Collatz conjecture
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From: TCL on 2 Jun 2010 14:53 Let n be an odd integer. Let m be the largest integer such that 2^m divides 3n+1. Define T(n) = (3n+1)/(2^m). Thus T(1)=1, T(5)=1, T(27)=41, T(17)=13 etc. The Collatz conjecture is equivalent to: for any odd integer n, T^k (n)=1 for some positive integer k. Does anyone have a counterexample to the following statement? "T^(2n) (n)=1 for all odd integers n." Here T^k denotes the kth iteration of T. Of course, the validity of this statement solves the Collatz conjecture. -TCL
From: Craig Feinstein on 3 Jun 2010 17:23 On Jun 2, 2:53 pm, TCL <tl...(a)cox.net> wrote: > Let n be an odd integer. Let m be the largest integer such > that 2^m divides 3n+1. Define > > T(n) = (3n+1)/(2^m). > > Thus T(1)=1, T(5)=1, T(27)=41, T(17)=13 etc. > The Collatz conjecture is equivalent to: for any odd integer > n, T^k (n)=1 for some positive integer k. > > Does anyone have a counterexample to the following statement? > > "T^(2n) (n)=1 for all odd integers n." > > Here T^k denotes the kth iteration of T. Of course, the validity > of this statement solves the Collatz conjecture. > -TCL n=27 might be a counterexample. It takes a long time for it to reach 1.
From: TCL on 3 Jun 2010 18:07 On Jun 3, 5:23 pm, Craig Feinstein <cafei...(a)msn.com> wrote: > On Jun 2, 2:53 pm, TCL <tl...(a)cox.net> wrote: > > > > > > > Let n be an odd integer. Let m be the largest integer such > > that 2^m divides 3n+1. Define > > > T(n) = (3n+1)/(2^m). > > > Thus T(1)=1, T(5)=1, T(27)=41, T(17)=13 etc. > > The Collatz conjecture is equivalent to: for any odd integer > > n, T^k (n)=1 for some positive integer k. > > > Does anyone have a counterexample to the following statement? > > > "T^(2n) (n)=1 for all odd integers n." > > > Here T^k denotes the kth iteration of T. Of course, the validity > > of this statement solves the Collatz conjecture. > > -TCL > > n=27 might be a counterexample. It takes a long time for it to reach 1.- Hide quoted text - > > - Show quoted text - No. For n=27, it takes 41 iterates to reach 1.
From: christian.bau on 3 Jun 2010 19:03 On Jun 2, 7:53 pm, TCL <tl...(a)cox.net> wrote: > Let n be an odd integer. Let m be the largest integer such > that 2^m divides 3n+1. Define > > T(n) = (3n+1)/(2^m). > > Thus T(1)=1, T(5)=1, T(27)=41, T(17)=13 etc. > The Collatz conjecture is equivalent to: for any odd integer > n, T^k (n)=1 for some positive integer k. > > Does anyone have a counterexample to the following statement? > > "T^(2n) (n)=1 for all odd integers n." > > Here T^k denotes the kth iteration of T. Of course, the validity > of this statement solves the Collatz conjecture. n = 1 is the first counterexample. 1 -> 4 -> 2. n = 3 is the second counterexample: 3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2. n = 5 is the third counterexample: 5 -> 16 -> 8 -> 4 -> 2 -> 1 -> 4 -> 2 -> 1 -> 4 -> 2. So you say it's true for all odd integers and you're looking for a counter example, and the first three odd integers are counterexamples. Sad.
From: Rick Decker on 3 Jun 2010 20:45
On 6/3/10 7:03 PM, christian.bau wrote: > On Jun 2, 7:53 pm, TCL<tl...(a)cox.net> wrote: >> Let n be an odd integer. Let m be the largest integer such >> that 2^m divides 3n+1. Define >> >> T(n) = (3n+1)/(2^m). >> >> Thus T(1)=1, T(5)=1, T(27)=41, T(17)=13 etc. >> The Collatz conjecture is equivalent to: for any odd integer >> n, T^k (n)=1 for some positive integer k. >> >> Does anyone have a counterexample to the following statement? >> >> "T^(2n) (n)=1 for all odd integers n." >> >> Here T^k denotes the kth iteration of T. Of course, the validity >> of this statement solves the Collatz conjecture. > > n = 1 is the first counterexample. 1 -> 4 -> 2. > n = 3 is the second counterexample: 3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2. > n = 5 is the third counterexample: 5 -> 16 -> 8 -> 4 -> 2 -> 1 -> 4 -> > 2 -> 1 -> 4 -> 2. > > So you say it's true for all odd integers and you're looking for a > counter example, and the first three odd integers are counterexamples. > Sad. Not so sad (you missed the "divide out the 2-factors" condition): T(1) = 4/4 = 1, T^2(1) = 1 T(3) = 10/2 = 5, T^2(3) = T(5) = 16/16 = 1, ... , T^6(3) = 1 T(5) = 16/16 = 1, ... , T^10(5) = 1 T(7) = 22/2 = 11, T^2(7) = 34/2 = 17, T^3(7) = 52/4 = 13, T^4(7) = 40/8 = 5, T^5(7) = 16/16 = 1, ... , T^14(7) = 1 It holds for the first four odd numbers, and for these the "order" is quite a bit less than 2n. Regards, Rick |