What is the smallest number in the form \(2 \uparrow\uparrow n\), which is not apocalyptic number and n>4? Ikosarakt1 (talk) 20:45, February 16, 2013 (UTC)
- Interesting question, but I don't think it's possible to solve it efficiently :/ This is my algorithm:
function apocalyptic(n):
while n > 0:
m := n mod 10
n := floor(n / 10)
if the last 3 m's were (6, 6, 6):
return true
return false
- FB100Z • talk • contribs 22:07, February 16, 2013 (UTC)
- Wait, I got it. \(2 \uparrow\uparrow BOX\_\widetilde{M}\) is not apocalyptic. FB100Z • talk • contribs 22:08, February 16, 2013 (UTC)
Surprisingly, there are finitely many such numbers. We know that as height of tower grows, some digits "frozes" by the tetrational law of last digits convergence. Thus, block of "666" can be frozen and from some n, \(2 \uparrow\uparrow x\), where x>=n, cannot be non-apocalyptic. I will try to find this "n" using a special program. Ikosarakt1 (talk) 22:43, February 16, 2013 (UTC)
I found calculator that works faster than my program. Now I can say that any integer tetration larger than \(2 \uparrow\uparrow 1213\) will be 100% apocalyptic. Last digits of \(2 \uparrow\uparrow 1213\) I put here. Ikosarakt1 (talk) 21:34, February 18, 2013 (UTC)
- Cool beans, now we have an upper bound. Only 1309 numbers to search :P FB100Z • talk • contribs 23:31, February 18, 2013 (UTC)
IDEA:n-pocalyptic number[]
A number of form \(n^{m}\) (m is natural) which has a 666 in their decimal expression.Supyovalk (talk) 12:43, April 23, 2020 (UTC)