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.

1, 3, 175, 7, 47, 431, 87791, 15, 743151, 111, 22255, 943, 751, 218863, 175087, 31, 5871, 1791727, 1431279, 239, 191, 55023, 44015, 1967, 11917039, 1775, 3515647479163389605506303638875119, 481007, 382703, 437231, 108202665749908974283165422824431, 63, 95803119

Offset: 1

Author

N. J. A. Sloane, Feb 01 2007

Keywords

• nonn
• look
• base

Examples

			6 -> 3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1, so a(3) is the decimal equivalent of 10101111, which is 175.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..10000
• Index entries for sequences related to 3x+1 (or Collatz) problem

Crossrefs

Cf. A006577, A125710, A125626.

Programs

Maple

b:= proc(n) option remember; `if`(n=1, [0$2], `if`(n::even,
     (p-> p+[1, 2^p[1]])(b(n/2)), (p-> p+[1, 0])(b(3*n+1))))
    end:
a:= n-> b(2*n)[2]:
seq(a(n), n=1..33);  # Alois P. Heinz, Apr 02 2025

Mathematica

f[x_] := If[EvenQ[x], x/2, 3x + 1];g[n_] := FromDigits[Mod[Most[NestWhileList[f, 2n, # > 1 &]], 2, 1] - 1, 2];Table[g[n], {n, 40}] (* Ray Chandler, Feb 02 2007 *)

Extensions

Extended by Ray Chandler, Feb 02 2007