Oh... I didn't realize I'vnt posted anything this year. Anyway, here's my first post of the year - my thoughts on the Divisor function.
The Divisor functions are well known and have been studied extensively in Mathematics. Their relation to the Partition function is one of best identities I've seen. I was playing around with them as part of my effort to solve a problem and came up with some identities that I can't find anywhere on the Web. So this post will be about them.
The Divisor function is known to be multiplicative, but not completely multiplicative. But I needed an some formulas where I wanted to find the value of the Divisor functions of product of two numbers that are not co-prime. After some extensive calculations, I ended with the following:
Let $m=p_1^{m_1}p_2^{m_2}p_3^{m_3}\cdots$ and $n=p_1^{n_1}p_2^{n_2}p_3^{n_3}\cdots$. Then,
$\displaystyle\frac{\sigma_k(m\cdot n)}{\sigma_k(n)}=\prod_{p_j|m}\left(1+\frac{p_j^k+p_j^{2k}+p_j^{3k}+\cdots+p_j^{m_jk}}{1+p_j^{-k}+p_j^{-2k}+p_j^{-3k}+\cdots+p_j^{-n_jk}}\right)$
We can easily check a couple of easy cases. For example, when $n=1$, all of the $n_j$'s become $0$ and the above formulae reduces to the definition of Divisor function. Similarly, when $m$ and $n$ are co-prime, for any prime that divides $m$, the exponent of that prime in $n$ will be $0$, again reducing to a form that shows that divisor functions are multiplicative.
I found the case $k=0$ especially very interesting. We'll see why in a short while. First of all, when $k=0$, the above formula reduces to,
$\displaystyle\frac{\sigma_0(m\cdot n)}{\sigma_0(n)}=\left(1+\frac{m_1}{n_1+1}\right)\left(1+\frac{m_2}{n_2+1}\right)\left(1+\frac{m_3}{n_3+1}\right)\cdots$
As $\sigma_0(n)=(1+n_1)(1+n_2)(1+n_3)\cdots$, we can write
$\sigma_0(m\cdot n)=\sigma_0(n)+m_1\sigma_{0,p_1}(n)+m_2\sigma_{0,p_2}(n)+\cdots+m_1m_2\sigma_{0,p_1p_2}(n)+\cdots+m_1m_2\cdots m_k\sigma_{0,p_1p_2\cdots p_k}(n)+\cdots$
where $\sigma_{0,B}(n)$ is (defined only for square-free $B$) a function that counts the number of divisors of $n$ that are not divisible by any of the prime factors of $B$. Note that we have eliminated everything related to prime factorization of $n$.
Actually this is why I said the case $k=0$ is interesting. Had $k$ been greater than $0$, we would have those annoying prime factors of $n$ in the above expression. Now with what we have above we can make an interesting Dirichlet Generating function.
Given $m=p_1^{m_1}p_2^{m_2}p_3^{m_3}\cdots$, we have
$\displaystyle\sum_{n=1}^\infty\frac{\sigma_0(m\cdot n)}{n^s}=\zeta^2(s) \prod_{p_j|m}\left(1+m_j(1-p_j^{-s})\right)$
Even better, for a given $m$, if we define a function $f(n)$ as
$f(n)=\displaystyle\sum_{d|m}\sigma_0(d\cdot n)$
again we end up with a nice DGF. This we have,
$\displaystyle\sum_{n=1}^\infty\frac{f(n)}{n^s}=\sigma_0(m)\zeta^2(s)\prod_{p_j|m}\left(1+\frac{m_j}{2}(1-p_j^{-s})\right)$
All of the above DGFs in combination with Dirichlet's Hyperbola method can be used to find their respective summatory function.
We can see how this generalizes in the next post. See ya soon.
Till then
Yours Aye,
Me
The Divisor functions are well known and have been studied extensively in Mathematics. Their relation to the Partition function is one of best identities I've seen. I was playing around with them as part of my effort to solve a problem and came up with some identities that I can't find anywhere on the Web. So this post will be about them.
The Divisor function is known to be multiplicative, but not completely multiplicative. But I needed an some formulas where I wanted to find the value of the Divisor functions of product of two numbers that are not co-prime. After some extensive calculations, I ended with the following:
Let $m=p_1^{m_1}p_2^{m_2}p_3^{m_3}\cdots$ and $n=p_1^{n_1}p_2^{n_2}p_3^{n_3}\cdots$. Then,
$\displaystyle\frac{\sigma_k(m\cdot n)}{\sigma_k(n)}=\prod_{p_j|m}\left(1+\frac{p_j^k+p_j^{2k}+p_j^{3k}+\cdots+p_j^{m_jk}}{1+p_j^{-k}+p_j^{-2k}+p_j^{-3k}+\cdots+p_j^{-n_jk}}\right)$
We can easily check a couple of easy cases. For example, when $n=1$, all of the $n_j$'s become $0$ and the above formulae reduces to the definition of Divisor function. Similarly, when $m$ and $n$ are co-prime, for any prime that divides $m$, the exponent of that prime in $n$ will be $0$, again reducing to a form that shows that divisor functions are multiplicative.
I found the case $k=0$ especially very interesting. We'll see why in a short while. First of all, when $k=0$, the above formula reduces to,
$\displaystyle\frac{\sigma_0(m\cdot n)}{\sigma_0(n)}=\left(1+\frac{m_1}{n_1+1}\right)\left(1+\frac{m_2}{n_2+1}\right)\left(1+\frac{m_3}{n_3+1}\right)\cdots$
As $\sigma_0(n)=(1+n_1)(1+n_2)(1+n_3)\cdots$, we can write
$\sigma_0(m\cdot n)=\sigma_0(n)+m_1\sigma_{0,p_1}(n)+m_2\sigma_{0,p_2}(n)+\cdots+m_1m_2\sigma_{0,p_1p_2}(n)+\cdots+m_1m_2\cdots m_k\sigma_{0,p_1p_2\cdots p_k}(n)+\cdots$
where $\sigma_{0,B}(n)$ is (defined only for square-free $B$) a function that counts the number of divisors of $n$ that are not divisible by any of the prime factors of $B$. Note that we have eliminated everything related to prime factorization of $n$.
Actually this is why I said the case $k=0$ is interesting. Had $k$ been greater than $0$, we would have those annoying prime factors of $n$ in the above expression. Now with what we have above we can make an interesting Dirichlet Generating function.
Given $m=p_1^{m_1}p_2^{m_2}p_3^{m_3}\cdots$, we have
$\displaystyle\sum_{n=1}^\infty\frac{\sigma_0(m\cdot n)}{n^s}=\zeta^2(s) \prod_{p_j|m}\left(1+m_j(1-p_j^{-s})\right)$
Even better, for a given $m$, if we define a function $f(n)$ as
$f(n)=\displaystyle\sum_{d|m}\sigma_0(d\cdot n)$
again we end up with a nice DGF. This we have,
$\displaystyle\sum_{n=1}^\infty\frac{f(n)}{n^s}=\sigma_0(m)\zeta^2(s)\prod_{p_j|m}\left(1+\frac{m_j}{2}(1-p_j^{-s})\right)$
All of the above DGFs in combination with Dirichlet's Hyperbola method can be used to find their respective summatory function.
We can see how this generalizes in the next post. See ya soon.
Till then
Yours Aye,
Me
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