Ratio Mathematica Volume 40, 2021, pp. 225-246 225 About two countable families in the finite sets of the Collatz Conjecture Michele Ventrone* Abstract With 𝒕 ∈ β„• we define the sets 𝑲𝒕 and 𝑲𝒕 βˆ— containing all positive integers that converge to 1 in t iterations in the form of Collatz algorithm. The following are the properties of the {𝑲𝒕}π’•βˆˆβ„• and {𝑲𝒕 βˆ—}π’•βˆˆβ„•: countability, empty intersection between the elements of the same family, and - at the end of the work - we conjecture that both of the two families are a partition of β„•πŸŽ. We demonstrate also that each set 𝑲𝒕 and 𝑲𝒕 βˆ— is the union of two sets, a set includes even positive integers, the other, if it is non-empty, includes odd positive integers different from 1 and we go on proving that the maximum of each set 𝑲𝒕 and 𝑲𝒕 βˆ— is 2t and that 𝑲𝒕 ∩ 𝑲𝒕 βˆ— = {πŸπ’•}. Keywords: Collatz conjecture, Syracuse problem, 3n+1 problem, Hailstone numbers. 2010 Mathematics Subject Classification: 11A25, 11N56, 11M06. * Received on 2021-03-30. Accepted on 2021-05-30, 2021. Published on 2021-06-30. doi: 10.23755/rm.v39i0.5xx. ISSN: 1592-7415. eISSN: 2282-8214. doi: 10.23755/rm.v40i1.585 Β©The Authors. This paper is published under the CC-BY licence agreement. micheleventrone57@gmail.com. About two countable families in the finite sets of the Collatz Conjecture 226 1 Introduction Let us consider the Collatz conjecture (Leggerini, 2004), also known as the 3𝑛 + 1 problem. We start from a positive integer n, if it is even we divide it by two, if it is odd we multiply it by three and add one to it, then we start over by applying the same rules on the number obtained. For example, starting from 3 the sequence is generated: 3, 10, 5, 16, 8, 4, 2, 1. In the second form of the algorithm of 3𝑛 + 1 we calculate 3𝑛+1 2 if n is odd. With 3 we obtain the sequence 3, 5, 8, 4, 2, 1. It is conjectured that, from any positive integer we start, the sequences always arrive at 1 in a finite number of steps. It seems that all trajectories fall into the banal cycle 4, 2, 1 if n > 2. The conjecture has not yet been proven and many mathematicians believe the question be undecidable (Conway, J. H, 1972). By applying the algorithm to a positive integer n, a sequence of integers is generated which we will call a sequence or trajectory of n which we will denote with 𝑇(𝑛) (ΠΎr π‘‡βˆ—(𝑛) with the second form of the algorithm). For example 𝑇(5) = {5,16,8,4,2,1} and π‘‡βˆ—(3) = {3,5,8,4,2,1}. Let β„• = {0, 1, 2 … } and β„•0 = {1, 2, 3 … }. If 𝑖 ∈ β„• and 𝑛 ∈ β„•0, we denote by 𝑇𝑖 (𝑛) the element of place i in the trajectory 𝑇(𝑛). If 𝑖 = 0 we set 𝑇0(𝑛) = 𝑛. The same meaning will have 𝑇𝑖 βˆ—(𝑛). For example 𝑇0(5) = 5, 𝑇3(5) = 4, 𝑇2 βˆ—(3) = 8. We define convergent a trajectory that contains the number 1. In any trajectory containing 1 we will ignore the terms subsequent. If the trajectory generated by the integer n converges we will say that the number n converges. Any number of a trajectory will be treated as a positive integer. The term ”t-convergent” will be equivalent to ”convergent in t iterations”. We will call the number t the convergence time. The notation π‘˜π‘‘ will indicate that the positive integer k is t- convergent. In the following TC will be the set of convergence times of the converging positive integers. 2 The two forms of Collatz Conjecture First form. With 𝑛 ∈ β„•0 e 𝑖 ∈ β„•0 the algorithm is the iteration of the function: 𝑇𝑖 (𝑛) = { π‘‡π‘–βˆ’1(𝑛) 2 if π‘‡π‘–βˆ’1(𝑛) ≑ 0(π‘šπ‘œπ‘‘2) 3 β‹… π‘‡π‘–βˆ’1(𝑛) + 1 if π‘‡π‘–βˆ’1(𝑛) ≑ 1(π‘šπ‘œπ‘‘2) (2.1) with 𝑇0(𝑛) = 𝑛 if 𝑖 = 0. Second form. With 𝑛 ∈ β„•0 and 𝑖 ∈ β„•0, the algorithm is the iteration of the function: Michele Ventrone 227 𝑇𝑖 βˆ—(𝑛) = { π‘‡π‘–βˆ’1 βˆ— (𝑛) 2 if π‘‡π‘–βˆ’1 βˆ— (𝑛) ≑ 0(π‘šπ‘œπ‘‘ 2) 3π‘‡π‘–βˆ’1 βˆ— (𝑛)+1 2 if π‘‡π‘–βˆ’1 βˆ— (𝑛) ≑ 1(π‘šπ‘œπ‘‘ 2) (2.2) with 𝑇0 βˆ—(𝑛) = 𝑛 𝑖𝑓 𝑖 = 0. 