In the "3x+1" problem, let 0 denote a halving step and 1 denote an x->3x+1 step. Then a(n) is obtained by writing the sequence of steps needed to reach 1 from 2n+1 and reading it as a decimal number.
A125710
In the "3x+1" problem, let 0 denote a halving step and 1 denote an x->3x+1 step. Then a(n) is obtained by writing the sequence of steps needed to reach 1 from 2n+1 and reading it as a decimal number.
Terms
- a(0) =4a(1) =80a(2) =16a(3) =43280a(4) =305424a(5) =10512a(6) =272a(7) =87056a(8) =2320a(9) =665872a(10) =64a(11) =21520a(12) =4860176a(14) =141584a(16) =38414608a(17) =5136a(18) =1091856a(19) =11358841104
External references
- oeis: A125710