(08/19/2014, 07:56 PM)sheldonison Wrote: How large a value of "k" have you calculated A(k) for, and how long does it take? I know you were working on a logarithmic algorithm in terms of memory requirements... but I thought the time was still proportional to "k". I didn't follow where you got to after that. Your chart shows 2^240 which is pretty big ... impressive!

Also, I think , but I'm only using k=800,000, along with some questionable accelerator techniques to estimate the function, also assuming that it converges to a particular periodic function which I posted earlier, fairly quickly.

Then I took the inverse of the zerocount function at 16, 15.5, 15, 14.5. 14, 13.5, 13, 12.5 and averaged all of the alpha_1 values with a binomial weighting. I estimated the error term by varying the starting point from 16 to 15.5. Of course, it may be an illusion that this sum converges to alpha_1... I could try more terms though since I'm only using k=800000

I need to take a look at your acceleration trick, because it's remarkably accurate, for only going up to about 2^20 terms. I've taken my calculations out to 2^432 terms so far, and I'm only getting a couple more digits than you are, using simple quadratic interpolation.

Code:

`log_2(k) C_k = A(k)/f(k) a_1(k) = f(k)/A(k) C (weighted avg.) a_1 (weighted avg.)`

309.113942364489 0.923305180556966373 1.08306551404462924

309.616329012193 0.923309674118625365 1.08306024298357084 0.923307427587674182 1.08306287821457317

310.118712300961 0.923305181556479624 1.08306551287216931 0.923307427588546353 1.08306287821355009

310.621091079723 0.923309673122600798 1.08306024415192708 0.923307427587672253 1.08306287821457543

311.123466524098 0.923305182549007791 1.08306551170790309 0.923307427588535283 1.08306287821356308

311.625837484441 0.923309672133524754 1.08306024531213255

...

430.098420027089 0.923309589752691560 1.08306034194646453

430.600126880573 0.923305265669410344 1.08306541420511103 0.923307427587743738 1.08306287821449158

431.101831191477 0.923309589259462704 1.08306034252503169 0.923307427587438262 1.08306287821484991

431.603534017818 0.923305266161417297 1.08306541362797178

The values of log_2(k) and C_k were found by linear regression. From there, in order to determine the approximate value of C, take 3 consecutive extrema (min/max/min or max/min/max), and take a weighted average, giving double weight to the middle value. Doing so, for the values above, gives the values in the last two columns.

Here's a graphical extrapolation, based on a grid of m*2^n, 2049 <= m <= 4096, 298 <= n <= 300, giving a range of about 2^309..2^312:

Here's a graph illustrating how I derived a quadratic fitting curve using linear least squares regression on a sample of 6 points near the apex (or 5 points, if the central point is very close to the apex), in order to get a reasonable estimate of the location of the local minimum/maximum:

Finally, after a few more hours of number crunching, I came up with a grid of points of of m*2^n, 2049 <= m <= 4096, 419 <= n <= 420, giving a range of about 2^430..2^432:

~ Jay Daniel Fox