Collatz conjecture 37895 226108786 2008-07-16T21:23:16Z Greg Kuperberg 139643 /* Iterating on rational numbers with odd denominators */ [[Image:Collatz-graph-all-30-no27.svg|thumb|100px|Directed [[graph (mathematics)|graph]] showing the orbits of small numbers under the Collatz map. The Collatz conjecture is equivalent to the statement that all paths eventually lead to 1.]] {{in-text citations}} The '''Collatz conjecture''' is an unsolved [[conjecture]] in [[mathematics]]. It is named after [[Lothar Collatz]], who first proposed it in 1937. The conjecture is also known as the '''3''n'' + 1 conjecture''', as the '''Ulam conjecture''' (after [[Stanislaw Ulam]]), as the '''Syracuse problem''', as the '''hailstone sequence''' or '''hailstone numbers''', or as '''wondrous numbers''' per ''[[Gödel, Escher, Bach]]''. In some Computer Programming publications<ref>Advanced Logo: A Language for Learning By Michael Friendly, 1988</ref> it is referred to as '''HOTPO''', which stands for "Half Or Triple Plus One". It asks whether a certain kind of number sequence always ends in the same way, regardless of the starting number. [[Paul Erdős]] said about the Collatz conjecture: "Mathematics is not yet ready for such problems." He offered $500 for its solution. (Lagarias 1985) == Statement of the problem == Consider the following operation on an arbitrary positive [[integer]]: * If the number is even, divide it by two. * If the number is odd, triple it and add one. For example, if this operation is performed on 3, the result is 10; if it is performed on 28, the result is 14. In [[modular arithmetic]] notation, define the [[function (mathematics)|function]] f as follows: : <math> f(n) = \begin{cases} n/2 &\mbox{if } n \equiv 0 \pmod{2}\\ 3n+1 & \mbox{if } n\equiv 1 \pmod{2}.\end{cases} </math> [[Image:Totalstoppingtime.png|thumb|300px|left|Numbers from 2 to 9999 and their corresponding total stopping time.]] Now, form a sequence by performing this operation repeatedly, beginning with any positive integer, and taking the result at each step as the input at the next. In notation: : <math> a_i = \begin{cases}n & \mbox{for } i = 0 \\ f(a_{i-1}) & \mbox{for } i > 0. \end{cases}</math> The Collatz conjecture is: ''This process will eventually reach the number 1, regardless of which positive integer is chosen initially.'' That smallest ''i'' such that the above holds is called the '''total stopping time''' of ''n''. The conjecture asserts that every ''n'' has a well-defined stopping time. If, for some ''n'', such an ''i'' doesn't exist, we say that ''n'' has infinite total stopping time and the conjecture is false. If the conjecture is false, it can only be because there is some starting number which gives rise to a sequence which does not contain 1. Such a sequence might enter a repeating cycle that excludes 1, or increase without bound. No such sequence has been found. == Examples == For instance, starting with ''n'' = 6, one gets the sequence 6, 3, 10, 5, 16, 8, 4, 2, 1. Starting with ''n'' = 11, the sequence takes longer to reach 1: 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1. If the starting value ''n'' = 27 is chosen, the sequence takes 111 steps, climbing above 9,000 before descending to 1. { 27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1 } [[Image:Collatz5.svg|800px|center]] The number less than 100 million with the longest total stopping time is 63728127, with 949 steps. == Program to calculate Collatz sequences == A specific Collatz sequence can be easily computed, as is shown by this [[pseudocode]] example: '''function''' collatz(n) '''while''' n > 1 '''show''' n '''if''' n is odd '''set''' n to 3n + 1 '''else''' '''set''' n to n / 2 '''show''' n This program halts when the sequence reaches 1, in order to avoid printing an endless cycle of 4, 2, 1. If the Collatz conjecture is true, the program will always halt