# Abstract

This work strongly relates to Lambert series, but uses an alternative kernel in the transformation. Many results are shown from experimental techniques and tabulated below.

# Main

From the product representation of $$e^z$$ we can write $x = \sum_{k=1}^\infty \frac{-\mu(k)}{k} \log(1-x^k)$ or perhaps more nicely $x = \sum_{k=1}^\infty \frac{\mu(k)}{k} \log\left(\frac{1}{1-x^k}\right)$ I then wonder what other functions can be written in terms of this, for example $e^x -1 = \sum_{k=1}^\infty \frac{a_k}{k!}\log(1-x^k)$ where the $$a_k$$ starting from $$a_1$$ are $$-1,0,1,5,23,59,719,839$$ which appear to be A253901 minus $$1$$, an then we have $e^x \approx 1+\sum_{k=1}^\infty \frac{1}{k!}\left(\prod_{d|k}(d-1)!^{\mu\left(\frac{k}{d}\right)}-1\right)\log(1-x^k)$ but this seems to be wrong for the $$12^{th}$$ power and some others, so this is not the full story! We seem to have $x^2 = \sum_{k=1}^\infty \frac{-\mu(k)}{k} \log(1-x^{2k})$ in fact it seems $x^n = \sum_{k=1}^\infty \frac{-\mu(k)}{k} \log(1-x^{nk})$ which is fairly obvious from taking the power of the argument, however we could also have squared the entire sum, so we know that $\sum_{k=1}^\infty \frac{-\mu(k)}{k} \log(1-x^{2k}) = \left(\sum_{k=1}^\infty \frac{-\mu(k)}{k} \log(1-x^{k})\right)^2$ and more generally $\sum_{k=1}^\infty \frac{-\mu(k)}{k} \log(1-x^{nk}) = \left(\sum_{k=1}^\infty \frac{-\mu(k)}{k} \log(1-x^{k})\right)^n$ interestingly we can write $\log(1-x^m)= \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{\mu(k)}{nk}\log(1-x^{mnk})$ and we must have the expansion $e^x =1 -\sum_{n=1}^\infty \sum_{k=1}^\infty \frac{\mu(k)}{n!k}\log(1-x^{nk})$ we can get from this the generating function of the Bell Numbers $e^{e^x-1}= \prod_{n=1}^\infty \prod_{k=1}^\infty (1-x^{nk})^{\frac{-\mu(k)}{kn!}}$ of course this is all for $$x<1$$. The product seems to converge for constants for example $$e^{e^{1/2}-1}$$. Other interesting results $\frac{x}{(1-x)} = \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{-\mu(k)}{k}\log(1-x^{nk})$ $\frac{x}{(1-x)^2} = \sum_{n=1}^\infty \sum_{k=1}^\infty -n\frac{\mu(k)}{k}\log(1-x^{nk})$ $O\left[\frac{1}{\zeta(s-1)}\right] = \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{n\mu(n)\mu(k)}{k}\log\left(\frac{1}{1-x^{nk}}\right)$ $\sum_{n=1}^\infty A101035(n)\frac{x^n}{n} = \sum_{n=1}^\infty \sum_{k=1}^\infty \mu(n)\mu(k)\log\left(\frac{1}{1-x^{nk}}\right)$ \begin{aligned} -\log(1-x) = \sum_{n=1}^\infty \sum_{k=1}^\infty \mu(n)\log\left(\frac{1}{1-x^{nk}}\right)\\ -\log(1-x) = \sum_{n=1}^\infty \sum_{k=1}^\infty \mu(k)\log\left(\frac{1}{1-x^{nk}}\right)\\ -\log(1-x) = \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{\mu(n)}{nk}\log\left(\frac{1}{1-x^{nk}}\right)\\ -\log(1-x) = \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{\mu(k)}{nk}\log\left(\frac{1}{1-x^{nk}}\right)\\\end{aligned} $\sum_{n=1}^\infty \sum_{d|n} d\tau(d)x^n = \sum_{n=1}^\infty \sum_{k=1}^\infty \log\left(\frac{1}{1-x^{nk}}\right)\\$