3 Construction of the sets K Let us put in the same set 𝐾𝑑 the totality of positive integers t-convergent with the algorithm in the first form: βˆ€π‘‘ ∈ β„•, 𝐾𝑑 = {π‘˜ ∈ 𝑁0: π‘˜ = π‘˜π‘‘ }. (3.1) For example, applying the algorithm in the first form: 𝐾0 = {1} because 1 converges to 1 in zero iterations; 𝐾1 = {2} because 2 converges to 1 in an iteration; 𝐾2 = {4} because 4 converges to 1 in two iterations; … … … … If the Collatz algorithm is used in the second form, in (3.1) we will add to 𝐾𝑑 and its elements the symbol βˆ—, that is: βˆ€π‘‘ ∈ β„•, 𝐾𝑑 βˆ— = {π‘˜βˆ— ∈ β„•0: π‘˜ βˆ— = π‘˜π‘‘ βˆ—}. (3.2) Proposition 3.1. (Basic) βˆ€π‘‘ ∈ β„•, 𝐾𝑑 e 𝐾𝑑 βˆ— are non-empty. Proof. Trivially: whatever 𝑑 ∈ β„•, the number 2𝑑 converges to 1 in t iterations, hence in 𝐾𝑑 there is at least 2 𝑑 . For the same reason π‘˜π‘‘ βˆ— is also non-empty. β–‘ We consider the set TC of all times of convergence. Since each 𝑑 ∈ β„• can be associated with a 𝐾𝑑 and a 𝐾𝑑 βˆ— by means of 2𝑑 and vice versa, we can state that 𝑇𝐢 = β„• and that the families {𝐾𝑑 }π‘‘βˆˆβ„• and {𝐾𝑑 βˆ—}π‘‘βˆˆβ„• are countable. The following corollaries then hold. Corollary 3.2. Any positive or null integer is a time of convergence. About two countable families in the finite sets of the Collatz Conjecture 228 Corollary 3.3. Each of the families {𝐾𝑑 }π‘‘βˆˆβ„• and {𝐾𝑑 βˆ—}π‘‘βˆˆβ„• is countable. Proposition 3.4. If 𝑑1 π‘Žπ‘›π‘‘ 𝑑2, with 𝑑1 β‰  𝑑2, are in the set TC, then it results: i) 𝐾𝑑1 ∩ 𝐾𝑑2 = βˆ… i*) 𝐾𝑑1 βˆ— ∩ 𝐾𝑑2 βˆ— = βˆ… . Proof. i) Algorithm in the first form. By Proposition 3.1, 𝐾𝑑1 and 𝐾𝑑2 are non-empty. Assume that 𝐾𝑑1 ∩ 𝐾𝑑2 β‰  βˆ… , with 𝑑1 β‰  𝑑2. If π‘˜ ∈ 𝐾𝑑1 ∩ 𝐾𝑑2 then k must converge in the same number of iterations, so 𝑑1 = 𝑑2, against the hypothesis. Therefore 𝐾𝑑1 ∩ 𝐾𝑑2 = βˆ… . β€’ i*) Algorithm in the second form. The proof is similar to the previous one: just insert the asterisk to the sets 𝐾𝑑 . β–‘ Each family {𝐾𝑑 }π‘‘βˆˆβ„• and {𝐾𝑑 βˆ—}π‘‘βˆˆβ„• divides β„•0 into classes that we cannot consider at the moment of equivalence. 4 Decomposition of sets K Let 𝑑 ∈ β„•0. Applying the first form of the Collatz algorithm we will prove that each set 𝐾𝑑 is formed by a set 𝐴𝑑 and a set 𝐡𝑑, that is 𝐾𝑑 = 𝐴𝑑 βˆͺ 𝐡𝑑 with 𝐴𝑑 containing only even numbers and 𝐡𝑑 empty or containing only odd numbers different from 1. Applying the second form of the Collatz algorithm we will prove that 𝐾𝑑 βˆ— = 𝐴𝑑 βˆ— βˆͺ 𝐡𝑑 βˆ— with 𝐴𝑑 βˆ— containing only even numbers and 𝐡𝑑 βˆ— empty or containing only odd numbers different from 1. We will also prove that the elements of 𝐾𝑑 can be obtained from all the elements of πΎπ‘‘βˆ’1 and the elements of 𝐾𝑑 βˆ— can be obtained from all elements of πΎπ‘‘βˆ’1 βˆ— . If t = 0 it is 𝐾0 = 𝐾0 βˆ— = {1} and therefore 2𝐾0 = 2𝐾0 βˆ— = {2} = 𝐾1 = 𝐾1 βˆ—. Some B sets are empty such as sets 𝐡1, 𝐡1 βˆ—, 𝐡2, 𝐡2 βˆ—, 𝐡3, 𝐡3 βˆ—, 𝐡4, 𝐡4 βˆ—, 𝐡6, 𝐡8, 𝐡10. I don't know if there are other empty B sets. In this study 𝐴0 = {1} , 𝐴0 βˆ— = {1} , 𝐡0 = βˆ… and 𝐡0 βˆ— = βˆ… . Let t ∈ β„•. Here we will assume that 𝐾𝑑+1 is made up of two sets of numbers: 1) by the doubles of the numbers of 𝐾𝑑; 2) from the integers 𝑏 β‰  1 which are odd solutions in β„•0 of the equation 3𝑏 + 1 = π‘˜π‘‘ , with π‘˜π‘‘ ∈ 𝐾𝑑 , π‘˜π‘‘ even and π‘˜π‘‘ β‰  4; and that 𝐾𝑑+1 βˆ— is formed by two sets of numbers: Michele Ventrone 229 1*) by the doubles of the numbers of 𝐾𝑑 βˆ—; 2*) from the integers π‘βˆ— β‰  1 which are odd solutions in β„•0 of the equation 3π‘βˆ—+1 2 = π‘˜π‘‘ βˆ— , with π‘˜π‘‘ βˆ— ∈ 𝐾𝑑 βˆ— and π‘˜π‘‘ βˆ— β‰  2. Called P the set of even positive integers, we denote by 2𝐾𝑑 (set of even derivatives of the first type or set of even derivatives of 𝐾𝑑 or set of doubles of the first type) the set obtained by doubling all the numbers of 𝐾𝑑: βˆ€π‘‘ ∈ β„•, 2𝐾𝑑 = {π‘Ž ∈ 𝑃: π‘Ž = 2π‘˜π‘‘ , π‘˜π‘‘ ∈ 𝐾𝑑 }. (4.1) We denote by 2𝐾𝑑 βˆ— (set of even derivatives of the second type or set of even derivatives of 𝐾𝑑 βˆ— or set of doubles of the second type) the set obtained by doubling all the numbers of 𝐾𝑑 βˆ—: βˆ€π‘‘ ∈ β„•, 2𝐾𝑑 βˆ— = {π‘Žβˆ— ∈ 𝑃: π‘Žβˆ— = 2π‘˜π‘‘ βˆ—, π‘˜π‘‘ βˆ— ∈ 𝐾𝑑 βˆ—} . (4.2) We denote by 𝐡𝑑+1 (set of odd derivatives of 𝐾𝑑 or set of odd derivatives of the first type) the numbers with the property 2) and by 𝐡𝑑+1 βˆ— (set of odd derivatives of 𝐾𝑑 βˆ— to set of odd derivatives of second type) numbers with the property 2βˆ—). Called D the set of