For example, if $n$ is an odd number (with finite trajectory) and $(k, a_1, a_2, … , a_k+1)$ is its associated (reduced map) tuple, then $(k, a_1, a_2 + 2, a_3 + 2, … a_{k+1} + 2)$ is another solution, namely $4n+1$. Here we are adding 2 to each exponent other than $a_1 = 0$. $4n+1$ then has the same trajectory as $n$ (after the seed value). Indeed, you can add any even integer $2m$ the exponents to obtain the $m’th$ iterate of $n$ under $4n+1$.

One can also add integer multiples of $phi(3^k)$ to $a_{k+1}$ to obtain new solutions. Here, $phi$ denotes Euler’s Totient function.

]]>*[Corrected, thanks - T.]*

N/4 is touchable for sure but you just need to look at one very particular string of odd numbers. this particular string of odd numbers to be exact 1+8n for all N. ex. 1,9,17,……

ill even make it easier for you with a simple proposition.

3(1 + 8N) +1 = (1 + 6M) 2^2 iff N is any Natural number and M is dependent on N. meaning there is only one M for each N.

]]>HuenYK (dr)

cosmolog92@gmail.com ]]>

Set X = Odds 2,5,8,11,14,17,….etc…..infinity

Set – = odds 3,6,9,12,15,18,…etc….infinity

Set + = Odds 4,7,10,13,16,….etc…..infinity

Set ???? = 1 and ONLY 1.

now the true point of these set designations is this. due to prime factorization and the fact that every odd number is factorized by a unique set of odd prime factors and all evens are just adding a string of multiplications of 2s on each unique odd numbers factorization. essentially what I’m saying is every even integer is composed of an odd base and multiplication of a string of 2′s that each even number has a particular unique number of 2s In its factorization and all even numbers if stripped of 2s will descend to a particular odd number and that number as well as the movements leading to that number are predetermined because due to he rules every number can ONLY move one preset direction. a few rules involving these sets are this

if any element of set X has ANY amount of 2′s in is prime factorization no whole positive odd integer X can ever equal that value as a result of (X*3)+1

if any element of set – has any odd number of 2′s in its prime factorization then One and only one X will satisfy for X*3+1. that value is predetermined in the following ways.

if an element of set – is multiplied by 2^1 then the value that produces that even as an odd within the collatz system is the element of -’s actual numerical value minus its cardinality within set – multiplied by 2. for 5 which is the first element of this set it would be for example 5 – 2*1 and the next number in set x is 11 and the value that creates it times 2 is 11 – 2*2 and so on so forth into infinity.

the other infinite set + acts directly parallel to set – except that it does this by adding its cardinality multiplied by 2 instead of subtracting. also where for set – all of the odd numbers of twos in is prime factorization gets switched with evens for this one. and the minimal case where the producing value is directly related to the odd is that odd multiplied by 4 for example 7 is the first number in this set and 7 *4 is equal to (7+(2*1)*3+1 and 13 which is the second number in this set times 4 is equal to 13+(2*2)*3+1

this is for the most part everything involving how to split up and see the predictable nature. but one will notice that logically it seems that you could have 1 be part of set + based on the progression of those numbers. BUT it cannot for it would throw off every other numbers cardinality and causes 1 to be left alone as 1 *4 = 1*3(+1) and is the only number that causes a number hat breaks down to itself on immediate removal of all 2′s on successive division of two.

also for all Odds numbers X. if X * 2^n = n*3+1 then X*2^n+2 = n*4+1 or n+x*2^n both of these give the exact same values.

also another reason 1 mus be in a set of itself is another way of splitting odds hat requires it to be so. it basically falls on my concept of the way a number must move is a transform stage. so in my notation for this part x—–>y means that X becomes Y under all circumstances and Xx then x——>1.

if element of set E——>X then X———>Y and y is always Greater than E for ALL E’s

if element of set O———->X then X———–Y and y is always less than E for all E’s

also if N———>X————–>Y if N is any odd Y will be odd and NO Y will EVER be an Element of set X from the previous groups.(no number divisible by 3 will occur as Y and if y——–>z——->f,,,, so on so forth and no number will ever be divisible by 3 after a value divisible by three has occurred.

this last part is the one area i haven’t gone too much into. I’m sure there might be a real rule that pinpoints exactly how much larger or smaller a number will be on second transform but i haven’t really felt like going into this any further as i am currently working on a new problem. Hopefully this gives news answers and may be the part needed to prove this “Hard” problem. i just think its more absurd to think any other number would cause a cycle based on the two different clearly explained reasons why one causes an exception when working with the set of all odd numbers.

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