no matter what positive starting integer is given to it. (See [[Halting problem#Can humans solve the halting problem?]] for a discussion of the relationship between open-ended computer programs and unsolved mathematics problems.) == Supporting arguments == Although the conjecture has not been proven, most mathematicians who have looked into the problem believe intuitively that the conjecture is true. Here are two reasons for expecting this. === Experimental evidence === The conjecture has been checked by computer for all starting values up to 10 &times; 2<sup>58</sup>&nbsp;≈&nbsp;2.88{{e|18}}[http://www.ieeta.pt/~tos/3x+1.html]. While impressive, such computer evidence should be interpreted cautiously. More than one important conjecture has been found false, but only with very large counterexamples. (See for example the [[Pólya conjecture]], the [[Mertens conjecture]] and the [[Skewes' number]].) It is also known that {4,2,1} is the only cycle with fewer than 35400 terms.<ref>L. E. Garner (1981) ''Proc. Amer. Math. Soc.'' '''82''', 19-22.</ref> === A probabilistic heuristic === If one considers only the ''odd'' numbers in the sequence generated by the Collatz process, then on average the next odd number is ¾ of the previous one on average [http://www.cecm.sfu.ca/organics/papers/lagarias/paper/html/node3.html#SECTION00021000000000000000]. (More precisely, the geometric mean of the ratios of outcomes is ¾.) This yields a heuristic argument that every Collatz sequence should decrease in the long run, although this is not evidence against other cycles, only against divergence. The argument is not a proof because it pretends that Collatz sequences are assembled from uncorrelated probabilistic events. (It does rigorously establish that the 2-adic extension of the Collatz process has 2 division steps for every multiplication step for almost all 2-adic starting values.) == Other ways of approaching the problem== === In reverse === There is another approach to prove the conjecture, which considers the bottom-up method of growing the so called ''Collatz graph''. The ''Collatz graph'' is a [[graph (mathematics)|graph]] defined by the inverse [[relation (mathematics)|relation]] <math> R(n) = \begin{cases} 2n & \mbox{if } n\equiv 0,1,2,3,5 \\ 2n, (n-1)/3 & \mbox{if } n\equiv 4 \end{cases} \pmod{6}. </math> So, instead of proving that all natural numbers eventually lead to 1, we can prove that 1 leads to all natural numbers. For any integer ''n'', 3''n'' + 1 ≡4 (mod 6) [[iff]] ''n'' ≡1 (mod 2) and thus ''n'' ≡1, 3 or 5 (mod 6). Also, the inverse relation forms a tree except for the 1-2-4 loop (the inverse of the 1-4-2 loop of the unaltered function ''f'' defined in the statement of the problem above). When the relation 3''n'' + 1 of the function ''f''(''n'') is replaced by the common substitute "shortcut" relation (3''n'' + 1)/2 (see Optimizations below), the Collatz graph is defined by the inverse relation, <math> R(n) = \begin{cases} 2n & \mbox{if } n\equiv 0,1 \\ 2n, (2n-1)/3 & \mbox{if } n\equiv 2 \end{cases} \pmod{3}. </math> This inverse relation forms a tree except for a 1-2 loop (the inverse of the 1-2 loop of the function ''f''(''n'') revised as indicated above). === As rational numbers === The natural numbers can be converted to rational numbers in a certain way. To get the rational version, find the highest power of two less than or equal to the number, use it as the denominator, and subtract it from the original number for the numerator (527 → 15/512). To get the natural version, add the numerator and denominator (255/256 → 511). The Collatz conjecture then says that the numerator will eventually equal zero. The Collatz function changes to: : <math> f(n, d) = \begin{cases} (3n + d + 1)/2d & \mbox{if } 3n + d + 1 < 2d \\ (3n - d + 1)/4d & \mbox{if } 3n + d + 1 \ge 2d \end{cases} </math> (''n'' = numerator; ''d'' = denominator). This works because 3''x'' + 1 = 3(''d'' + ''n'') + 1 = (2''d'') + (3''n'' + ''d'' + 1) = (4''d'') + (3''n'' - ''d'' + 1). Reducing a rational before every operation is required to get ''x'' as an odd. === As an abstract machine ...