integers odd positive, the set of odd derivatives of 𝐾𝑑 we have: βˆ€π‘‘ ∈ β„•, 𝐡𝑑+1 = {𝑏 ∈ 𝐷 βˆ’ {1}: 3𝑏 + 1 = π‘˜π‘‘ , π‘˜π‘‘ ∈ 𝐾𝑑 ∩ 𝑃, π‘˜π‘‘ β‰  4} (4.3) while the set of odd derivatives of 𝐾𝑑 βˆ— is βˆ€π‘‘ ∈ β„•, 𝐡𝑑+1 βˆ— = {π‘βˆ— ∈ 𝐷 βˆ’ {1}: 3π‘βˆ—+1 2 = π‘˜π‘‘ βˆ—, π‘˜π‘‘ βˆ— ∈ 𝐾𝑑 βˆ—, π‘˜π‘‘ βˆ— β‰  2} . (4.4) Theorem 4.1. (Theorem of the inclusion of doubles) The even derivative of 𝐾𝑑 (𝐾𝑑 βˆ—) is contained in 𝐾𝑑+1(𝐾𝑑+1 βˆ— ), that is: a) βˆ€π‘‘ ∈ β„•, 2𝐾𝑑 βŠ† 𝐾𝑑+1 b) βˆ€π‘‘ ∈ β„•, 2𝐾𝑑 βˆ— βŠ† 𝐾𝑑+1 βˆ— . (4.5) Proof. By Corollary 3.2 every t is a time of convergence. Given 𝑑 ∈ β„•, we consider 𝐾𝑑 (which is non-empty by Proposition 3.1). Trivially: βˆ€π‘˜π‘‘ ∈ 𝐾𝑑, the trajectory 𝑇(π‘˜π‘‘ ) = {π‘˜π‘‘ , … ,4,2,1} is contained in the trajectory 𝑇(2π‘˜π‘‘ ) = {2π‘˜π‘‘ , π‘˜π‘‘ , … ,4,2,1}. This means that 2π‘˜π‘‘ is (t+1)-convergent, so 2π‘˜π‘‘ ∈ 𝐾𝑑+1. β€’ About two countable families in the finite sets of the Collatz Conjecture 230 If 𝐾𝑑+1 is devoid of odd numbers, only the sign of equality holds. To prove it, let's suppose that 𝐾𝑑+1 is devoid of odd numbers and that, absurdly, it contains an even number π‘Žπ‘‘+1 which does not is double of any number of 𝐾𝑑. Since the even π‘Žπ‘‘+1 is also is (t+1)-convergent, the trajectory 𝑇(π‘Žπ‘‘+1) = {π‘Žπ‘‘+1, π‘Žπ‘‘+1 2 , … ,4,2,1} will contain the trajectory 𝑇 ( π‘Žπ‘‘+1 2 , ) = { π‘Žπ‘‘+1 2 , … ,4,2,1} so π‘Žπ‘‘+1 2 is t-convergent, that is π‘Žπ‘‘+1 2 ∈ 𝐾𝑑 , against our hypothesis. It follows that 2𝐾𝑑 coincides with 𝐾𝑑+1 if this is devoid of odd. Then the relation a) of (4.5) holds for the arbitrariness of t. β€’ In the case of 2𝐾𝑑 βˆ— proceed in the same way, mutatis mutandis. β–‘ Theorem 4.2. (Odd derivative theorem of the first type) Let π‘˜π‘‘ be even and π‘˜π‘‘ β‰  4 . If there is a positive integer b satisfying the equation 3𝑏 + 1 = π‘˜π‘‘ (4.6) then 𝑏 = π‘˜π‘‘βˆ’1 3 (4.7) belongs to 𝐡𝑑+1. Proof. Let b and 𝐾𝑑 satisfy the hypotheses. Since 𝑏 is odd and different from 1, its successor is π‘˜π‘‘, because to b is applied (2.1), so the trajectory 𝑇(𝑏) = 𝑇 ( π‘˜π‘‘βˆ’1 3 ) = { π‘˜π‘‘βˆ’1 3 , π‘˜π‘‘ , … ,4,2,1} contains the trajectory 𝑇(π‘˜π‘‘ ) = {π‘˜π‘‘ , … ,4,2,1}. This means that b converges in t + 1 iterations, that is 𝑏 ∈ 𝐡𝑑+1. β–‘ Recall that an odd derivative 𝐡𝑑 either is empty or is formed only by odd positive different from 1. Theorem 4.3. (Theorem of strict inclusion of odd derivatives of the first type) The odd derivative of 𝐾𝑑 (𝐾𝑑 βˆ—) is strictly contained in 𝐾𝑑+1(𝐾𝑑+1 βˆ— ), that is: a) βˆ€π‘‘ ∈ β„•, 𝐡𝑑+1 βŠ‚ 𝐾𝑑+1 b) βˆ€π‘‘ ∈ β„•, 𝐡𝑑+1 βˆ— βŠ‚ 𝐾𝑑+1 βˆ— . (4.8) Proof. By Proposition 3.1 every 𝐾𝑑+1(𝐾𝑑+1 βˆ— ) is non-empty because it contains at least the even number 2𝑑+1, therefore 𝐡𝑑+1 even if it were empty could not coincide with 𝐾𝑑+1(𝐾𝑑+1 βˆ— ). β–‘ Michele Ventrone 231 Theorem 4.4. (Theorem of the union of even and odd derivatives of the first type) The set 𝐾𝑑+1 is the union of the set of doubles of 𝐾𝑑 and of the odd derivative of 𝐾𝑑, that is: βˆ€π‘‘ ∈ β„•, 𝐾𝑑+1 = 2𝐾𝑑 βˆͺ 𝐡𝑑+1. (4.9) (remarkable equality, algorithm in first form) Proof. Let us consider 𝐾𝑑, with 𝑑 ∈ β„•. It is necessary to demonstrate that 1) there are no other even integers (t+1)-convergent beyond those of 2𝐾𝑑; 2) the odd numbers (t+1)-convergent are only those of 𝐡𝑑+1. We prove 1). We denote by 𝐴𝑑+1 the totality of even positive integers converging in t + 1 iterations that we know to be non-empty (each 𝐴𝑑 contains at least 2𝑑 ). It immediately turns out that βˆ€π‘‘ ∈ β„•, 2𝐾𝑑 βŠ† 𝐴𝑑+1. We show that βˆ€π‘‘ ∈ β„•, 2𝐾𝑑 = 𝐴𝑑+1. (4.10) If for a fixed 𝑑 ∈ β„• there were an even π‘Žπ‘‘+1 ∈ 𝐴𝑑+1 that was not double of any positive integer of 𝐾𝑑, it would be absurd because the trajectory 𝑇(π‘Žπ‘‘+1) = {π‘Žπ‘‘+1, π‘Žπ‘‘+1 2 , … 4,2,1} would contain the trajectory 𝑇 ( π‘Žπ‘‘+1 2 ) = { π‘Žπ‘‘+1 2 , … ,4,2,1} whose seed at π‘Žπ‘‘+1 2 ∈ 𝐾𝑑 and whose double π‘Žπ‘‘+1 is in 𝐴𝑑+1, against the hypothesis. Hence the strict inclusion cannot hold and, by the arbitrariness of t, (4.10) is true. β€’ We prove 2). With the same fixed 𝑑 ∈ β„•, we denote by 𝛽𝑑+1 the totality of the odd positive integers converging in t + 1 iterations . Obviously we have 𝐡𝑑+1 βŠ† 𝛽𝑑+1. We show that βˆ€π‘‘ ∈ β„•, 𝐡𝑑+1 = 𝛽𝑑+1 . (4.11) If for the fixed t, 𝛽𝑑+1 = βˆ…, then also 𝐡𝑑+1 = βˆ… and therefore 𝐾𝑑+1 = 2𝐾𝑑, that is (4.9) for the arbitrariness of t. Otherwise, for fixed t, let 𝛽𝑑+1 β‰  βˆ… . If there was an 𝑏𝑑+1 ∈ 𝛽𝑑+1 not coming by any even of 𝐾𝑑, that is such that 𝑏𝑑+1 βˆ‰ 𝐡𝑑+1 , then an absurdity would follow because the trajectory 𝑇(𝑏𝑑+1) = {𝑏𝑑+1, 3𝑏𝑑+1 + 1, … ,4,2,1} would contain the trajectory 𝑇(3𝑏𝑑+1 + 1) = {3𝑏𝑑+1 + 1, … ,4,2,1} whose even seed 3𝑏𝑑+1 + 1 = π‘˜π‘‘ ∈ 𝐾𝑑 , therefore, by Theorem 4.2, 𝑏𝑑+1 ∈ 𝐡𝑑+1 against the hypothesis. For this reason strict inclusion cannot be valid and, due to the arbitrariness of t (4.11) is true. β€’ About two countable families in the finite sets of the Collatz Conjecture 232 From 1) and 2) follows the remarkable equality (4.9). β–‘ By (4.10), (4.9) becomes βˆ€π‘‘ ∈ β„•, 𝐾𝑑+1 = 𝐴𝑑+1 βˆͺ 𝐡𝑑+1 . (4.12) (remarkable equality, algorithm in the first form) If for a given t the derivative 𝐡𝑑+1 of 𝐾𝑑 is empty, we have 𝐾𝑑+1 = 𝐴𝑑+1. (4.13) We now find the numbers of 𝐡𝑑+1 βˆ— . Theorem 4.5. (Theorem of the odd derivative of the second type) Let π‘˜π‘‘ βˆ— ∈ β„•0, π‘˜π‘‘ βˆ— β‰  2 . If there exists the positive integer b satisfying the equation 3π‘βˆ— + 1 = 2π‘˜π‘‘ βˆ— (4.14) then π‘βˆ— = 2π‘˜π‘‘ βˆ—βˆ’1 3 (4.15) belongs to 𝐡𝑑+1 βˆ— . Proof. Let π‘˜π‘‘ βˆ— and π‘βˆ— satisfy the hypotheses. Since π‘βˆ— is odd and different from 1, its successor is π‘˜π‘‘ βˆ—, because (2.2) is applied to π‘βˆ—, so the trajectory 𝑇(π‘βˆ—) = 𝑇 ( 2π‘˜π‘‘ βˆ—βˆ’1 3 ) = { 2π‘˜π‘‘ βˆ—βˆ’1 3 , π‘˜π‘‘ βˆ—, … ,4,2,1} contains the trajectory 𝑇(π‘˜π‘‘ βˆ—) = {π‘˜π‘‘ βˆ—, … ,4,2,1} . This means that π‘βˆ— converges in t + 1 iterations, that is π‘βˆ— ∈ 𝐡𝑑+1 βˆ— . β–‘ Recall that an odd derivative 𝐡𝑑 βˆ— o is either empty or is formed only by odd positive integers different from 1. As shown for (4.10) it results βˆ€π‘‘ ∈ β„•, 2𝐾𝑑 βˆ— = 𝐴𝑑+1 βˆ— (4.16) Michele Ventrone 233 where 𝐴𝑑+1 βˆ— is the totality of the even positive integers (t+1)-convergent, that is of the doubles of the numbers of 𝐾𝑑 βˆ—. Equation (4.16) is demonstrated how it is done for the first part of the proof of the Theorem 4.4 by adding the asterisk βˆ— to the 2𝐾𝑑 and 𝐴𝑑 sets. Equation (4.16) will occur in the proof of first part of Theorem 4.6. Theorem 4.6. (Theorem of the union of even and odd derivatives of the second type) The set 𝐾𝑑+1 βˆ— is the union of the set of doubles of 𝐾𝑑 βˆ— and the odd derivative of 𝐾𝑑 βˆ—, that is: βˆ€π‘‘ ∈ β„•, 𝐾𝑑+1 βˆ— = 2𝐾𝑑 βˆ— βˆͺ 𝐡𝑑+1 βˆ— . (4.17) (remarkable equality, algorithm in the second form) Proof. We proceed as in the proof of Theorem 4.4 adding the asterisk βˆ— to all the sets and considering, in the second part, ( 3𝑏𝑑+1 βˆ— +1 2 ) as successor of 𝑏𝑑+1 βˆ— ∈ 𝛽𝑑+1 βˆ— . β–‘ By (4.16), (4.17) can be written βˆ€π‘‘ ∈ β„•, 𝐾𝑑+1 βˆ— = 𝐴𝑑+1 βˆ— βˆͺ 𝐡𝑑+1 βˆ— (4.18) (remarkable equality, algorithm in the second form) and if, for a certain t, the derivative 𝐡𝑑+1 βˆ— of 𝐾𝑑+1 βˆ— it is empty, then 𝐾𝑑+1 βˆ— = 𝐴𝑑+1 βˆ— . (4.19) 5 Examples To obtain the set 𝐾𝑑+1 it will be necessary to double all the numbers π‘˜π‘‘ of 𝐾𝑑 in order to have 𝐴𝑑+1 and it will be necessary to determine all the numbers 𝑏 ∈ 𝐡𝑑+1 starting from the even numbers π‘˜π‘‘ of 𝐾𝑑, that is, it will be necessary to verify if π‘˜π‘‘ βˆ’ 1 is divisible by three when π‘˜π‘‘ is even with π‘˜π‘‘ β‰  4 (Theorem 4.2 and definition of 𝐡𝑑+1 in (4.3)). β–Ί We determine the sets 𝐾8 and 𝐾9. 𝐾8 We use the set 𝐾7 = {3,20,21,128} . We have 𝐴8 = 2𝐾7 = {6,40,42,256} . It turns out 𝐡8 = βˆ… since none of the equations About two countable families in the finite sets of the Collatz Conjecture 234 (1) 3b + 1 = 20 (2) 3b + 1 = 128 has solutions in β„•0. Hence 𝐾8 = 𝐴8 βˆͺ βˆ… = {6,40,42,256}. 