Re: [Jprogramming] new years eve puzzle

2017-12-30 Thread Jimmy Gauvin
Still thinking about the puzzle. Some tools to explore the properties of the conjecture: NB. pretty print toca =: '0123456789' {~ ] tobi =: 3 : '(2#~1+>.2^.>./y)#:y' todc =: 3 : '(10#~>.10^.>./y)#:y' todcc =: 3 : '(1+>.10^.>./y)":,.y' NB. divisor formula from Roger Hui, http://cod

[Jprogramming] new years eve puzzle

2017-12-30 Thread R.E. Boss
From http://list.seqfan.eu/pipermail/seqfan/2017-December/018222.html " Consider the divisors of 136 (A018299): 1, 2, 4, 8, 17, 34, 68, 136. 136 in binary is 10001000. Reading off successive bits from the left we have: 1 1 10 2 1004 1000 8 10001 17 10001034 1