08/11/2014, 05:19 PM

(08/11/2014, 04:51 PM)sheldonison Wrote: Nice post. What is the most accurate version of the 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 found that 0.9233 by accident, searching the OEIS to see if they had an integer sequence for the 2^m terms of A(n). Interestingly, I see some of my ideas listed on the page for that integer sequence, so I suppose that means I'm on the right track? I'm just a few decades late to the party.

The code I posted earlier will generate arbitrary terms in the sequence with indexes of 2^m, in approximately O(m^5) time.

I.e., I can generate terms A(1), A(2), A(4), A(8 ), ..., A(2^m)

The code can be modified to calculate an arbitrary term, e.g., A(12345678987665321385) or whatever I'd like, with only a modest increase in running time. However, it becomes far less flexible (it's a one-shot deal, all that effort to calculate a single term), so I haven't implemented it yet in code. (I've used it in Excel for modestly-sized terms, so I know it works.)

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.

BTW, the locations of the local maxima and minima follow a recurrence similar to locations of the zeroes. I don't have the numbers in front of me to post, so I will try to find which spreadsheet I buried them in and post the details later. I'll be working at a factory location today, so it'll probably be at least a day or two before I can update the group.

~ Jay Daniel Fox