𝐾9 We use the set 𝐾8 = {6,40,42,256}. We have 𝐴9= 2𝐾8 = {12, 80, 84, 512}. We solve in β„•0 the following equations: (1) 3b + 1 = 6 (2) 3b + 1 = 40 (3) 3b + 1 = 42 (4) 3b + 1 = 256. The first and third equations have no solutions in β„•0. The second and fourth equations have as solutions in β„•0 13 and 85 respectively, therefore 𝐡9 = {13, 85}. Thus 𝐾9 = 𝐴9 βˆͺ 𝐡9 = {12,80,84,512} βˆͺ {13,85} = {12,13,80,84,85,512} . In the same way they are obtained 𝐾10 = {4 βˆ’ 26 βˆ’ 160 βˆ’ 168 βˆ’ 170 βˆ’ 1024} 𝐾11 = {48 βˆ’ 52 βˆ’ 53 βˆ’ 320 βˆ’ 336 βˆ’ 340 βˆ’ 341 βˆ’ 2048} 𝐾12 = {17 βˆ’ 96 βˆ’ 104 βˆ’ 106 βˆ’ 113 βˆ’ 640 βˆ’ 672 βˆ’ 680 βˆ’ 682 βˆ’ 4096} … The underlined numbers are the odd derivatives of the previous set. To obtain the set 𝐾𝑑+1 βˆ— it will be necessary to double all the numbers π‘˜π‘‘ βˆ— of 𝐾𝑑 βˆ— in order to have 𝐴𝑑+1 βˆ— and it will be necessary to determine all the numbers π‘βˆ— ∈ 𝐡𝑑+1 βˆ— starting from each π‘˜π‘‘ βˆ— of 𝐾𝑑 βˆ—, that is, it will be necessary to verify whether 2π‘˜π‘‘ βˆ— βˆ’ 1 is divisible by three when π‘˜π‘‘ βˆ— β‰  2 (Theorem 4.5 and definition of 𝐡𝑑+1 βˆ— in (4.4)). β–Ί We determine the sets 𝐾5 βˆ— and 𝐾6 βˆ—. 𝐾5 βˆ— We consider 𝐾4 βˆ— = {5,16} . Its even derivative is 𝐴5 βˆ— = {10,32}. Of the two equations Michele Ventrone 235 (1) 3π‘βˆ—+1 2 = 5 (2) 3π‘βˆ—+1 2 = 16 only the first admits in β„•0 the solution 𝑏 βˆ—= 3 therefore 𝐡5 βˆ— = {3} Π΅ 𝐾5 βˆ— = 𝐴5 βˆ— βˆͺ 𝐡5 βˆ— = {3,10,32}. 𝐾6 βˆ— We consider 𝐾5 βˆ— = {3,10,32} . Its even derivative is 𝐴6 βˆ— = {6,20,64} . Of the three equations (1) 3π‘βˆ—+1 2 = 3 (2) 3π‘βˆ—+1 2 = 10 (3) 3π‘βˆ—+1 2 = 32 only the third has solution π‘βˆ— = 21 in β„•0. Hence 𝐡6 βˆ— = {21} Π΅ 𝐾6 βˆ— = 𝐴6 βˆ— βˆͺ 𝐡6 βˆ— = {6,20,21,64} . In the same way they are obtained 𝐾7 βˆ— = {12 βˆ’ 13βˆ— βˆ’ 40 βˆ’ 42 βˆ’ 128} 𝐾8 βˆ— = {24 βˆ’ 26 βˆ’ 80 βˆ’ 84 βˆ’ 85βˆ— βˆ’ 256} 𝐾9 βˆ— = {17βˆ— βˆ’ 48 βˆ’ 52 βˆ’ 53βˆ— βˆ’ 160 βˆ’ 168 βˆ’ 170 βˆ’ 512} 𝐾10 βˆ— = {11βˆ— βˆ’ 34 βˆ’ 35βˆ— βˆ’ 96 βˆ’ 104 βˆ’ 106 βˆ’ 113βˆ— βˆ’ 320 βˆ’ 336 βˆ’ 340 βˆ’ 341βˆ— βˆ’ 1024} ... The numbers with an asterisk are the odd derivatives of the previous set. About two countable families in the finite sets of the Collatz Conjecture 236 6 The maxima of 𝑲𝒕 and 𝑲𝒕 βˆ— By examining the sets K, we can suppose that the number 2𝑑 is the maximum of every set 𝐾𝑑 and of every 𝐾𝑑 βˆ—. This is confirmed by the subsequent Theorem 6.2. The following Lemma 6.1 contains some obvious conclusions. Lemma 6.1. i) If π‘˜π‘‘ ∈ 𝐾𝑑 then 2π‘˜π‘‘ ∈ 𝐴𝑑+1, βˆ€π‘‘ ∈ β„• ii) If π‘Žπ‘‘ ∈ 𝐴𝑑 then 2π‘Žπ‘‘ ∈ 𝐴𝑑+1, βˆ€π‘‘ ∈ β„•0 iβˆ—) If π‘˜π‘‘ βˆ— ∈ 𝐾𝑑 βˆ— then 2π‘˜π‘‘ βˆ— ∈ 𝐴𝑑+1 βˆ— , βˆ€π‘‘ ∈ β„• iiβˆ—) If π‘Žπ‘‘ βˆ— ∈ 𝐴𝑑 βˆ— then 2π‘Žπ‘‘ βˆ— ∈ 𝐴𝑑+1 βˆ— , βˆ€π‘‘ ∈ β„•0 . Proof. Recall that (4.10) and (4.16) hold. i) Let π‘˜π‘‘ ∈ 𝐾𝑑, with 𝑑 ∈ β„•. The trajectory 𝑇(π‘˜π‘‘ ) is contained in the trajectory 𝑇(2π‘˜π‘‘ ) = {2π‘˜π‘‘ , π‘˜π‘‘ , … ,4,2,1} because 2π‘˜π‘‘ is an even that converges in t + 1 iterations, that is 2π‘˜π‘‘ ∈ 𝐴𝑑+1. β€’ ii) Let π‘Žπ‘‘ ∈ 𝐴𝑑, with t ∈ β„•0. Since 𝐴𝑑 βŠ† 𝐾𝑑 is also π‘Žπ‘‘ ∈ 𝐾𝑑. Applying i) it follows that 2π‘Žπ‘‘ ∈ 𝐴𝑑+1 βˆ€π‘‘ ∈ β„•0. β€’ The iβˆ—) and iiβˆ—) prove to be the i) and ii) respectively, just asterisking the sets 𝐾𝑑, 𝐴𝑑 and their elements. β–‘ Theorem 6.2. (Maxima theorem of 𝐾𝑑 and 𝐾𝑑 βˆ—) 𝑖) βˆ€π‘‘ ∈ β„•0, π‘šπ‘Žπ‘₯(𝐾𝑑 ) = 2 𝑑 i*) βˆ€π‘‘ ∈ β„•0, π‘šπ‘Žπ‘₯(𝐾𝑑 βˆ—) = 2𝑑 . Proof. i) We will proceed by induction using the remarkable equality (4.12). If 𝑑 = 1 then max π‘šπ‘Žπ‘₯(𝐾1) = 2 1 = 2. Let us fix a 𝑑 > 1 and let, by inductive hypothesis π‘šπ‘Žπ‘₯(𝐾𝑑 ) = 2 𝑑 . (6.1) We will prove that it is also π‘šπ‘Žπ‘₯(𝐾𝑑+1) = 2 𝑑+1. To do this, it will be necessary to prove that 1) max π‘šπ‘Žπ‘₯(𝐴𝑑+1) = 2 𝑑+1 and 2) every number of 𝐡𝑑+1 is less than 2 𝑑+1. Michele Ventrone 237 First part 1) We show that every number of 𝐴𝑑+1 is less than or equal to 2 𝑑+1 and that 2𝑑+1is in 𝐴𝑑+1. Let π‘˜π‘‘ ∈ 𝐾𝑑. Then, by hypothesis (6.1) βˆ€π‘˜π‘‘ ∈ 𝐾𝑑 , π‘˜π‘‘ ≀ 2 𝑑 . (6.2) By the i) of Lemma 6.1 2π‘˜π‘‘ ∈ 𝐴𝑑+1. (6.3) From (6.2) it follows that βˆ€π‘˜π‘‘ ∈ 𝐾𝑑 , 2π‘˜π‘‘ ≀ 2 𝑑+1. (6.4) Since for the inductive hypothesis (6.1) it is 2𝑑 ∈ 𝐾𝑑 , then, for the remarkable equality (4.12), we have 2𝑑 ∈ 𝐴𝑑, from which, for the ii) of Lemma 6.1, it follows that 2𝑑+1 ∈ 𝐴𝑑+1. (6.5) From (6.3), (6.4) and (6.5) we obtain that π‘šπ‘Žπ‘₯(𝐴𝑑+1) = 2 𝑑+1. β€’ Second part 2) If 𝐡𝑑+1 = βˆ… from (4.12) it follows that 𝐾𝑑+1 = 𝐴𝑑+1 and from π‘šπ‘Žπ‘₯(𝐴𝑑+1) = 2𝑑+1 (First part) it follows that