Trifurcate Collatz

	A	B	C	D	E	F	G	H	I	J	K	L	M	N	O	P
1	Start with the first six members of the 4(n)+1 sequence (on C9:H9, this sheet.)
2		While the process is demonstrated here using only six members, it could be extended indefinately.
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4	to be continued		3			19	19-3=	16
5	on row 10 of Sheet2		5			29	29-5=	24
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8	to row 9, apply	(2(n)-1)/3	3	8.333333333	13.66666667	19	24.33333333	29.66666667
9	increment by	8	5	13	21	29	37	45	yields-->	24	21	45	69	93	117	141
10	to row 9, apply	(4(n)-1)/3	6.333333333	17	27.66666667	38.33333333	49	59.66666667
11														The series is indefinately extendable.
12													Thrid iteration of this particular permutation.
13	to be continued			13			37	37-13=	24			Second iteration of this particular permutation.
14	on row 28 of Sheet2			17			49	49-17=	32		First iteration of this particular permutation.
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16											The particular varient is where the 4(n)+1 is itself alread a 3(n)
17	At each stage break the 4(n)+1 sequence into the following three sub-sets:
18		where applying (2(n)-1)/3 results in a set of whole number values.
19		where applying (4(n)-1)/3 results in a set of whole number values.
20		where neither formula results in a set of whole number values, terminating the process for the extendable portion of the series of those particular 4(n+1) sequences.
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23		3(odd)	3	9	15	21	27	33	39	45	51	57	63	69	75	81
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25			Each unique 3(n) is connected to an equally unique 4(n)+1, either directly, as illustrated on this sheet (K9:P9), or by selectively applying the 3n+1 rule, isolating results which become odd in either one or two subsequent divisions by 2
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27		4(odd)+1	5	13	21	29	37	45	53	61	69	77	85	93	101	109
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29			As shown on Sheet1b of Recursive Collatz Aspect *, applying 3n+1 to a 4(n)+1 results in an even number requiring three or more subsequent divisions by 2 to reduce it back to an odd number.
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31				* Recursive Collatz Aspect:		https://docs.google.com/spreadsheets/d/1z-yl5gg_Tz0CoXTU2vVVK6lziyqtrfba15v0hjcEOfM/edit#gid=0
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