How can I find the terminating digits of Moser's number, which is 2 in a "mega"-gon in Steinhaus-Moser notation (a mega being 2 in a pentagon)?

I was able to calculate the last 6 digits by analyzing the last digits of much smaller numbers in the 2 in a polygon sequence (2 in a hexagon, 2 in a heptagon, etc.), but is it possible to calculate, say, the last 10 digits?

How many terminating digits of the Moser can be practically calculated?

  • 1
    $\begingroup$ Are you really sure about the last $6$ digits of Moser ? And if yes, why doesn't this approach work for $10$ digits then ? $\endgroup$
    – Peter
    Aug 30, 2019 at 9:42
  • $\begingroup$ Yes, I am pretty sure that the Moser ends in ...301056. Trying the same approach for 10 digits, though, would take me a lot of time without a computer program for this stuff (which I don't currently have). Also, I would need the last 10 digits of 2 in a hexagon, heptagon, etc. $\endgroup$
    – Allam A.
    Aug 31, 2019 at 2:45
  • $\begingroup$ Since Moser is not a power tower, how can we calculate the last digits at all ? In the case of Mega this works because the number of iterations is still small. But how can we do it with Moser ? It would help a lot if you demonstrate how you did that. $\endgroup$
    – Peter
    Aug 31, 2019 at 9:47
  • $\begingroup$ I was able to find the last 6 digits of Moser by analyzing the last digits of 2 in a pentagon, hexagon, heptagon, octagon, nonagon, and decagon, and looking for patterns in those. $\endgroup$
    – Allam A.
    Aug 31, 2019 at 15:20
  • 1
    $\begingroup$ @Peter Last digits tend to be cyclic e.g. $2^2$ is $4$ and the last digit of $4^4$ is $6$, and last digit of $6^n$ is $6$, so we are stuck there. Though it would seem that repeating this for more digits is a very lengthy task indeed... $\endgroup$ Sep 3, 2019 at 13:56


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