A212653 Number of steps to reach 1 in the Collatz (3x+1) problem starting with 3^n + 1.

1, 2, 6, 18, 110, 21, 95, 32, 75, 74, 42, 134, 133, 132, 131, 143, 204, 128, 189, 139, 94, 93, 260, 427, 90, 257, 393, 330, 254, 253, 389, 388, 387, 461, 460, 459, 458, 457, 456, 455, 454, 453, 452, 500, 499, 449, 497, 496, 751, 494, 493, 492, 747, 490, 745

Offset: 0

Author

Michel Lagneau, Feb 14 2013

Keywords

• nonn

Comments

It is interesting to note that the quantity 3^k + 1 appears in the formula: A006577(n + 2^A006666(n)) = A006577(n) + A006577(1 + 3^A006667(n)) where A006577 is the n number of halving and tripling steps to reach 1 in '3x+1' problem, A006666 is the number of halving steps to reach 1 and A006667 the number of tripling steps to reach 1.

For example with n = 19, A006577(19 + 2^14) = A006577(19) + A006577(1 + 3^6) => 115 = 20 + 95.

Examples

			a(2) = 6 because 3^2 + 1 = 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1 with 6 iterations.
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 0..1000
• Wikipedia, Collatz conjecture

Crossrefs

Cf. A006577, A006666, A006667, A034472, A003462, A075487.

Programs

Mathematica

f[n_] := Module[{a=3^n+1, k=0}, While[a!=1, k++; If[EvenQ[a], a=a/2, a=a*3+1]]; k]; Table[f[n], {n, 100}]
Table[Length[NestWhileList[If[EvenQ[#],#/2,3#+1]&,3^n+1,#!=1&]]-1,{n,0,60}] (* Harvey P. Dale, Sep 26 2015 *)

Formula

a(n) = A075487(n) - 1.