In the "3x+1" problem, let 1 denote a halving step and 0 denote an x->3x+1 step. Then a(n) is obtained by writing the sequence of steps needed to reach 1 from 2n and reading it as a decimal number.
A125711
In the "3x+1" problem, let 1 denote a halving step and 0 denote an x->3x+1 step. Then a(n) is obtained by writing the sequence of steps needed to reach 1 from 2n and reading it as a decimal number.
Terms
- a(0) =1a(1) =3a(2) =175a(3) =7a(4) =47a(5) =431a(6) =87791a(7) =15a(8) =743151a(9) =111a(10) =22255a(11) =943a(12) =751a(13) =218863a(14) =175087a(15) =31a(16) =5871a(17) =1791727a(18) =1431279a(19) =239a(20) =191a(21) =55023a(22) =44015a(23) =1967a(24) =11917039a(25) =1775a(27) =481007a(28) =382703a(29) =437231a(31) =63
External references
- oeis: A125711