Take any number...

Take any number
if it is even, halve it;
if it is odd, multiply it by 3 and add 1 and then halve it.
repeat these operations until you reach the number 1.

For example, starting with the number 5:
5 is odd; 5 x 3 = 15; add 1 : 16
16 ÷ 2 = 8 (step 1)
8 is even; halve it : 4 (step 2)
4 is even; halve it : 2 (step 3)
2 is even; halve it: 1, and stop. (step 4)

Taking it for granted that we will always reach the number 1 in the end, is there a connection between the number of steps required to reach the number 1 and the number we started with?

Here, if S(n) stands for the number of steps when we start with n, S(5) = 4.

So far all I've come up with is a pattern for the powers of two : S(2ⁿ) = n
e.g. 16 = 2⁴, so S(16)=4.
We can ignore the even numbers, by the way, as s(k x 2ⁿ) = s(k) + n

Table 1: Some results k<145)		Table 2: Numbers matched by number of links

n	S(n)	n	S(n)	n	S(n)	n	S(n)	n	S(n)
3	5	35	10	67	19	99	18	131	20
5	4	37	15	69	11	101	18	133	20
7	11	39	23	71	65	103	56	135	28
9	13	41	69	73	73	105	26	137	58
11	10	43	20	75	11	107	64	139	28
13	7	45	12	77	16	109	72	141	12
15	12	47	66	79	24	111	45	143	66
17	9	49	17	81	16	113	10	145
19	14	51	17	83	70	115	23	147
21	6	53	9	85	8	117	15	149
23	11	55	71	87	21	119	23	151
25	16	57	22	89	21	121	61	153
27	70	59	22	91	59	123	31	157
29	13	61	14	93	13	125	69	159
31	67	63	68	95	67	127	31	161
33	18	65	19	97	75	129	77	163

s(k)	k	s(k)	k	s(k)	k
3	-	31	123	59	91
4	5	32		60
5	3	33		61	121
6	21	34		62
7	13	35		63
8	85	36		64	107
9	17, 53	37	127	65
10	11, 35, 113	38		66	47, 143
11	7, 23, 69, 75	39		67	31, 95
12	15, 45, 141	40		68	63
13	9, 29, 93	41		69	41, 125
14	19, 61	42		70	27, 83
15	37, 117	43		71	55, 65
16	25, 77, 81	44		72	109
17	49, 51	45	111	73	73
18	33, 99, 101	46		74
19	65, 67	47		75	97
20	43, 131, 133	48		76
21	87, 89	49		77	129
22	57, 59	50		78
23	39, 115, 119	51		79
24		52		80
25		53		81
26	105	54		82
27		55		83
28	135, 139	56	103	84
29		57		85
30		58	137	86

That's as far as we've got to date, the results for k=99 to k=143 copied from a post by Dan in the Dozensonline Forum (Feb25)