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December 31st, 2011, 11:48 PM   #1
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Collatz conjecture, questions

This is probably wrong, but here goes. (http://en.wikipedia.org/wiki/Collatz_conjecture)

Since the conjecture has been verified upto a certain limit, proof by contradiction?

Let C be the smallest number to contradict the Collatz conjecture. C1 would mean the next term in the Collatz sequence, C2, and so on.

If C is even, C=2N. C1=N. As C is the smallest contradiction to the conjecture, C1 has a sequence to 1. Therefore C must be odd.

Let C=2N+1

Quote:
If N=2n,
C=4n+1
C1=3(4n+1)+1=12n+4=6n+2=3n+1
3n+1 < 4n+1, therefore 3n+1 travels to 1.
Quote:
If N=2n+1,
C=2(2n+1)+1=4n+3
C1=3(4n+3)+1=12n+10
C2=6n+5
C3=3(6n+5)+1=18n+16
C4=9n+8
[quote:35ksggzg]
Let n=2t
C=4n+3=8t+3
C4=18t+8
C5=9t+4
[quote:35ksggzg]
Let t=2m
C=16m+3
C5=18m+4
C6=9m+2
9m+2 < 16m+3, hence leads to 1.
Quote:
Let t=2m+1
C=16m+11
C5=18m+13
However, C5=3(6m+4)+1
6m+4 < 16m+11. As 6m+4 is not a contradiction, 18m+13 must lead to 1.
Done!
[/quote:35ksggzg]
Quote:
Let n=2t+1
C=4n+3=8t+7
C4=9(2t+1)+8=18t+17
C5=3(18t+17)+1=54t+52
C5=27t+26
[quote:35ksggzg]
Let t=2g
C=16g+7
C5=54g+26
C6=27g+13
C6=3(9g+4)+1
9g+4 < 16g+7. Yep.
Quote:
Let t=2g+1
C=8(2g+1)+7=16g+15
C5=27(2g+1)+26=54g+53
C6=162g+160
C7=81g+80

...contd....
[/quote:35ksggzg]
[/quote:35ksggzg]

Quote:
t=2g+1 continued, 3 quotes embedding limit ;0
[quote:35ksggzg]
Let g=2p
C=32p+15
C7=162g+80
C7=81p+40
81p+40=3(27p+13)+1
27p+13 < 32p+15. Whadya say
Quote:
Let g=2p+1
C=32p+31
C7=81(2p+1)+80
C8=162p+161
C9=486p+484
C10=243p+242
[quote:35ksggzg]
Let p=2f
C=64f+31
C10=486f+242
C11=243f+121
243f+121=3(81f+40)+1
81f+40=3(27f+13)+1
27f+13 < 64f+31
Done done done!
Quote:
p=2f+1
C=32(2f+1)+31=64f+63
C10=243(2f+1)+242=486f+485
C11=3(486f+485)+1=1458f+1456
C12=729f+728

To preserve remnants of sanity, I stop here
[/quote:35ksggzg]
[/quote:35ksggzg]

After any substitution, we get the case C= (Multiple of 2^x)+(2^x-1) and can't simplify without another substition. (They are in bold)
If it is shown above cases always have a sequence, would it be a proof?

Also please share your attempts at this fun 'little' problem!

PS: Still learning LaTeX
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January 1st, 2012, 09:11 AM   #2
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Re: Collatz conjecture, questions

You can always find residues that work mod 2^n for any n. The number of residues increase at some point---I'm not sure if your calculations are right up to this point or not (I checked only about half, which looked good) but it does go to
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