where $$O$$ hypothetically maps a Dirichlet generating function to an ordinary generating function. The last one is a statement that $\prod_{n=1}^\infty e^{\sum_{d|n} d \tau(d) x^n}= \prod_{n=1}^\infty \prod_{k=1}^\infty \frac{1}{1-x^{n k}}$ this has a few similarities to the Euler product of the zeta function $\zeta(s) = \prod_{\text{primes}} \frac{1}{1-p^{-s}}$ if we let $$x=1/p$$, then we get $\prod_{n=1}^\infty e^{\sum_{d|n} d \tau(d) p^{-n}}= \prod_{n=1}^\infty \prod_{k=1}^\infty \frac{1}{1-p^{-n k}}$ $\prod_{n=1}^\infty \prod_{d|n}e^{d \tau(d) p^{-n}}= \prod_{n=1}^\infty \prod_{k=1}^\infty \frac{1}{1-p^{-n k}}$ this then is a relation $\prod_{n=1}^\infty \prod_{d|n}e^{d \tau(d) p^{-n}}= \prod_{n=1}^\infty \frac{1}{(\frac{1}{p^n};\frac{1}{p^n})_\infty}$ interestingly we could write $\prod_{n=1}^\infty \prod_{d|n}e^{d \tau(d) p^{-n}}= \left(\prod_{k=1}^\infty \frac{1}{1-p^{-k}}\right)\prod_{n=2}^\infty \prod_{k=1}^\infty \frac{1}{1-p^{-n k}}$ then take the product over all primes $\prod_{\text{primes}}\frac{\prod_{n=1}^\infty \prod_{d|n}e^{d \tau(d) p^{-n}}}{\prod_{n=2}^\infty \prod_{k=1}^\infty \frac{1}{1-p^{-n k}}}= \prod_{\text{primes}}\left(\prod_{k=1}^\infty \frac{1}{1-p^{-k}}\right)$ $\prod_{\text{primes}}\frac{\prod_{n=1}^\infty \prod_{d|n}e^{d \tau(d) p^{-n}}}{\prod_{n=2}^\infty \prod_{k=1}^\infty \frac{1}{1-p^{-n k}}}= \prod_{k=1}^\infty\zeta(k)$ of course this is probably nonsense

# Other Interesting Sums

we can change $$\mu(n)$$ to be something else for example we seem to have $\sum_{n=1}^\infty\sum_{k=1}^\infty \frac{\varphi(k)}{k}\log\left(\frac{1}{1-x^{kn}}\right) = \sum_{n=1}^\infty d(n)x^n$ $\sum_{n=1}^\infty\sum_{k=1}^\infty \frac{\varphi(k)}{nk}\log\left(\frac{1}{1-x^{kn}}\right) = \sum_{n=1}^\infty \frac{\sigma_1(n)}{n}x^n$ $\sum_{n=1}^\infty\sum_{k=1}^\infty \frac{\varphi(k)}{1}\log\left(\frac{1}{1-x^{kn}}\right) = \sum_{n=1}^\infty \frac{\sigma_2(n)}{n}x^n$ for Euler totient function $$\varphi(n)$$ and divisors $$d(n)$$. $\sum_{n=1}^\infty\sum_{k=1}^\infty k\log\left(\frac{1}{1-x^{kn}}\right) = \sum_{n=1}^\infty \sum_{d|n}\frac{d \sigma_1(d)}{n}x^n$ $\sum_{n=1}^\infty\sum_{k=1}^\infty \sigma_0(k)\log\left(\frac{1}{1-x^{kn}}\right) = \sum_{n=1}^\infty \sum_{d|n}\frac{d \sigma_1(\frac{n}{d})\tau(d)}{n}x^n$ $\sum_{n=1}^\infty\sum_{k=1}^\infty k^2\log\left(\frac{1}{1-x^{kn}}\right) = \sum_{n=1}^\infty \sum_{d|n}\frac{d^3 \sigma_1(\frac{n}{d})}{n}x^n$ $\sum_{n=1}^\infty\sum_{k=1}^\infty \frac{1}{k}\log\left(\frac{1}{1-x^{kn}}\right) = \sum_{n=1}^\infty \sum_{d|n}\frac{\sigma_1(d)}{n}x^n$ other results are,$$k/n$$ transforms to $$A007433$$,$$(k-1)/k$$ transforms to $$A069914/n$$.

We seem to have $\sum_{n=1}^\infty\sum_{k=1}^\infty \frac{1}{n}\left(\sum_{d|n}\mu(d)\mu\left(\frac{n}{d}\right)\right)\log\left(\frac{1}{1-x^{kn}}\right) = \sum_{n=1}^\infty \frac{\varphi(n)}{n}x^n$ $\sum_{n=1}^\infty\sum_{k=1}^\infty \left(\sum_{d|n}\mu(d)\mu\left(\frac{n}{d}\right)\right)\log\left(\frac{1}{1-x^{kn}}\right) = \sum_{n=1}^\infty \frac{1}{n}\left(\sum_{d|n}d \mu(d)\right)x^n$ $\sum_{n=1}^\infty\sum_{k=1}^\infty \lambda(n)\log\left(\frac{1}{1-x^{kn}}\right) = \sum_{n=1}^\infty \frac{\text{sum of square divisors}(n)}{n}x^n$

# Other

$\sum_{n=1}^\infty\sum_{k=1}^\infty \lambda_3(n)\log\left(\frac{1}{1-x^{kn}}\right) = -\log(1-x)+\sum_{n=2}^\infty x^{n^3} + \sum_{k=2}^\infty \sum_{n=2}^\infty \frac{x^{k^3 n}}{n}$

where $$\lambda_3(n)$$ is the equivalent of the Liouville function for squares but for cubes.