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

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 Tags collatz, conjecture, questions Search tags for this page

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