Monday, August 11, 2014

3x+1 over the weekend

You will recall from my previous 3x+1 posts that using

T^n(q2^n + r) = q3^j + T^n(r)

one could build an evaluation matrix for each 0 <= r < 2^n.  For a long time, I had strong circumstantial evidence that the proportion of evaluation matrix rows satisfying T^n(k) > k tended to shrink as n grew.  It's so unsatisfying to merely feel something has to be true...

I am happy now, though.  After reading Concrete Mathematics for several hours, I managed a tentative proof showing the proportion of rows satisfying the growth inequality tends to zero as n goes to infinity.  If you are interested, send me a note --- there is no way I am typesetting that heavy math in this blog post!

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