=== Repeated applications of the Collatz function can be represented as an [[abstract machine]] that handles [[string (computer science)|string]]s of [[bit]]s. The machine will perform the following two steps on any odd number until only one "1" remains: # Add the original with a "1" appended to the end to the original (interpreting the string as a binary integer), i.e. <math>3n+1 = (2n+1) + n</math> # Remove all trailing "0"s. === ...which is equivalent to Base Two arithmetic === Another way to examine the 3n+1 conjecture is through the base two system. An example would go as follows: Ex. We will use the number 7, so in base two it is written thus: 111 111 <u>1111</u> 1011<strike>0</strike> <u>10111</u> 10001<strike>0</strike> <u>100011</u> 1101<strike>00</strike> <u>11011</u> 101<strike>000</strike> <u>1011</u> 1<strike>0000</strike> The method used to iterate any number in base two is write the initial number, then beneath it write the same number with an additional 1 on the right side, then add these two numbers. Any zeros that result on the right side can be crossed out and the process is repeated until the number iterates to 1. '''Comparison of the Abstract Machine to equivalent Base 2 arithmetic''' # # Python # import re # regular expressions import gmpy # base 2 math library def abstract_machine(s): # define Truth Tables for the Full Adder sum_tt = {'000':'0','001':'1','010':'1','011':'0','100':'1','101':'0','110':'0','111':'1'} carry_tt = {'000':'0','001':'0','010':'0','011':'1','100':'0','101':'1','110':'1','111':'1'} print s while s != '1': if s[-1]=='1': # it's odd s = '00' + s # operands must be same length, so prepend with MS 0 ss = '0' + s + '0' # shift left (append LS 0) and prepend (MS 0) to allow carry t = "".join(reversed(s)) # iterating is L->R, so temporarily reverse tt = "".join(reversed(ss)) carry = '1' # preset carry (the '1' of '3n+1') answer = "" # initialize answer for i,j in enumerate(t): # walk through operands one char at a time the_input = carry + j + tt[i] # assemble input from previous carry & two operands the_sum = sum_tt[the_input] # look up sum out in sum Truth Table carry = carry_tt[the_input] # look up carry out in carry Truth Table answer = answer + the_sum # append sum to answer (carry used on next iteration) final_answer = "".join(reversed(answer)) # un-reverse answer if final_answer[0]=='0': # if the last pad caharacter didn't receive a carry, final_answer = final_answer[1:] # ...get rid of it print final_answer # show result before stripping LS 0's else: # it's even final_answer = s m = re.search('(.*1)(0*$)',final_answer) # remove all LS 0's in one fell swoop s = "".join(m.groups()[0]) # reassemble answer to do next iteration print s def base_2(n): while n>1: f = gmpy.scan1(n,0) # find position of LS 1-bit if f>0: # it's even print gmpy.digits(n,2) # print n in base 2 n = n >> f # remove all LS 0's in one fell swoop else: # it's odd print gmpy.digits(n,2) # print n in base 2 n = (n << 1) + n + 1 # multiply by 3 and add 1 print gmpy.digits(n,2) # print n in base 2 # main print 'test of abstract machine:' print abstract_machine('111') print print print 'test of base 2:' print base_2(7) ## test of abstract machine: ## ## 111 ## 10110 ## 1011 ## 100010 ## 10001 ## 110100 ## 1101 ## 101000 ## 101 ## 10000 ## 1 ## ## ## test of base 2: ## ## 111 ## 10110 ## 1011 ## 100010 ## 10001 ## 110100 ## 1101 ## 101000 ## 101 ## 10000 ## 1 ## === As a parity sequence === For this section, consider the Collatz function in the slightly modified form : <math> f(n) = \begin{cases} n/2 &\mbox{if } n \equiv 0 \\ (3n +1)/2 & \mbox{if } n \equiv 1 \end{cases} \pmod{2}.