π‘šπ‘Žπ‘₯(𝐾𝑑+1) = 2 𝑑+1. Let 𝐡𝑑+1 β‰  βˆ…. We show that every element 𝑏𝑑+1 of 𝐡𝑑+1 is less than 2 𝑑+1. The numbers of 𝐡𝑑+1 are the odd numbers of the form (4.7): 𝑏𝑑+1 = π‘˜π‘‘βˆ’1 3 π‘π‘œπ‘› π‘˜π‘‘ ∈ 𝐾𝑑 π‘Žπ‘›π‘‘ π‘˜π‘‘ even (6.6) but, from π‘˜π‘‘ βˆ’ 1 < π‘˜π‘‘ we get that π‘˜π‘‘βˆ’1 3 < π‘˜π‘‘ (6.7) then from (6.6), (6.7) and (6.1) it follows that 𝑏𝑑+1 < π‘˜π‘‘ ≀ 2 𝑑 (6.8) About two countable families in the finite sets of the Collatz Conjecture 238 and therefore: βˆ€π‘π‘‘+1 ∈ 𝐡𝑑+1, 𝑏𝑑+1 < 2 𝑑+1, that is 2). From the first and the second part it follows that all the numbers of 𝐾𝑑+1 are less than or equal to 2𝑑+1and this proves the i). β€’ i*) We will proceed by induction using the remarkable equality (4.18). If 𝑑 = 1 then π‘šπ‘Žπ‘₯(𝐾1 βˆ—) = 21=2. Let, by inductive hypothesis, be π‘šπ‘Žπ‘₯(𝐾𝑑 βˆ—) = 2𝑑 con 𝑑 > 1 . (6.9) We will prove that it is also π‘šπ‘Žπ‘₯(𝐾𝑑+1 βˆ— ) = 2𝑑+1. To do this, it will be necessary to prove that 1*) π‘šπ‘Žπ‘₯(𝐴𝑑+1 βˆ— ) = 2𝑑+1 and 2*) every number of 𝐡𝑑+1 βˆ— is less than 2𝑑+1 . First part βˆ— 1*) The proof is similar to that of the first part of i), just adding the asterisk βˆ— to the sets 𝐴𝑑+1, 𝐾𝑑+1 and their elements. Therefore 2 𝑑+1 is the maximum of 𝐴𝑑+1 βˆ— and 1*) is proved. β€’ Second part βˆ— 2*) If 𝐡𝑑+1 βˆ— β‰  βˆ… from 4.18) it follows that 𝐾𝑑+1 βˆ— = 𝐴𝑑+1 βˆ— and from π‘šπ‘Žπ‘₯(𝐴𝑑+1 βˆ— ) = 2𝑑+1 (First part*) it follows that π‘šπ‘Žπ‘₯(𝐾𝑑+1 βˆ— ) = 2𝑑+1. Let 𝐡𝑑+1 βˆ— β‰  βˆ…. We show that every 𝑏𝑑+1 βˆ— of 𝐡𝑑+1 βˆ— is less than 2𝑑+1. The numbers of 𝐡𝑑+1 βˆ— are the odd numbers of the form (4.15): 𝑏𝑑+1 βˆ— = 2π‘˜π‘‘ βˆ—βˆ’1 3 , with π‘˜π‘‘ βˆ— ∈ 𝐾𝑑 βˆ— (6.10) but, from 2π‘˜π‘‘ βˆ— βˆ’ 1 < 2π‘˜π‘‘ βˆ— we get that 2π‘˜π‘‘ βˆ—βˆ’1 3 < 2π‘˜π‘‘ βˆ— (6.11) and for the inductive hypothesis (6.9) we also have that βˆ€π‘˜π‘‘ βˆ— ∈ 𝐾𝑑 βˆ—, 2π‘˜π‘‘ βˆ— ≀ 2𝑑+1 . (6.12) Michele Ventrone 239 Finally for (6.10), (6.11), (6.12) we can write that 𝑏𝑑+1 βˆ— < 2π‘˜π‘‘ βˆ— ≀ 2𝑑+1, then βˆ€π‘‘ > 1 all numbers 𝑏𝑑+1 βˆ— di 𝐡𝑑+1 βˆ— are less than 2𝑑+1. The 2*) is thus proved. β€’ From 1*) and from 2*) it follows that all integers of 𝐾𝑑+1 βˆ— are less than or equal to 2𝑑+1 and so i*) is also proved. β–‘ The following corollaries immediately follow from Theorem 6.2. Corollary 6.3. i) βˆ€π‘‘ ∈ β„•, π‘šπ‘Žπ‘₯(2𝐾𝑑 ) = 2 𝑑+1 iβˆ—) βˆ€π‘‘ ∈ β„•, π‘šπ‘Žπ‘₯(2𝐾𝑑 βˆ—) = 2𝑑+1 . Corollary 6.4. i) βˆ€π‘‘ ∈ β„•0, π‘šπ‘Žπ‘₯(𝐴𝑑 ) = 2 𝑑 iβˆ—) βˆ€π‘‘ ∈ β„•0, π‘šπ‘Žπ‘₯(𝐴𝑑 βˆ—) = 2𝑑 . Theorem 6.2 provides indications on the type of numbers contained in the sets K: either there is only 2𝑑 or there are positive integers less than or equal to 2𝑑 and this means that each set K is finite. Therefore, the following corollary can also be stated. Corollary 6.5. βˆ€π‘‘ ∈ β„•, 𝐾𝑑 π‘Žπ‘›π‘‘ 𝐾𝑑 βˆ— are finite. Each set K is formed by the finite numerical sets A and B. It follows that if B is non-empty then it has a maximum. Therefore the following corollary holds. Corollary 6.6. i) If for 𝑑 ∈ β„•0 is 𝐡𝑑 β‰  βˆ… then βˆƒ π‘šπ‘Žπ‘₯(𝐡𝑑 ) i*) If for 𝑑 ∈ β„•0 is 𝐡𝑑 βˆ— β‰  βˆ… then βˆƒ π‘šπ‘Žπ‘₯(𝐡𝑑 βˆ—) . In the following paragraph 7 we will investigate the maxima of the sets B. About two countable families in the finite sets of the Collatz Conjecture 240 7 On the maxima of the sets B We give a strict increase of the maxima of the sets B. Proposition 7.1. i) If 𝐡𝑑+1 β‰  βˆ… then βˆƒπ‘˜π‘‘ ∈ 𝐾𝑑 , π‘˜π‘‘ β‰  4, π‘˜π‘‘ even : π‘šπ‘Žπ‘₯(𝐡𝑑+1) < π‘˜π‘‘βˆ’1 2 i*) If 𝐡𝑑+1 βˆ— β‰  βˆ… then βˆƒπ‘˜π‘‘ βˆ— ∈ 𝐾𝑑 βˆ—, π‘˜π‘‘ βˆ— β‰  2: π‘šπ‘Žπ‘₯(𝐡𝑑+1 βˆ— ) < 2π‘˜π‘‘ βˆ—βˆ’1 2 . Proof. i) If 𝐡𝑑+1 β‰  βˆ…, then by definition of 𝐡𝑑+1 in correspondence of every odd 𝑏𝑑+1 ∊ 𝐡𝑑+1 will exist an even number π‘˜π‘‘ ∈ 𝐾𝑑 with π‘˜π‘‘ β‰  4 such that 𝑏𝑑+1 = π‘˜π‘‘βˆ’1 3 but π‘˜π‘‘βˆ’1 3 < π‘˜π‘‘βˆ’1 2 , then 𝑏𝑑+1 < π‘˜π‘‘βˆ’1 2 . Then, in particular, i) holds also for the maximum of 𝐡𝑑+1. β€’ i*) If 𝐡𝑑+1 βˆ— β‰  βˆ…, then by definition of 𝐡𝑑+1 βˆ— in correspondence of every odd 𝑏𝑑+1 βˆ— ∈ 𝐡𝑑+1 βˆ— will exist an even number π‘˜π‘‘ βˆ— ∈ 𝐾𝑑 βˆ— with π‘˜π‘‘ βˆ— β‰  2 such that 𝑏𝑑+1 βˆ— = 2π‘˜π‘‘ βˆ—βˆ’1 3 but 2π‘˜π‘‘ βˆ—βˆ’1 3 < 2 π‘˜π‘‘ βˆ—βˆ’1 2 , then 𝑏𝑑+1 βˆ— < 2 π‘˜π‘‘ βˆ—βˆ’1 2 . Then, in