User blog:Allam948736/Finding last digits of Triton

Recently, I began computing last digits of the great mega (or triton, whichever you prefer), which is equal to 3 in a pentagon in Steinhaus-Moser notation. I had already known that the last 2 digits are ...27, since the number of times the triangle operator is applied to 3 is odd.

To find the last 3 digits, I simply observed that the number of triangles is of the form 3 + ...7 + ...3 = ...3. This means that the last 3 digits of 3[5] are the same as those of 3[4] (or ...627).

After finding the last 5 digits, I started using Wolfram Mathematica to compute the last digits. I easily obtained ...721627 as the last 6 digits of the Triton, and obtained ...3721627 for the last 7 digits. I computed all the way up to the last 10 digits, but after that, the amount of time I estimate the computation would take would exceed the computation time in the online version of Mathematica.

To continue, I observed that 3[3]1250 ends in ...6843000003 and 3[3]2500 ends in ...3686000003. Each time you multiply the number of triangles by 10, the digits ...6843 would shift one place to the left, and thus 3[3]1250000 ends in ...1843000000003, and 3[3]2500000 ends in ...3686000000003, and each time you add another 1250000 triangles, 1843 is added to the digits to the left of the zeroes. By taking advantage of this, I eventually found that the last 14 digits are ...85706013721627.

The difficulty in computing the last digits of most numbers defined with Steinhaus-Moser notation doesn't have to do with resources, but rather, time. Even finding the last 14 digits of 3[5] would have required billions of operations if I didn't use the nifty shortcut that I mentioned in the above paragraph, and finding, say, 30 digits would probably take more time than the age of the universe! The same goes for pretty much all other numbers defined using Steinhaus-Moser notation, such as 2 in a hexagon. (The Mega is an exception though because the number of iterations is small).