</math> This can be done because when ''n'' is odd, 3''n'' + 1 is always even. If P(…) is the parity of a number, that is P(2''n'') = 0 and P(2''n'' + 1) = 1, then we can define the Collatz parity sequence for a number ''n'' as ''p<sub>i</sub>'' = P(''a<sub>i</sub>''), where ''a''<sub>0</sub> = ''n'', and ''a''<sub>''i''+1</sub> = ''f''(''a''<sub>''i''</sub>). Using this form for ''f''(''n''), it can be shown that the parity sequences for two numbers ''m'' and ''n'' will agree in the first ''k'' terms if and only if ''m'' and ''n'' are equivalent modulo 2<sup>''k''</sup>. This implies that every number is uniquely identified by its parity sequence, and moreover that if there are multiple Collatz cycles, then their corresponding parity cycles must be different. The proof is simple: it is easy to verify by hand that applying the ''f'' function ''k'' times to the number ''a'' 2<sup>''k''</sup>+''b'' will give the result ''a'' 3<sup>''c''</sup>+''d'', where ''d'' is the result of applying the ''f'' function ''k'' times to ''b'', and ''c'' is how many odd numbers were encountered during that sequence. So the parity of the first ''k'' numbers is determined purely by ''b'', and the parity of the (''k''+1)th number will change if the least significant bit of ''a'' is changed. The Collatz Conjecture can be rephrased as stating that the Collatz parity sequence for every number eventually enters the cycle 0 → 1 → 0. === As a tag system === For the Collatz function in the form : <math> f(n) = \begin{cases} n/2 &\mbox{if } n \equiv 0 \\ (3n +1)/2 & \mbox{if } n \equiv 1 \end{cases} \pmod{2}</math> the Collatz sequences are computed by the extremely simple [[Tag system#Example: Computation of Collatz sequences|2-tag system]] whose production rules are : <math> \begin{cases} a \rarr bc \\ b \rarr a \\ c \rarr aaa \end{cases}</math> and in which a positive integer ''n'' is represented by a string of ''n'' <math>a</math>'s, with iteration of the tag operation halting on any word of length less than 2. (Adapted from De Mol.) The Collatz conjecture can be rephrased as stating that this tag system, with an arbitrary finite string of <math>a</math>'s as the initial word, eventually halts. See the linked article for a worked example. ==Extensions to larger domains== ===Iterating on ''all'' integers=== For '''any''' integer n, rather than just positive integers, we map it to the integer f(n), where f(n) = 3n + 1 if n is odd; f(n) = n/2 if n is even. Interestingly, there are in this case a total of '''5''' known cycles, which all integers seem to eventually fall into under iteration of f. These cycles are listed here, starting with the well-known cycle for positive n. To save steps, we list ''only the '''odd''' numbers of each cycle'' (except for the trivial cycle {0}). Each odd number n, when f is applied repeatedly, will next reach an odd number at (3n+1) / (the [[2-order|largest power of 2 that divides]] 3n+1); each cycle is listed with its member of least absolute value first. We follow each cycle with the size of the '''full''' cycle (in parentheses): the number of members, odd or even, belonging to a cycle, counted without repetition. a) 1 → 1 (size 3) b) 0 → 0 (size 1) c) -1 → -1 (size 2) d) -5 → -7 → -5 (size 5) e) -17 → -25 → -37 → -55 → -41 → -61 → -91 → -17 (size 18) We may define the '''Generalized Collatz Conjecture''' as the assertion that every integer, under iteration by f, eventually falls into one of these five cycles a), b), c), d), or e). === Iterating with odd denominators or 2-adic integers === The standard Collatz map can be extended to (positive or negative) rational numbers which have odd denominators when written in lowest terms. The number is taken to be odd or even according to whether its numerator is odd or even. A closely related fact is that the Collatz map extends to the ring of 2-adic integers, which