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 Number theoretic formula for hyper operators (-oo, 2] at prime numbers JmsNxn Ultimate Fellow Posts: 895 Threads: 110 Joined: Dec 2010 06/04/2012, 08:05 PM (This post was last modified: 06/04/2012, 08:36 PM by JmsNxn.) It took some thinking but I recovered a closed form expression for hyper operators at prime numbers. The result is interesting to say the least. It is not analytic; and in fact the iteration may not even be continuous but discrete. We start by considering a function $\psi: \mathbb{N} \to \mathbb{N}$ $\forall x \forall y\,\,\,x ,y \in \mathbb{N}$ $x = \prod_i \rho_i^{\mu_{x;i}}$ $\psi(x) = \sum_i \mu_{x;i} \cdot \rho_i$ $\forall \rho\,\,\rho \in \mathbb{P}$ $\psi(\rho) = \rho$ $\psi(1) = 0$ $\psi(xy) = \psi(x) + \psi(y)$ $\psi(x^y) = y \psi(x)$ But the fundamental theorem of arithmetic applies for $\psi$ as well. $\psi(x) = \prod_i \rho_i^{\kappa_{x;i}}$ Iterating we get a formula: $\psi^{\circ s}(x) = \sum_i \mu_{\psi^{\circ s-1}(x);i} \rho_i$ This allows us to develop an expression for integer arguments of $s$. We will now make some claims about the structure of iterating $\psi$. Claim I: $\psi^{\circ s}$ is one to one for all complex iterates that exist and satisfy $|\Re(s)| < 1$. This qualifies as the reason operators in between $\Re(s) \in (1,2]$ are not commutative. The reason $\psi$ isn't one to one is because it turns $\psi(\rho_i^{\rho_j}) = \rho_j \cdot \rho_i = \psi(\rho_j^{\rho_i})$. However; since $\Re(s) \in (0,1]$ we have $\psi^{\circ s}(\rho_i^{\rho_j}) = \rho_i \,\,\Delta_{2-s}\,\,\rho_j \neq \psi^{\circ s}(\rho_j^{\rho_i})$ because $\Delta_{2-s}$ is not commutative. Claim II: $\psi^{\circ s}$'s range is closed under multiplication Claim III: $\psi^{\circ s}(x) = \prod_i \rho_i^{\eta_{x;i}}$ for some sequence $\eta_{x;i} \in \mathbb{A}$. This claim implies II with some algebra, but as well determines some more powerful results. This factorization into primes with algebraic exponents is unique by Baker's theorem. This would be the main restriction determining for what values of $s$ we can iterate $\psi$. With these we define hyper operators by the following formula. $s = \sigma + it$ $x \Delta_s y = \left\{ \begin{array} \psi^{\circ 1-s}(\psi^{\circ s-1}(x)\psi^{\circ s-1}(y)) & \sigma \in (0,1]\\ \psi^{\circ 1-s}(\psi^{\circ s-1}(x) y) & \sigma \in [1,2]\\ \end{array}$ And we extend the operators recursively to the negative domain by the following law, $-s \not \in \mathbb{N}$ and $0 \le \Re(q) < 1$ $\psi^{\circ q}(x \Delta_s y) = \psi^{\circ q}(x) \Delta_{s-q} \psi^{\circ q}(y)$ These operators form the tremendous result, for $\sigma < 1$ of $\{\mathbb{D}, \Delta_{s}, \Delta_{s-1}\}$ forming a commutative semi-ring! But on top of that: Recursion is satisfied for all prime numbers. $x\,\, \Delta_{s+1} \,\,\rho = x_1\,\, \Delta_{s}\,\, x_2\,\, \Delta_s\,\, ...\,\, \Delta_s\,\, x_{\rho}$ The question is, of course, how do we define iteration on $\psi$. And on top of that, where and when is an iterate defined? These operators have some really interesting results involving unique factorization. Although, all these further results rely on iterating $\psi$ with the restriction of claims I-III. So I'm wondering if anyone knows of any techniques that would be required? I'm thinking the iteration would require not a continuous expression but a discrete expression involving the unique coefficients of $\psi^{\circ s}(x)$. This can work in fact for every natural number given a few tweaks that I'll explain later. tommy1729 Ultimate Fellow Posts: 1,645 Threads: 369 Joined: Feb 2009 06/29/2012, 10:09 PM i awaited the second post ( of james ) ... but it did not come. in fact i wonder what and if anyone would reply. i think there are no replies because james confused ppl , maybe including himself. in fact , i dont agree. james uses the same P_i everywhere , but the factorization of a sum has other prime factors than the product of prime factors. for instance , take p and q odd primes and J(n) the function james proposes then J(p^2 * q) = 2*p + q. but the primefactors of 2*p + q are not easily expressed and certainly not a multiple of p or q. i then see no way to extend this idea to something as "complicated " as number theory and/or tetration ... regards tommy1729 JmsNxn Ultimate Fellow Posts: 895 Threads: 110 Joined: Dec 2010 07/17/2012, 02:12 AM (This post was last modified: 07/18/2012, 06:05 AM by JmsNxn.) I apologizing for confusing you. I should have you know you're probably right. I was just trying really hard to look for a solution. I'm approaching from every angle. I will try to reserve my posts to a more refined quality. « Next Oldest | Next Newest »

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