A Lambert like Transformation with a Logarithmic Kernel

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.

[Someone else is editing this]