contains the ring of rationals with odd denominators as a subring. The parity sequences as defined above are no longer unique for fractions. However, it can be shown that any possible parity cycle is the parity sequence for exactly one fraction: if a cycle has length ''n'' and includes odd numbers exactly ''m'' times at indices ''k''<sub>0</sub>, …, ''k''<sub>''m''-1</sub>, then the unique fraction which generates that parity cycle is :<math>\frac{3^{m-1} 2^{k_0} + ... + 3^0 2^{k_{m-1}}}{2^n - 3^m}</math>. For example, the parity cycle (1 0 1 1 0 0 1) has length 7 and has 4 odd numbers at indices 0, 2, 3, and 6. The unique fraction which generates that parity cycle is :<math>\frac{3^3 2^0 + 3^2 2^2 + 3^1 2^3 + 3^0 2^6}{2^7 - 3^4} = \frac{151}{47}</math>. The complete cycle being: 151/47 → 250/47 → 125/47 → 211/47 → 340/47 → 170/47 → 85/47 → 151/47 Although the cyclic permutations of the original parity sequence are unique fractions, the cycle is not unique, each permutation's fraction being the next number in the loop cycle: :(0 1 1 0 0 1 1) → <math>\frac{3^3 2^1 + 3^2 2^2 + 3^1 2^5 + 3^0 2^6}{2^7 - 3^4} = \frac{250}{47}</math> <br /> :(1 1 0 0 1 1 0) → <math>\frac{3^3 2^0 + 3^2 2^1 + 3^1 2^4 + 3^0 2^5}{2^7 - 3^4} = \frac{125}{47}</math> <br /> :(1 0 0 1 1 0 1) → <math>\frac{3^3 2^0 + 3^2 2^3 + 3^1 2^4 + 3^0 2^6}{2^7 - 3^4} = \frac{211}{47}</math> <br /> :(0 0 1 1 0 1 1) → <math>\frac{3^3 2^2 + 3^2 2^3 + 3^1 2^5 + 3^0 2^6}{2^7 - 3^4} = \frac{340}{47}</math> <br /> :(0 1 1 0 1 1 0) → <math>\frac{3^3 2^1 + 3^2 2^2 + 3^1 2^4 + 3^0 2^5}{2^7 - 3^4} = \frac{170}{47}</math> <br /> :(1 1 0 1 1 0 0) → <math>\frac{3^3 2^0 + 3^2 2^1 + 3^1 2^3 + 3^0 2^4}{2^7 - 3^4} = \frac{85}{47}</math> Also, for uniqueness, the parity sequence should be "prime", i.e., not partitionable into identical sub-sequences. For example, parity sequence (1 1 0 0 1 1 0 0) can be partitioned into two identical sub-sequences (1 1 0 0)(1 1 0 0). Calculating the 8-element sequence fraction gives :(1 1 0 0 1 1 0 0) → <math>\frac{3^3 2^0 + 3^2 2^1 + 3^1 2^4 + 3^0 2^5}{2^8 - 3^4} = \frac{125}{175}</math> But when reduced to lowest terms {5/7}, it is the same as that of the 4-element sub-sequence :(1 1 0 0) → <math>\frac{3^1 2^0 + 3^0 2^1}{2^4 - 3^2} = \frac{5}{7}</math> And this is because the 8-element parity sequence actually represents two circuits of the loop cycle defined by the 4-element parity sequence. In this context, the Collatz conjecture is equivalent to saying that (0 1) is the only cycle which is generated by positive whole numbers (i.e. 1 and 2). === Iterating on real or complex numbers === [[Image:CobwebCollatz2.PNG|right|thumb|300px|[[Cobweb plot]] of the orbit 10-5-8-4-2-1-2-1-2-1-etc. in the real extension of the Collatz map (optimized by replacing "3''n'' + 1" with "(3''n'' + 1)/2" )]] The Collatz map can be viewed as the restriction to the integers of the smooth real and complex map :<math>f(z):=\frac 1 2 z \cos^2\left(\frac \pi 2 z\right)+(3z+1)\sin^2\left(\frac \pi 2 z\right)</math>, which simplifies to <math>\frac{1}{4}(2 + 7z - (2 + 5z)\cos(\pi z))</math>. If the standard Collatz map defined above is optimized by replacing "3''n'' + 1" with "(3''n'' + 1)/2" (see Optimizations below), it can be viewed as the restriction to the integers of the smooth real and complex map :<math>f(z):=\frac 1 2 z \cos^2\left(\frac \pi 2 z\right)+\frac 1 2 (3z+1)\sin^2\left(\frac \pi 2 z\right)</math>, which simplifies to <math>\frac{1}{4}(1 + 4z - (1 + 2z)\cos(\pi z))</math>. [[Iterated function|Iterating]] the above optimized map in the complex plane produces the Collatz [[fractal]]. <div style="clear:both;"></div> [[Image:CollatzFractal.png|thumb|center|500px|Collatz map [[fractal]] in a neighbourhood of the real line]] == Optimizations == The "parity" section above gives a way to speed up simulation of the sequence. To jump ahead ''k'' steps on each