A fast inverse for f derives by observing that f is equivalent to: f3 =: - -/@:>.@:(2 6 3&%:)
That is, f3 n is n less the perfect squares and perfect cubes bounded by n, plus the perfect 6-th powers. The function g3a then is very near to the inverse to f3 : g3a=: + -/@:>.@:(2 6 3&%:) For example: f3 1e9 999967409 g3a 999967409 1000000000 However: g3a 112 125 f3 125 111 The details of deriving g3, the inverse for f3, are left as an exercise for the reader. ----- Original Message ----- From: Björn Helgason <gos...@gmail.com> Date: Sunday, December 26, 2010 11:48 Subject: [Jprogramming] finding number To: Programming forum <programming@jsoftware.com> > I have a small task that looks like this > > f=: [: # i. -. (2 ^~ i.) , 3 ^~ i. > f 87 > 75 > > I want to use tacit to find what number gives me 75 numbers (the > answerbeing 87) > > Could be asking what argument y do I need to find x numbers > according to the > formula above ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
