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Re:A ride that takes 10^20k years to complete in Roller Coas... by nitrogen on Monday August 03, 2020 @05:05PM (#24043408)
I'm not 100% confident in my explanation below, but maybe it's something like this, if we don't have prior knowledge of the asymptotic behavior of random walks:

The measured values are empirically exponential with a factor of 1.424 per indentation. If we assume that as an upper bound (which I don't think is quite right), then we have O(1.424^N) where N is the number of indentations.

Any exponent can be written in terms of any other base by multiplying by a constant. Since we often use base 2, we can choose to rewrite it in base 2 as O(2^(K*N)), where K is approximately 0.51.

Constant terms can be ignored in big O notation, so we can just drop the K to get O(2^N) (I am less sure about dropping a constant within the exponent).

Re:A ride that takes 10^20k years to complete in Roller Coas... by gopiandcode on Monday August 03, 2020 @05:19PM (#24043592)
You can't drop constant multiplicative terms in an exponent - O(2^n) is not equivalent to O(2^(2n)) as O(4^n) \= O(2^n) (the first one grows a lot faster).
Re:A ride that takes 10^20k years to complete in Roller Coas... by karatinversion on Monday August 03, 2020 @05:12PM (#24043508)
You cannot drop constant terms from exponents in big-O notation.