My Math Forum Mif (minimum figure) function

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 March 8th, 2011, 01:13 PM #1 Newbie   Joined: Jan 2011 Posts: 2 Thanks: 0 Mif (minimum figure) function x=10000000000 For writing x we used one 1 and ten 0... So we used 11 figures for writing x. But we can also write x = 10 ^ 11 . On this writing we only used three 1, and a 0... We expressed x with only four figures. Lets define a mif function. When x is a positive integer, mif(x) means minimum figure number need when we want t o write x with +(sum), -(subtract), *(multiply), /(divide) , ^(power) . So mif(x) = 4 mif(221861106E17) = mif(12^17 - 1) = 5 mif(179216043E16) = mif(13 ^11 + 4) = 5 Lets think mif(1), mif(2), mif(3), mif(4), mif(5), mif(6), .... as series. Is this convergent or divergent? How can we draw graphics? Sorry for the neantherdal english...
 March 8th, 2011, 02:27 PM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Mif (minimum figure) function This is an integer complexity function. There are many such sequences in the OEIS, though I'm not sure if yours is one. Other examples include A005245 and A003313. Your sequence is divergent. More is true: not only are there some subsequences with only finitely many members below an arbitrary N, *all* subsequences have this property. A relevant concept is Kolmogorov complexity: the smallest number of characters needed to express a program to display a given number.

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