User talk:Sbiis Saibian

Welcome
Hi, welcome to ! Thanks for your edit to the User:Sbiis Saibian page.

Please leave a message on my talk page if I can help with anything! -- FB100Z (Talk) 05:14, January 12, 2013

I've finally decided to join the googology wiki. I see that there is an aweful lot of inactive members. Since this is basically the only social hub for large numbers, I figured I might as well join. -- Sbiis Saibian

Oh my! It's E100# himself! I want more clouds! 12:27, January 14, 2013 (UTC)

Hi, Cloudy176. I think this site was a great idea. It's a great place for googologist's to collaborate. It's also a great way to establish what's cannon, and catalog everything. It's difficult to keep track of all the various systems around and how they relate. I can't keep up with all the salad numbers people come up with on a daily basis, but even the more professional stuff is difficult to keep straight. Sbiis Saibian (talk) 15:34, January 14, 2013 (UTC)

Last digits of Mega
Hi, Sbiis Saibian.

I saw your article about Mega on your site and developed a better method than you used:

Firstly, consider \(256[3]_1\). It is \(256^{256} = 2^{2048} = ...30656\) Then \(256[3]_2 = (2^{2048})^{(2^{2048})} = 2^{2048 \times 2^{2048}} = 2^{2048 \times ...30656} = 2^{...83488}\)

Then we need to compute last digits of \(2^{83488}\). For this I suggest to drop \(2^{83488}\) into factors of "powers of powers" of 2: \(2^{83488} = 2^{65536} \times 2^{16384} \times 2^{1024} \times 2^{512} \times 2^{32} \). Find last digits of \(2^{2^n}\) by the following way:

Let stage 1 = 4 stage 2 = 16 stage 3 = 256

stage 4 = 65536

stage 5 = ...67296

stage 6 = ...51616

In general, we can obtain the last digits of "stage x" if we take last digits of "stage x-1" and square it.

Then \(2^{2^n}\) = stage n.

Therefore, we can get last digits of each factor of \(2^{83488}\) fairly easy, and hence, of \(2^{83488}\) (if we multiply them). Specifically, using this, I found that \(256[3]_2 = ...65056\)

Now look at \(256[3]_3 = (2^{...83488})^{(2^{...83488})} = (2^{...83488})^{...65056} = 2^{...83488 \times ...65056} = 2^{...33568}\). You can see the pattern. I showed it only at 5 digits, but as number of digits grows, number of calculations don't grows exponentially, and, if you still interested to find more digits of Mega, you can write the program based on my algorithm.

Ikosarakt1 (talk) 11:22, January 14, 2013 (UTC)

Hi, Ikosarakt1

I noticed your comment on this method on the Mega article talk page. It is an intriguing alternative, and I did actually attempt using it once on paper to obtain ...656 as the last 3 digits of Mega. However I'm not entirely sure this gets completely around the issues I encountered when trying to compute the last 14 digits of Mega. Since this method depends on squares, a table of squares needs to be constructed, and one needs to know how long it takes to return to the starting value. Unless this is known in advance you would actually need to carry out to Stage n, and that could well be impossible if n>E20. While the method works, it would still rely on the problem of memory vs. computation that I encountered. You can reduce the amount of computation by loading up the memory, but that can quickly get used up when we get to around 10-14 digits of accuracy, because of the very long cycle times. Even with my 3GB ram laptop, I wasn't able to get past 12 digits without overloading the memory. One can go the other extreme and compute everything, but every digit seems to increase the computation time about 5 fold, so that within just a few digits you go from minutes, to hours, to days, to weeks, to months of computation. At that point it becomes impractical for me to run the program that long. I would have to break it up into sessions, and the results would be difficult to verify.

None the less, I like your method, I might well try to implement it and see what benefits it offers. It was my intention to get more than 14 digits. I wanted to at least reach 16 digits which was the maximum I could achieve with my programs current configuration. I was originally going to run a program to get the 15th digit, but the way it was set up it was going to take me about a month to compute, broken up into nightly run sessions. I actually ran it for a night or two, but I stopped because I had thought of a better way that would reduce the run time considerably to perhaps a few days. I never got around to implementing it though.

When I get a chance to revisit the topic I'll see if I can get more digits. There might be a method to reduce the exponential run time, but even so it would still be problematic to reach 100 digits because of the limited data-types I have to work with. The largest integer I can use is a 64-bit integer which is still way too small (It would be roughly E19 I think). To get around that limitation I would have to write my own routine, probably in something like C#. I'm sure it can be done but I need to learn more programming first.

Keep Counting

Sincerely,

Sbiis Saibian (talk) 15:23, January 14, 2013 (UTC)

Paper
A while ago I happened across a philosophical/metamathematical paper that may be of interest to you (and other GWikians): Warning Signs of a Possible Collapse of Contemporary Mathematics. It's certainly relevant to googology, and in particular our discussion on Talk:Xi function. FB100Z &bull; talk &bull; contribs 00:06, February 7, 2013 (UTC)

HELP ME!!!
HELP!!! I was exploring your web page, and I fell into an infinite loop! I am stuck on the page [ https://sites.google.com/site/largenumbers/home/a-1/infinite_numbers/infinite_numbers_infinitely], so please come and rescue me! &mdash; I want more clouds! 09:15, February 7, 2013 (UTC)