iteration (using the ''f'' function from that section), break up the current number into two parts, ''b'' (the ''k'' least significant bits, interpreted as an integer), and ''a'' (the rest of the bits as an integer). The result of jumping ahead ''k''+''c''[b] steps can be found as: :''f'' <sup>''k''+''c''[b]</sup>(''a'' 2<sup>''k''</sup>+''b'') = ''a'' 3<sup>''c''[b]</sup>+''d''[b] The ''c'' and ''d'' arrays are precalculated for all possible ''k''-bit numbers ''a'', where ''d'' [a] is the result of applying the ''f'' function ''k'' times to ''b'', and ''c'' [a] is the number of odd numbers encountered on the way. For example, if k=5, you can jump ahead 5 steps on each iteration by separating out the 5 least significant bits of a number and using: : ''c'' [0...31] = {0,3,2,2,2,2,2,4,1,4,1,3,2,2,3,4,1,2,3,3,1,1,3,3,2,3,2,4,3,3,4,5} : ''d'' [0...31] = {0,2,1,1,2,2,2,20,1,26,1,10,4,4,13,40,2,5,17,17,2,2,20,20,8,22,8,71,26,26,80,242} For the special purpose of searching for a counterexample to the Collatz conjecture, this precomputation leads to an even more important acceleration which is due to Tomás Oliveira e Silva and is used in the record confirmation of the Collatz conjecture. The first counterexample, if it exists, is clearly odd because 2''n'' → ''n'' in one step. By the same reasoning, the first counterexample is 3 mod 4, because 4''n'' + 1 → 3''n'' + 1 in three steps. Working modulo 2<sup>''j''</sup>, more and more congruences can be eliminated as ''j'' increases. Indeed, for each starting value ''a'' which is not a counterexample to the Collatz conjecture, there are integers ''j'', ''k'', and ''b'' such that 2<sup>''j''</sup>''n'' + ''a'' → 3<sup>''k''</sup>''n'' + ''b'' in ''j''+''k'' steps, and such that the target value is less than the starting value. As ''j'' increases, the search only needs to check those residues ''a'' that are not eliminated by lower values of ''j''. For example, the only surviving residues mod 32 are 7, 15, 27, and 31; and only 573162 residues survive mod 2<sup>25</sup> = 33554432. == Syracuse function == If ''k'' is an odd integer, then 3''k'' + 1 is even, so we can write 3''k'' + 1 = 2<sup>a</sup>''k''&prime;, with ''k''' odd and a ≥ 1. We define a function ''f'' from the set <math>I</math> of odd integers into itself, called the ''Syracuse Function,'' by taking&nbsp;''f''&nbsp;(''k'')&nbsp;=&nbsp;''k''&prime; {{OEIS|id=A075677}}. Some properties of the Syracuse function are: * ''f'' (4''k'' + 1) = ''f'' (''k'') for all ''k'' in <math>I</math>. * For all ''p ≥ 2'' and ''h'' odd, ''f'' <sup>''p'' - 1</sup>(2 <sup>''p''</sup> ''h'' - 1) = 2 3 <sup>''p'' - 1</sup>''h'' - 1 (see [[function composition#Functional powers|here]] for the notation). * For all odd ''h'', ''f'' (2''h'' - 1) ≤ (3''h'' - 1)/2 The Syracuse Conjecture is that for all ''k'' in <math>I</math>, there exists an integer ''n'' ≥ 1 such that ''f'' <sup>''n''</sup>(''k'') = 1. Equivalently, let ''E'' be the set of odd integers ''k'' for which there exists an integer ''n'' ≥ 1 such that ''f'' <sup>''n''</sup>(''k'') = 1. The problem is to show that ''E'' = <math>I</math>. The following is the beginning of an attempt at a proof by induction: 1, 3, 5, 7, and 9 are known to exist in ''E''. Let ''k'' be an odd integer greater than 9. Suppose that the odd numbers up to and including ''k'' - 2 are in ''E'' and let us try to prove that ''k'' is in ''E''. As ''k'' is odd, ''k'' + 1 is even, so we can write ''k'' + 1 = 2<sup>''p''</sup>''h'' for ''p'' ≥ 1, ''h'' odd, and ''k'' = 2<sup>''p''</sup>''h''-1. Now we have: * If ''p'' = 1, then ''k'' = 2''h'' - 1. It is easy to check that ''f'' (''k'') < ''k '', so ''f'' (''k'') ∈ ''E''; hence ''k'' ∈ ''E''. * If ''p'' ≥ 2 and ''h'' is a multiple of 3, we can write ''h'' = 3''h&prime;''. Let ''k&prime;'' = 2<sup>''p'' + 1</sup>''h&prime;'' - 1; we have ''f'' (''k&prime;'') = ''k'' , and as ''k&prime;'' < ''k'' , ''k&prime;'' is in ''E''; therefore ''k'' = ''f'' (''k&prime;'') ∈ ''E''. * If ''p'' ≥ 2 and ''h'' is not a multiple of 3 but ''h'' ≡ (-1)<sup>''p''</sup> mod 4, we can still show that ''k'' ∈ ''E''. ([http://www.geocities.com/dwakeel/syracuse.htm Cf.]) The problematic case is that where ''p'' ≥ 2 , ''h'' not multiple of 3 and ''h'' ≡ (-1)<sup>''p+1''</sup> mod 4. Here, if we manage to show that for every odd integer ''k''&prime;, 1 ≤ ''k''&prime; ≤ ''k''-2 ; 3''k''&prime; ∈ ''E'' we are done. ([http://www.geocities.com/dwakeel/syracuse.htm Cf.]). == See also == * [[Residue class-wise affine groups]] * [[Modular arithmetic]] == References and external links == * [[Jeffrey Lagarias|Jeffrey C. Lagarias]]. [http://arxiv.org/abs/math.NT/0309224 ''The 3x + 1 problem: An annotated bibliography (1963–2000)'']. * Jeffrey C. Lagarias. [http://arxiv.org/abs/math.NT/0608208 ''The 3x + 1 problem: An annotated bibliography, II (2001–)'']. * Jeffrey C. Lagarias. [http://www.cecm.sfu.ca/organics/papers/lagarias/ ''The 3x + 1 problem and its generalizations''], American Mathematical Monthly Volume 92, 1985, pp. 3–23. * {{SpringerEOM | urlname=S/s110330 | title=Syracuse problem | author=Jeffrey C. Lagarias}} * Günther J. Wirsching. The Dynamical System Generated by the <math>3n+1</math> Function. Number 1681 in Lecture Notes in Mathematics. Springer-Verlag, 1998. * An ongoing [[distributed computing]] [http://www.ericr.nl/wondrous/index.html project] by Eric Roosendaal verifies the Collatz conjecture for larger and larger values. * Another ongoing [[distributed computing]] [http://www.ieeta.pt/~tos/3x+1.html project] by Tomás Oliveira e Silva continues to verify the Collatz conjecture (with fewer statistics than Eric Roosendaal's page but with further progress made). * {{MathWorld | urlname=CollatzProblem | title=Collatz Problem}} * {{PlanetMath | urlname=CollatzProblem | title=Collatz Problem}} * [http://members.aol.com/mensanator666/collatz/hailstone.htm Hailstone Patterns] discusses different resonators along with using important numbers in the problem (like 6 and 3^5) to discover patterns. * [http://www.numbertheory.org/3x+1/ Keith Matthews' 3x+1 page: Review of progress, plus various programs] * Ohira, Reiko & Yamashita, Michinori [http://www.ris.ac.jp/yamasita/open/p-col.pdf A generalization of the Collatz problem] * URATA, Toshio [http://auemath.aichi-edu.ac.jp/~turata/Fall.files/CTZVI.pdf Some Holomorphic Functions connected with the Collatz Problem] * Matti K. Sinisalo: [http://www.geocities.com/mattiksinisalo/collatz.doc On the minimal cycle lengths of the Collatz sequences], Preprint, June 2003, University of Oulu * Paul Stadfeld: [http://home.versatel.nl/galien8/blueprint/blueprint.html Blueprint for Failure: How to Construct a Counterexample to the Collatz Conjecture] * Liesbeth De Mol: [http://logica.ugent.be/liesbeth/TagColOK.pdf "Tag systems and Collatz-like functions"], Preprint (Nr. 314), to appear in ''Theoretical Computer Science''. * Kawasaki Hiroyuki: [http://www.crt.or.jp/~kokochi/kaknodK/kaknodKe.htm] Claimed proof of Collatz conjecture *{{youtube|id=BhR-iIjhTNM|title=Collatz Iterations on the Ulam Spiral grid}} * [http://demonstrations.wolfram.com/CollatzPaths/ Collatz Paths] by Jesse Nochella, [[The Wolfram Demonstrations Project]]. ==Footnotes== {{reflist}} [[Category:Conjectures]] [[Category:Number theory]] [[Category:Unsolved problems in mathematics]] [[ar:حدسية سيراكيز]] [[bn:কোলাজ অনুমান]] [[da:Collatz-formodningen]] [[de:Collatz-Problem]] [[es:Conjetura de Collatz]] [[fa:حدس کولاتز]] [[fr:Conjecture de Syracuse]] [[ko:콜라츠 추측]] [[is:Tilgáta Collatz]] [[it:Congettura di Collatz]] [[he:השערת קולץ]] [[nl:Vermoeden van Collatz]] [[ja:コラッツの問題]] [[pl:Problem Collatza]] [[pt:Conjectura de Collatz]] [[sl:Collatzova domneva]] [[fi:Collatzin konjektuuri]] [[sv:Collatz antagande]] [[zh:考拉兹猜想]]