\exp(x)))$ $\text{zerocount}(f(-\ln(x))) \approx \frac{d}{dx}\ln(f(\exp(x)))$ Uh ? can't we use d/dx ln(g(x)) = g ' (x) / g(x) ? And further d/dx ln(g(exp(x)) = g ' (exp(x)) exp(x) / g(exp(x)) then since your LHS includes a ln I think we can substitute ln(x) = y or y = exp(x) and get zerocount(rho(x)) ~~ 2 f ' (x) / f(x) and zerocount(rho(-x)) ~~ f ' (x) / f(x) and similar for the half-iterate of exp. Im a bit tired so forgive if my calculus is bad today. regards tommy1729 Hey Tommy, I think your calculus is correct! Very nice. That simplifies the heck out of the equations .... by the way, that should be h(n) from the other post... kinda busy today and tomorrow morning though. - Sheldon jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 08/19/2014, 01:36 AM (08/11/2014, 05:19 PM)jaydfox Wrote: (08/11/2014, 04:51 PM)sheldonison Wrote: Nice post. What is the most accurate version of the $\alpha_1$ scaling constant that you know of? Earlier, you posted, "1.083063". I'm currently working on a version that would allow me to calculate all the terms of the form [(2^(n-1)+1)...(2^n)]*(2^m), for a given small n and arbitrary m, which should run in about O(2^n m^5) time. For example, for n=3, I could calculate 5..8 * 2^m, so let's say: 5, 6, 7, 8, 10, 12, 14, 16, 20, 24, 28, 32, 40, 48, 56, 64, ... 5*2^m, 6*2^m, 7*2^m, 8*2^m This would allow me to more accurately estimate a_1, because I would be able to monitor the peaks and troughs of the oscillations in the relative error. So hopefully, within a week or two, I should have an estimate of a_1 to at least 1-2 more digits than the current 1.083063. Who knows, I might soon have the most accurate estimate? Short of a closed form to calculate it, the only way I know to determine its value accurately is calculate very large terms in the sequence, isolate a local minimum and local maximum, and estimate the midpoint. I'm getting closer to finding a better approximation of a_1. So far, I think it's slightly less than 1.083063, perhaps in the neighborhood of 1.0830628-1.0830629. Time for more pictures. I finished the code I mentioned above. I calculated A(k) for k = m*2^n, where 513 <= m <= 1024, and n up to 150 (so the largest value was 1024 * 2^150, or 2^160). I have a calculation running in the background to get me up to 2^260, so I'll have new pictures tomorrow. Note that earlier, I defined a_1 as the limit of f(k)/A(k). However, to me it is more useful to graph A(k)/f(k), because this helps me remember that A(k) over estimates for even k, and underestimates for odd k. So the graph will be a graph of 1/a_1 in the limit as k goes to infinity. So I made a secondary axis on the right side of the graphs, to show the value of a_1(k). I'll start by graphing A(k)/f(k) versus k:     Then I have a few graphs versus log_2(k).             Finally, I have a graph versus log_2(log_2(k)), which I think helps show how quickly (or not) the series converges on f(x).     Stay tuned for more exciting pictures tomorrow (or sometime this week). Also, I'll try to post an updated version of the code later this week, which has a minor bug fix, as well as some shortcuts which cut the running time by about 40% for calculating the polynomials. ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 08/19/2014, 02:05 AM (08/19/2014, 01:36 AM)jaydfox Wrote: Stay tuned for more exciting pictures tomorrow (or sometime this week). Also, I'll try to post an updated version of the code later this week, which has a minor bug fix, as well as some shortcuts which cut the running time by about 40% for calculating the polynomials. The background job crashed, due to exceeding the default 4M stack in PARI/gp. So here's the output up to the point where it crashed (it had finished up to 1024*2^241).     And here's an extended version graphed versus log log k:     ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 08/19/2014, 05:31 PM (08/11/2014, 05:19 PM)jaydfox Wrote: (08/11/2014, 04:51 PM)sheldonison Wrote: Nice post. What is the most accurate version of the $\alpha_1$ scaling constant that you know of? Earlier, you posted, "1.083063". I'm not having much luck finding references to the constant online. The closest two I've seen are the 1.083063 that I found earlier through a google search (post #3 in the current discussion), and a reference to 1/a_1 of 0.9233, on this page: https://oeis.org/A002577 The latter has 4 sig-figs, though it's accurate to almost 5 (1/1.083063 ~= 0.923307)I just found another esimate for a_1 (well, 1/a_1). Check this out: http://www.sciencedirect.com/science/art...6503001237 Quote:Results on the number of binary partitions, Bn: In [8], Fröberg proves the following: Define $ F(n):= \sum_{k=0}^{\infty}\, \frac{n^k}{2^{\frac{k(k+1)}{2}}k!}$ Then Bn=Cn·F(n), where (Cn) is a sequence bounded between 0.63722

 Possibly Related Threads... Thread Author Replies Views Last Post My favorite theorem tommy1729 0 3,905 08/15/2015, 09:58 PM Last Post: tommy1729

Users browsing this thread: 1 Guest(s)