Re: [math-fun] Parenthesizing
Suppose ^ means exponentiation and we want to count the *different integers* that 2^2^...^2 is equal to (N 2's in all) using various parenthesizations. E.g. #(2^2) = #{4} = 1 #(2^2^2) = #{16} = 1 #(2^2^2^2) = #{64, 256, 65536} = 3 #(2^2^2^2^2) = ??? Of course, any 3 twos grouped together make 4^2 = 2^4 = 16. Is there an asymptotic formula? Or better, an exact one? —Dan
I'm repeating the link Neil provided to the Guy and Selfridge paper on this topic: http://oeis.org/A003018/a003018.pdf. Just skimming it, it looks like they answer most questions of this kind. (I don't think they discuss the case of i, though.) On Sat, Dec 2, 2017 at 2:15 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Suppose ^ means exponentiation and we want to count the *different integers* that
2^2^...^2
is equal to (N 2's in all) using various parenthesizations. E.g.
#(2^2) = #{4} = 1
#(2^2^2) = #{16} = 1
#(2^2^2^2) = #{64, 256, 65536} = 3
#(2^2^2^2^2) = ???
Of course, any 3 twos grouped together make 4^2 = 2^4 = 16.
Is there an asymptotic formula? Or better, an exact one?
—Dan
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (2)
-
Allan Wechsler -
Dan Asimov