A074472 Length of iteration sequence of Collatz-function (A006370) when initial value is 3^n (A000244) and final cycle is followed once.

1, 8, 20, 112, 23, 97, 34, 77, 76, 44, 136, 135, 134, 133, 145, 206, 130, 191, 141, 96, 95, 262, 429, 92, 259, 395, 332, 256, 255, 391, 390, 389, 463, 462, 461, 460, 459, 458, 457, 456, 455, 454, 502, 501, 451, 499, 498, 753, 496, 495, 494, 749, 492, 747, 490

Offset: 0

Author

Labos Elemer, Sep 19 2002

Keywords

• nonn

Examples

			n=2: initial value=3^2, list of iterates is {9,28,14,7,22,11,34,17,52,26,13,50,20,10,5,16,8,4,2,1} length=a(2)=20; Observe that consecutive powers of 3 as arguments frequently provide iteration-lengths of consecutive integers, for instance n=10,11,12,13 give L=136,135,134,133 or n=88-96 result in L=1278-1271.
		

Links

• T. D. Noe, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A000244, A006370, A008884, A075484-A075488.

Programs

Mathematica

f[x_] := (1-Mod[x, 2])*(x/2)+(Mod[x, 2])*(3*x+1); f[1]=1; Table[1+Length[FixedPointList[f, 3^w]], {w, 1, 100}]