Prime factorization theory - MathOverflow most recent 30 from http://mathoverflow.net2013-05-23T05:46:47Zhttp://mathoverflow.net/feeds/question/47495http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/47495/prime-factorization-theoryPrime factorization theoryJ Hato2010-11-27T10:09:21Z2010-11-27T23:03:24Z
<p>Firstly, let me divulge. I've been doing a lot of research on the summation of two coprime numbers and unfortunately have failed to come up with the properties I'm seeking; it is my hope that someone here might be of some help.</p>
<p>Let $(j, k)\in \mathbb{N}^2$ be coprime.</p>
<p>Can $\Omega(j + k)$ or $\omega(j + k)$ be expressed as some function of $\omega(j)$, $\omega(k)$, $\lambda(j)$, $\lambda(k)$, $ j$, and/or $k$?</p>
<p>If not, then maybe for the special case that $(j, k)$ are prime, or in particular, odd primes?</p>
<p>If the answer is still no, any information regarding this topic is much appreciated.</p>
<p>Note: The functions $\Omega$, $\omega$, and $\lambda$, are the total prime factors, distinct prime factors, and the Liouville function respectively.</p>
http://mathoverflow.net/questions/47495/prime-factorization-theory/47531#47531Answer by Chris Wuthrich for Prime factorization theoryChris Wuthrich2010-11-27T19:53:04Z2010-11-27T19:53:04Z<p>Obviously, $\Omega(j+k)$ is a function in $j$ and $k$, certainly it would be difficult to write it in a closed form other than $\Omega(j+k)$. Similar for $\omega(j+k)$. On the other hand neither of them can be a function in $\omega(j)$, $\omega(k)$, $\lambda(j)$, $\lambda(k)$ and $j$ only. For instance they take the same values for $(j,k)=(1,2)$ as for $(j,k)=(1,5)$, but $\omega(3)=\Omega(3)=1\neq \omega(6)=\Omega(6)=2$.</p>
http://mathoverflow.net/questions/47495/prime-factorization-theory/47548#47548Answer by Alastair Irving for Prime factorization theoryAlastair Irving2010-11-27T23:03:24Z2010-11-27T23:03:24Z<p>Suppose $p,q$ are prime. Thus $\omega,\Omega,\lambda$ are known constants at $p,q$ so we want to know if $\omega,\Omega(p+q)$ can be written as functions in $p,q$. To answer this you need to define what is meant by a function, as clearly $\omega(p+q)$ is a function in $p,q$.</p>
<p>Presumably you mean a function which is built from some class of elementary functions. Therefore, a possible approach would be to try and prove that certain classes of functions are not sufficient. For example, we cannot have $\omega(p+q)=Ap+Bq+C$, for constants $A,B,C$, as we can pick suitable combinations of primes to give a system of linear equations in $a,B,C$ with no solutions, (I haven't actually done this computation). It seems reasonable that for any degree $d$ we can find suitable pairs of primes to show that the function isn't a polynomial with degree $\leq d$, but I don't have a proof of this. Thus showing that the function isn't a polynomial might be a worthwhile first It will probably help to use that the function is symmetric in $p,q$ so if its a polynomial then its a polynomial in $p+q,pq$.</p>
<p>Once polynomials are dealt with, then try and extend the class of possible functions to something bigger.</p>