particular, also for the maximum of 𝐡𝑑+1 βˆ— holds i*). β–‘ From Proposition 7.1 follows the following corollary which gives a plus a bit more large of the maxima of the sets B. Corollary 7.2. i) If 𝐡𝑑+1 β‰  βˆ…, then π‘šπ‘Žπ‘₯(𝐡𝑑+1) < 2 𝑑 i*) If 𝐡𝑑+1 βˆ— β‰  βˆ…, then π‘šπ‘Žπ‘₯(𝐡𝑑+1 βˆ— ) < 2𝑑 . Proof. i) If 𝐡𝑑+1 β‰  βˆ…, then the inequality i) of Proposition 7.1 holds and also π‘˜π‘‘βˆ’1 2 < π‘˜π‘‘ but, by Theorem 6.2, the maximum of 𝐾𝑑 is 2 𝑑 , so π‘šπ‘Žπ‘₯(𝐡𝑑+1 βˆ— ) < 2𝑑 . β€’ i*) If 𝐡𝑑+1 βˆ— β‰  βˆ…, then the inequality i*) of Proposition 7.1 holds and also 2π‘˜π‘‘ βˆ—βˆ’1 2 < π‘˜π‘‘ βˆ— but, by Theorem 6.2, the maximum of 𝐾𝑑 βˆ— is 2𝑑 , so π‘šπ‘Žπ‘₯(𝐡𝑑+1 βˆ— ) < 2𝑑 . β–‘ Michele Ventrone 241 In some cases it is possible to determine the maximum of the sets B. Let's see how. The numbers of 𝐡𝑑+1and of 𝐡𝑑+1 βˆ— come from the integer solutions, if they exist, of the equations 𝑏𝑑+1 = π‘˜π‘‘βˆ’1 3 with π‘˜π‘‘ ∈ 𝐾𝑑 , π‘˜π‘‘ even and π‘˜π‘‘ β‰  4 (7.1) 𝑏𝑑+1 βˆ— = 2π‘˜π‘‘ βˆ—βˆ’1 3 with π‘˜π‘‘ βˆ— ∈ 𝐾𝑑 βˆ— and π‘˜π‘‘ βˆ— β‰  2 (7.2) by the Theorems, respectively, 4.2 and 4.5. In fact, the largest odd integer that can be obtained from (7.1), if we substitute the maximum of 𝐾𝑑 for π‘˜π‘‘, is 2π‘‘βˆ’1 3 , which is integer if 2𝑑 βˆ’ 1 is divisible by three. Likewise, the largest odd integer which can be obtained from (7.2), if we replace π‘˜π‘‘ βˆ— by the maximum of 𝐾𝑑 βˆ—, is 2𝑑+1βˆ’1 3 , which is integer if 2𝑑+1 βˆ’ 1 is divisible by 3. We can therefore state the following theorem. Theorem 7.3. i) If 2𝑑 βˆ’ 1 ≑ 0(π‘šπ‘œπ‘‘ 3), with 𝑑 ∈ β„•0 and 𝑑 > 2, then π‘šπ‘Žπ‘₯(𝐡𝑑+1) = 2π‘‘βˆ’1 3 i*) If 2𝑑+1 βˆ’ 1 ≑ 0(π‘šπ‘œπ‘‘ 3), with 𝑑 ∈ β„•0 and 𝑑 > 1, then π‘šπ‘Žπ‘₯(𝐡𝑑+1 βˆ— ) = 2𝑑+1βˆ’1 3 . SECOND DEMONSTRATION OF THE THEOREM 7.3 Proof. i) By hypothesis the number 2𝑑 is t-convergent and the equation 3𝑏 + 1 = 2𝑑 is satisfied by 𝑏 = 2π‘‘βˆ’1 3 which is different from 1 because 𝑑 > 2, therefore, by theorem 4.2 it is 𝑏 ∈ 𝐡𝑑+1. Assume that βˆƒπ›½ ∈ 𝐡𝑑+1: 𝑏 < 𝛽 that is, taking into account the form of b and Ξ², we suppose that it is 2π‘‘βˆ’1 3 < π‘˜π‘‘βˆ’1 3 with π‘˜π‘‘ ∈ 𝐾𝑑 e π‘˜π‘‘ even; from this it follows that 2 𝑑 < π‘˜π‘‘ , absurd thing because the maximum of 𝐾𝑑 is 2 𝑑 . Then it must turn out βˆ€π›½ ∈ 𝐡𝑑+1: 𝛽 ≀ 𝑏, that is the thesis. β€’ About two countable families in the finite sets of the Collatz Conjecture 242 i*) By hypothesis the number 2𝑑 is (t+1)-convergent and the equation 3𝑏 + 1 = 2𝑑+1 is satisfied by 𝑏 = 2𝑑+1βˆ’1 3 which is different from 1 because 𝑑 > 1, therefore, by theorem 4.5 it is π‘βˆ— ∈ 𝐡𝑑+1 βˆ— . Assume that βˆƒπ›½βˆ— ∈ 𝐡𝑑+1 βˆ— : π‘βˆ— < π›½βˆ— that is, taking into account the form of π‘βˆ— and π›½βˆ—, supposing it is 2𝑑+1βˆ’1 3 < 2π‘˜π‘‘ βˆ—βˆ’1 3 with π‘˜π‘‘ βˆ— ∈ 𝐾𝑑 βˆ—, from this it follows that 2𝑑 < π‘˜π‘‘ βˆ—, which is absurd because the maximum of 𝐾𝑑 βˆ— is 2𝑑 . Then it must turn out βˆ€π›½βˆ— ∈ 𝐡𝑑+1 βˆ— : π›½βˆ— ≀ π‘βˆ— that is the thesis. β–‘ For example: β€’ ... β€’ for 𝑑 = 14 risults 214 βˆ’ 1 ≑ 0(π‘šπ‘œπ‘‘3), then π‘šπ‘Žπ‘₯(𝐡15) = π‘šπ‘Žπ‘₯(𝐡14 βˆ— ) = 5461 β€’ for 𝑑 = 16 risults 216 βˆ’ 1 ≑ 0(π‘šπ‘œπ‘‘3), then π‘šπ‘Žπ‘₯(𝐡17) = π‘šπ‘Žπ‘₯(𝐡16 βˆ— ) = 21845 β€’ for 𝑑 = 18 risults 218 βˆ’ 1 ≑ 0(π‘šπ‘œπ‘‘3), then π‘šπ‘Žπ‘₯(𝐡19) = π‘šπ‘Žπ‘₯(𝐡18 βˆ— ) = 87381 β€’ .... 8 On the intersection of 𝑲𝒕 and 𝑲𝒕 βˆ— In this paragraph we will prove that the intersection of the sets 𝐾𝑑 and 𝐾𝑑 βˆ— is {2𝑑 }. Lemma 8.1. The intersection of the odd derivatives of the first type t-convergent and of the even derivatives of the second type t-convergent is empty, that is βˆ€π‘‘ ∈ β„•0, 𝐡𝑑 ∩ 𝐴𝑑 βˆ— = βˆ…. (8.1) Proof. Obviously, because an odd derivative either is empty or is made up of odd integers different from 1 and an even derivative contains only even numbers. β–‘ Lemma 8.2. The intersection of the odd derivatives of the second type t-convergent and of the even derivatives of the first type t-convergent is empty, that is βˆ€π‘‘ ∈ β„•0, 𝐡𝑑 βˆ— ∩ 𝐴𝑑 = βˆ…. (8.2) Michele Ventrone 243 Proof. Obviously, because an odd derivative either is empty or is made up of odd integers different from 1 and an even derivative contains only even