anyways, here are some of my results you guys may find interesting:

NB:

B(x) = B(x,x,x,x,x,x,x) repeating loops

B(X,Y) = B(X,Y,X,Y) repeating loops

B1(X,Y,Z) = B(X,Y,Z) 1 means first term first

B2(X,Y,Z) = B(Y,Z,X) 2 means second term first

B(2) = 1 (trivial cycle, 1,4,2,1…)

B(1) = -1 (true negative cycle, -1,-2,-1…)

B(1,2) = -5 (true negative cycle, -5,-14,-7,-20,-10,-5…)

B2(1,2) = -7 (true negative cycle above, -7,-20,-10,-5,-14,-7…)

B(1,1,1,4,1,1,2) = -17 (true negative cycle)

Non 3n+1 values that DO fit the formula:

B(-2,0,7) = 2

B(4,3,-1) = 5

B(1,6,-3) = -13

B(0,6,-3) = -4

I have 53 integers so far in ‘n = 3′ sets for outside collatz parameters.

]]>but i write a(k+1) as x1+…+xn, k as n and a1,a2, as x1,x1+x2…

But i’ve noticed that integer solutions pop up everywhere once you bend the rules. for example, plug in values for the accumulative totals of the exponent of 2′s that dont exist as a 3n+1 function, and put a zero or negative number in the mix and you get loops with even numbers where there should be odd, loops with a single number only being an integer. etc etc.

if the cumulative sum of x1 to xn is between n and 2n an integer solution can exist for the ‘n’ in in equation (1) (which i denote as a function of x1,x2…xn, representing how many times you divide by 2 between each time you multiply by 3 then add 1: as B(x1,…,xn) )

its essentially the same formula, but I use each value as a sum, instead of a1<a2…

]]>polymath prj, anyone?

am planning to crunch on this further but at a pause point, dont have next step in mind yet…. ]]>

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 ]]>