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December 31st, 2011, 10:48 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?

PS: Still learning LaTeX

 January 1st, 2012, 08:11 AM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 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 $\infty.$

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