numbers. β–‘ Lemma 8.3. The intersection of the odd derivatives of the first type t-convergent and of the odd derivatives of the second type t-convergent is empty, that is βˆ€π‘‘ ∈ β„•0, 𝐡𝑑 ∩ 𝐡𝑑 βˆ— = βˆ…. (8.3) Proof. Trivially, if 𝑑 = 1 the sets 𝐡1 and 𝐡1 βˆ— are both empty. Assume absurdly that for 𝑑 > 1 it results 𝐡𝑑 ∩ 𝐡𝑑 βˆ— β‰  βˆ… and consider every 𝑛𝑑 ∈ 𝐡𝑑 ∩ 𝐡𝑑 βˆ—. From 𝑛𝑑 ∈ 𝐡𝑑 = {𝑛𝑑 ∈ 𝐷 βˆ’ {1}: 3𝑛𝑑 + 1 = π‘˜π‘‘βˆ’1, π‘˜π‘‘βˆ’1 ∈ πΎπ‘‘βˆ’1 β‹‚ 𝑃 , π‘˜π‘‘βˆ’1 β‰  4, π‘˜π‘‘βˆ’1 βˆ’ 1 ≑ 0(π‘šπ‘œπ‘‘ 3)} follows that nt is an odd integer of the form (4.7), that is 𝑛𝑑 = π‘˜π‘‘βˆ’1βˆ’1 3 . From 𝑛𝑑 ∈ 𝐡𝑑 βˆ— = {𝑛𝑑 ∈ 𝐷 βˆ’ {1}: 3𝑛𝑑 + 1 2 = π‘˜π‘‘βˆ’1 βˆ— , π‘˜π‘‘βˆ’1 βˆ— ∈ πΎπ‘‘βˆ’1 βˆ— , π‘˜π‘‘βˆ’1 βˆ— β‰  2, 2π‘˜π‘‘βˆ’1 βˆ— βˆ’ 1 ≑ 0(π‘šπ‘œπ‘‘ 3)} it follows that nt is an odd integer of the form (4.15), that is 𝑛𝑑 = 2π‘˜π‘‘βˆ’1 βˆ— βˆ’1 3 . By equating the two expressions of 𝑛𝑑 we have π‘˜π‘‘βˆ’1βˆ’1 3 = 2π‘˜π‘‘βˆ’1 βˆ— βˆ’1 3 and therefore π‘˜π‘‘βˆ’1 = 2π‘˜π‘‘βˆ’1 βˆ— . (8.4) Equality (8.4) is manifestly absurd because π‘˜π‘‘βˆ’1 is (t-1)-convergent and 2π‘˜π‘‘βˆ’1 βˆ— is t-convergent. Therefore it makes no sense to suppose that the intersection 𝐡𝑑 ∩ 𝐡𝑑 βˆ— for 𝑑 > 1 is non-empty and (8.3) is proved. β–‘ Lemma 8.4. The intersection of 𝐾𝑑 and 𝐾𝑑 βˆ— is equal to the intersection of the even derivatives of πΎπ‘‘βˆ’1 and of the derivatives even of 𝐾𝑑 βˆ—, that is βˆ€π‘‘ ∈ β„•0, 𝐾𝑑 ∩ 𝐾𝑑 βˆ— = 𝐴𝑑 ∩ 𝐴𝑑 βˆ—. (8.5) Proof. We will use the notable equalities 4.12) and 4.18). We have βˆ€π‘‘ ∈ β„•0: About two countable families in the finite sets of the Collatz Conjecture 244 𝐾𝑑 ∩ 𝐾𝑑 βˆ— = (𝐴𝑑 βˆͺ 𝐡𝑑 ) ∩ (𝐴𝑑 βˆ— βˆͺ 𝐡𝑑 βˆ—) = = ((𝐴𝑑 βˆͺ 𝐡𝑑 ) ∩ 𝐴𝑑 βˆ—) βˆͺ ((𝐴𝑑 βˆͺ 𝐡𝑑 ) ∩ 𝐡𝑑 βˆ—) = = (𝐴𝑑 ∩ 𝐴𝑑 βˆ—) βˆͺ (𝐡𝑑 ∩ 𝐴𝑑 βˆ—) βˆͺ (𝐴𝑑 ∩ 𝐡𝑑 βˆ—) βˆͺ (𝐡𝑑 ∩ 𝐡𝑑 βˆ—). (8.6) The thesis follows by applying, in order, Lemmas 8.1, 8.2 and 8.3 to the second, third and fourth intersection in the last line of (8.6). β–‘ Lemma 8.5. The intersection of the even derivatives of the first and second type t-convergent is {2𝑑 }, that is: βˆ€π‘‘ ∈ β„•0, 𝐴𝑑 ∩ 𝐴𝑑 βˆ— = {2𝑑 }. (8.7) Proof. Applying the equalities (4.10) and (4.16) to the intersection 𝐴𝑑 ∩ 𝐴𝑑 βˆ— we have: βˆ€π‘‘ ∈ β„•0, 𝐴𝑑 ∩ 𝐴𝑑 βˆ— = 2πΎπ‘‘βˆ’1 ∩ 2πΎπ‘‘βˆ’1 βˆ— = 2(πΎπ‘‘βˆ’1 ∩ πΎπ‘‘βˆ’1 βˆ— ). (8.8) Applying Lemma 8.4 to the intersection in the last parenthesis of (8.8) we have βˆ€π‘‘ ∈ β„•0, 2(πΎπ‘‘βˆ’1 ∩ πΎπ‘‘βˆ’1 βˆ— ) = 2(π΄π‘‘βˆ’1 ∩ π΄π‘‘βˆ’1 βˆ— ) = = 2(2πΎπ‘‘βˆ’2 ∩ 2πΎπ‘‘βˆ’2 βˆ— ) = 22(πΎπ‘‘βˆ’2 ∩ πΎπ‘‘βˆ’2 βˆ— ). (8.9) Applying Lemma 8.4 again to the intersection in the last parenthesis of (8.9) and iterating, we obtain βˆ€π‘‘ ∈ β„•0, 2 2(πΎπ‘‘βˆ’2 ∩ πΎπ‘‘βˆ’2 βˆ— ) = 22(π΄π‘‘βˆ’2 ∩ π΄π‘‘βˆ’2 βˆ— ) = β‹― = 2π‘‘βˆ’1(𝐾1 ∩ 𝐾1 βˆ—). (8.10) Finally, applying Lemma 8.4 again to the intersection in the last parenthesis of (8.10), we have βˆ€π‘‘ ∈ β„•0, 2 π‘‘βˆ’1(𝐾1 ∩ 𝐾1 βˆ—) = 2π‘‘βˆ’1(𝐴1 ∩ 𝐴1 βˆ— ) = 2π‘‘βˆ’1(2𝐾0 ∩ 2𝐾0 βˆ—) = = 2𝑑 (𝐾0 ∩ 𝐾0 βˆ—) = 2𝑑 ({1} ∩ {1}) = {2𝑑 } . β–‘ (8.11) Theorem 8.6. The intersection between 𝐾𝑑 and 𝐾𝑑 βˆ— is equal to {2𝑑 }, that is βˆ€π‘‘ ∈ β„•0, 𝐾𝑑 ∩ 𝐾𝑑 βˆ— = {2𝑑 }. (8.12) Michele Ventrone 245 Proof. Applying Lemma 8.4 to the intersection 𝐾𝑑 ∩ 𝐾𝑑 βˆ—, we have (8.5). Applying the Lemma 8.5 at the intersection 𝐴𝑑 ∩ 𝐴𝑑 βˆ— we obtain (8.12). β–‘ 9 Conclusions Collatz's conjecture can be re-proposed using the sets K and their first properties. We have seen that the sets 𝐾𝑑 and 𝐾𝑑 βˆ— are non-empty (Basic 3.1) and they are also two by two disjoint (Corollary 3.4). So, if the following coverage equalities of β„•0 were also true: a) ⋃ 𝐾𝑑 +∞ 𝑑=0 = β„•0 , 𝑑 ∈ β„• b) ⋃ 𝐾𝑑 βˆ—+∞ 𝑑=0 = β„•0 , 𝑑 ∈ β„• we could say that each of the families {𝐾𝑑 }π‘‘βˆˆβ„• and {𝐾𝑑 βˆ—}π‘‘βˆˆβ„• is a partition of β„•0. In this case the Collatz conjecture would be proved. About two countable families in the finite sets of the Collatz Conjecture 246 References [1] Leggerini, S. (2004). β€œL’enigmatico pari e dispari che da cinquant’anni non fa dormire i matematici”. Newton, n.7, 2004. [2] Conway, J. H. "Unpredictable Iterations." Proc. 1972 Number Th. Conf., University of Colorado, Boulder, Colorado, pp. 49-52, 1972.