User:Gladamas/collatz

n is the starting number, s is the number of $$\frac{3n+1}{2^{v_2(3a_s + 1)}}$$ cycles to go through, $$a_s$$ is the ending number, and $$v_2(3a_k + 1)$$ is the 2-adic order of $$3a_k + 1$$.

For $$s \geq 2$$:

$$a_0 = n$$

$$a_{s+1} = \frac{3 a_s + 1}{2^{v_2(3a_s + 1)}}$$

$$a_{s+1} - (3*2^{-v_2(3a_s + 1)}) a_s = 2^{-v_2(3a_s + 1)}$$

$$\frac{a_{s+1}}{\prod_{k=0}^s (3*2^{-v_2(3a_k + 1)})} - \frac{(3*2^{-v_2(3a_s + 1)}) a_s}{\prod_{k=0}^s (3*2^{-v_2(3a_k + 1)})} = \frac{2^{-v_2(3a_s + 1)}}{\prod_{k=0}^s (3*2^{-v_2(3a_k + 1)})}$$

$$\left(\frac{2^{\sum_{k=0}^s v_2(3a_k + 1)}}{3^{s+1}}\right) a_{s+1} - \left(\frac{2^{\sum_{k=0}^{s-1} v_2(3a_k + 1)}}{3^s}\right) a_s = \frac{2^{\sum_{k=0}^{s-1} v_2(3a_k + 1)}}{3^{s+1}}$$

$$A_s = \left(\frac{2^{\sum_{k=0}^{s-1} v_2(3a_k + 1)}}{3^s}\right) a_s$$

$$A_{s+1} - A_s = \frac{2^{\sum_{k=0}^{s-1} v_2(3a_k + 1)}}{3^{s+1}}$$

$$\sum_{m=1}^{s-1}(A_{m+1} - A_m) = A_s - A_1 = \sum_{m=1}^{s-1}\frac{2^{\sum_{k=0}^{m-1} v_2(3a_k + 1)}}{3^{m+1}}$$

$$\left(\frac{2^{\sum_{k=0}^{s-1} v_2(3a_k + 1)}}{3^s}\right) a_s = A_1 + \sum_{m=1}^{s-1}\frac{2^{\sum_{k=0}^{m-1} v_2(3a_k + 1)}}{3^{m+1}}$$

$$a_s = {3^s}*\left({1/3} + n + {\sum_{m=1}^{s-1} \frac{2^{\sum_{k=0}^{m-1} v_2(3a_k + 1)}}{3^{m+1}}}\right)/\,{2^{\sum_{k=0}^{s-1} v_2(3a_k + 1)}}$$

$$a_s = [3^{s-1} + (n)(3^s) + (3^s)*{\sum_{m=1}^{s-1} (\frac{2^{\sum_{k=0}^{m-1} v_2(3a_k + 1)}}{3^{m+1}})}]/{2^{\sum_{k=0}^{s-1} v_2(3a_k + 1)}}$$

$$a_s = [3^{s-1} + (3n)(3^{s-1}) + \frac{1}{2}(3^{s-1} - 1){\sum_{m=0}^{s-2} ({2^{\sum_{k=0}^{m} v_2(3a_k + 1)}})}]/{2^{\sum_{k=0}^{s-1} v_2(3a_k + 1)}}$$

$$a_s = [(3n+1)(3^{s-1}) + (3^{s-1} - 1){\sum_{m=0}^{s-2} ({2^{({\sum_{k=0}^{m} v_2(3a_k + 1)}) - 1}})}]/{2^{\sum_{k=0}^{s-1} v_2(3a_k + 1)}}$$

When $$a_s = 1$$,

$$2^{\sum_{k=0}^{s-1} v_2(3a_k + 1)} = (3n+1)(3^{s-1}) + (3^{s-1} - 1){\sum_{m=0}^{s-2} ({2^{({\sum_{k=0}^{m} v_2(3a_k + 1)}) - 1}})}$$ One corollary of the Collatz conjecture states that $$\lim_{s \to +\infty} ([(3n+1)(3^{s-1}) + (3^{s-1} - 1){\sum_{m=0}^{s-2} ({2^{({\sum_{k=0}^{m} v_2(3a_k + 1)}) - 1}})}]/{2^{\sum_{k=0}^{s-1} v_2(3a_k + 1)}}) \neq \infty$$ for all positive integer n.