Collatz conjecture
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From: Rick Decker on 3 Jun 2010 21:32 On 6/3/10 8:45 PM, Rick Decker wrote: > 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." > FWIW, it holds for the first million odd numbers. Regards, Rick
From: TCL on 8 Jun 2010 21:44 On Jun 3, 9:32 pm, Rick Decker <rdec...(a)hamilton.edu> wrote: > On 6/3/10 8:45 PM, Rick Decker wrote:> 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." > > FWIW, it holds for the first million odd numbers. > > Regards, > > Rick I have verified it for the first two million odd numbers. In fact, my calculations show more: For each of these odd numbers n, If k is the least integer such that T^k (n)=1, then k/n <= 41/27, and equality holds only for n=27. -TCL
From: TCL on 8 Jun 2010 22:06 On Jun 8, 9:44 pm, TCL <tl...(a)cox.net> wrote: > On Jun 3, 9:32 pm, Rick Decker <rdec...(a)hamilton.edu> wrote: > > > > > > > On 6/3/10 8:45 PM, Rick Decker wrote:> 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." > > > FWIW, it holds for the first million odd numbers. > > > Regards, > > > Rick > > I have verified it for the first two million odd numbers. > In fact, my calculations show more: For each of these odd numbers n, > > If k is the least integer such that T^k (n)=1, then k/n <= 41/27, > and equality holds only for n=27. > > -TCL- Hide quoted text - > > - Show quoted text - Update: Statement above about 41/27 verified for the first 4 million odd numbers.
From: TCL on 9 Jun 2010 09:43 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 For an odd integer n, let k be the least integer such that T^k (n)=1. My computations strongly suggest that k/n can be greater than 1 only when n=27,31. Moreover, k/n --> 0 as n approaches infinity. -TCL
From: Gerry Myerson on 9 Jun 2010 19:15
In article <1cfa1738-fee4-4368-b1a0-baf43f43eabb(a)w31g2000yqb.googlegroups.com>, TCL <tlim1(a)cox.net> 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 > > For an odd integer n, let k be the least integer such that T^k (n)=1. Provided, of course, that any such integer exists. > My computations strongly suggest that k/n can be greater than 1 only > when n=27,31. > Moreover, k/n --> 0 as n approaches infinity. Have you looked at other people's computations? I believe they suggest (but don't prove) much stronger results on the rate of growth of k as a function of n. -- Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email) |