Consider identity
(n×3+1)=3(n×4+1)+1)÷4
Now consider n is even, n×3+1 will define a base odd (y) to which 3(n×4+1)+1)÷4 will fall to, if iteration of n×4+1 produces all odd numbers (n×4+1)k where k×3+1 will, through iteration of ÷2, fall to y, then the rules will demand y×3+1=m where m is 3(n×3+1)+1 and cannot sit on the same exponential route to y, as long as m reduces below n it cannot return to the same exponential route without some system delivering m to k. Im sorry if this makes no sense im just learning maths and enjoy the logic, but am struggling with notation.
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a=even
b=1 mod4
c=3 mod4
Rules:
If a:a+(a×2+1)…..(distributes over b&c geometrically)
If b:(b-1)÷4…….?
If c:c+((c+1)÷2)……(3+2,7+4,11+6,15+8…)
*b sit 1 above multiples of 4 so must always produce an integer a,b or c, ultimately this must bottom out at a or c.
C×4+1?
(C+1)÷3?
(b-1)÷4)+(((b-1÷4)+1)÷2)?
Okay so 0 mod4 will always produce b, lets call 0 mod4= A and 2 mod4=a for now
((A×3+1)-1)÷4=(A×3)÷4……choke point 4 mod8 return to odd multiples of 3, missing multiples of 3 in collatz is not doing anyone any favours, although odd multiples of 3 cant be returned to 10 can which would be equivalent to a return to 3.