Saturday, December 27, 2014

3x+1 and T^n(k)'s evaluation matrix

In previous posts I've been talking about the 3x+1 problem and the fabulous T(x) function that satisfies

T^n(k2^n + r) = k3^j + T^n(r)

If you let 0 <= r < 2^n, and evaluate T^m(x) for all the resulting values, and for 1 < m <= n, you get a so-called evaluation matrix.  In this evaluation matrix, such as the one below,
  • 16k + 0    8k + 0    4k + 0    2k + 0    1k + 0
  • 16k + 1    24k + 2    12k + 1    18k + 2    9k + 1
  • 16k + 2    8k + 1    12k + 2    6k + 1    9k + 2
  • 16k + 3    24k + 5    36k + 8    18k + 4    9k + 2
  • 16k + 4    8k + 2    4k + 1    6k + 2    3k + 1
  • 16k + 5    24k + 8    12k + 4    6k + 2    3k + 1
  • 16k + 6    8k + 3    12k + 5    18k + 8    9k + 4
  • 16k + 7    24k + 11    36k + 17    54k + 26    27k + 13
  • 16k + 8    8k + 4    4k + 2    2k + 1    3k + 2
  • 16k + 9    24k + 14    12k + 7    18k + 11    27k + 17
  • 16k + 10    8k + 5    12k + 8    6k + 4    3k + 2
  • 16k + 11    24k + 17    36k + 26    18k + 13    27k + 20
  • 16k + 12    8k + 6    4k + 3    6k + 5    9k + 8
  • 16k + 13    24k + 20    12k + 10    6k + 5    9k + 8
  • 16k + 14    8k + 7    12k + 11    18k + 17    27k + 26
  • 16k + 15    24k + 23    36k + 35    54k + 53    81k + 80
there are various k 2^a 3^b + r expressions.  I just proved that all such expressions satisfy 0 <= r < 2^a 3^b.  In other words, T^m(r) is the remainder of dividing T^m(x) by the corresponding 2^a 3^b.

Sunday, December 14, 2014

Smalltalks 2014 videos now available

All Smalltalks 2014 videos are now available here.  Enjoy, and happy holidays!