Interesting representation of $e^x$

We'll use $D(f(n),s)$ to denote the Dirichlet series of arithmetic function $f(n)$.

The recurrence relation obtained by marty cohen is $$\sum_{d|n}da_{d}(-1)^{\frac{n}{d}-1}= \begin{cases} 1 &\mbox{if } n=1,\\ 0&\mbox{if } n>1. \end{cases} $$ We can rewrite it as Dirichlet convolution $$ na_{n}*(-1)^{n-1}=I, $$ where $I$ is the multiplicative identity of Dirichlet convolution.

The convolution can be translated into the language of Dirichlet series as $$ D(na_{n},s)D((-1)^{n-1},s)=1. $$ On the other hand, we have $$ D((-1)^{n-1},s)=\eta(s)=(1-2^{1-s})\zeta(s), $$ where $\eta(s)$ is the Dirichlet eta function.

Thus we have $$\begin{align*} D(na_n,s)&=\frac{1}{1-2^{1-s}}\frac{1}{\zeta(s)}\\ &=\left(\sum_{n\ge 0}\frac{2^n}{(2^n)^s}\right)D(\mu(n),s).\\ \end{align*}.$$

Translating the product of Dirichlet series into the language of Dirichlet convolution, we have $$\begin{align*} a_n&=\frac{1}{n}\sum_{d|n}d[\log_{2}d\in \mathbb{N}]\mu(\frac{n}{d})\\ &=\frac{1}{n}\sum_{k=0}^{\nu_{2}(n)}2^k\mu(\frac{n}{2^k}),\\ \end{align*} $$ where $[P]$ denotes the Iverson bracket.

If $\mu(\frac{n}{2^{\nu_{2}(n)}})=0$, clearly we have $$a_n=0=\frac{\mu(\frac{n}{2^{\nu_{2}(n)}})}{\mathrm{rad}(n)}.$$

If $\mu(\frac{n}{2^{\nu_{2}(n)}})\neq 0$, we have $$ a_n=\frac{1}{n}\mu(n)=\frac{\mu(\frac{n}{2^{\nu_{2}(n)}})}{\mathrm{rad}(n)},$$ when $\nu_{2}(n)=0,$ and $$\begin{align*} a_n&=\frac{1}{n}\left(2^{\nu_{2}(n)}\mu(\frac{n}{2^{\nu_{2}(n)}})+2^{\nu_{2}(n)-1}\mu(2\frac{n}{2^{\nu_{2}(n)}})\right)\\ &=\frac{1}{n}\left(2^{\nu_{2}(n)}\mu(\frac{n}{2^{\nu_{2}(n)}})+\frac{1}{2}2^{\nu_{2}(n)}\mu(\frac{n}{2^{\nu_{2}(n)}})\mu(2)\right)\\ &=\frac{2^{\nu_{2}(n)}}{2n}\mu(\frac{n}{2^{\nu_{2}(n)}})\\ &=\frac{\mu(\frac{n}{2^{\nu_{2}(n)}})}{\mathrm{rad}(n)},\\ \end{align*} $$ when $\nu_{2}(n)\ge 1.$

Combining the results from all cases, we've proved your conjecture $$a_n=\frac{\mu(\frac{n}{2^{\nu_{2}(n)}})}{\mathrm{rad}(n)}.$$

Following Michael's suggestion, if $x =\sum_{n=1}^{\infty} a_n \ln(x^n+1) $, then $1 =\sum_{n=1}^{\infty} a_n \frac{nx^{n-1}}{x^n+1} $, so

$\begin{array}\\ 1 &=\sum_{n=1}^{\infty} a_n \frac{nx^{n-1}}{x^n+1}\\ &=\sum_{n=1}^{\infty} a_n nx^{n-1}\sum_{m=0}^{\infty} (-1)^m x^{mn}\\ \text{or}\\ x &=\sum_{n=1}^{\infty} a_n nx^{n}\sum_{m=0}^{\infty} (-1)^m x^{mn}\\ &=\sum_{n=1}^{\infty} a_n n\sum_{m=0}^{\infty} (-1)^m x^{mn+n}\\ &=\sum_{n=1}^{\infty} a_n n\sum_{m=0}^{\infty} (-1)^m x^{n(m+1)}\\ &=\sum_{n=1}^{\infty} a_n n\sum_{m=1}^{\infty} (-1)^{m-1} x^{nm}\\ &=\sum_{k=1}^{\infty}\sum_{d|k} a_d d (-1)^{k/d-1} x^{k}\\ &=\sum_{k=1}^{\infty}x^k\sum_{d|k} a_d d (-1)^{k/d-1}\\ \text{so that}\\ a_1 &=1\\ \text{and}\\ 0 &=\sum_{d|k} a_d d (-1)^{k/d-1} \qquad\text{for } k > 1\\ &=ka_k+\sum_{d|k, d<k} a_d d (-1)^{k/d-1} \text{or}\\ k a_k &=\sum_{d|k, d<k} a_d d (-1)^{k/d}\\ \text{or}\\ a_k &=\dfrac1{k}\sum_{d|k, d<k} a_d d (-1)^{k/d}\\ &=\dfrac1{k}\left((-1)^k+\sum_{d|k, 1<d<k} a_d d (-1)^{k/d}\right)\\ \end{array} $

In particular, if $p$ is prime, $a_p =\dfrac{(-1)^p}{p} $, so $a_2 = \frac12$ and $a_p = \dfrac{-1}{p}$ if $p$ is an odd prime.

If $k = 2^m$,

$\begin{array}\\ a_{2^m} &=\dfrac1{2^m}\left(1+\sum_{d|2^m, 1<d<2^m} a_d d (-1)^{2^m/d}\right)\\ &=\dfrac1{2^m}\left(1+\sum_{j=1}^{m-1} a_{2^j} 2^j \right)\\ \end{array} $

Therefore $a_4 =\frac14(1+2a_2) =\frac12 $.

If $a_{2^j} =\frac12 $ for $1 < j < m$, then

$\begin{array}\\ a_{2^m} &=\dfrac1{2^m}\left(1+\sum_{j=1}^{m-1} \frac12 2^j \right)\\ &=\dfrac1{2^m}\left(1+\sum_{j=0}^{m-2} 2^j \right)\\ &=\dfrac1{2^m}\left(1+2^{m-1}-1 \right)\\ &=\frac12\\ \end{array} $

for all $m$.

If $k = p^2$, $a_{p^2} =\dfrac1{p^2}(-1+pa_p(-1)^p) =\dfrac1{p^2}(-1+p\frac{-1}{p}(-1)) =0 $.

If $k = p^m$ where $m > 2$, suppose $a_{p^j} = 0 $ for $1 < j < m$. This is true for $m=3$.

Then

$\begin{array}\\ a_{p^m} &=\dfrac1{p^m}\left(1+\sum_{d|p^m, 1<d<p^m} a_d d (-1)^{p^m/d}\right)\\ &=\dfrac1{p^m}\left(1+\sum_{j=1}^{m-1} a_{p^j} p^j \right)\\ &=\dfrac1{p^m}\left(1+a_p p +\sum_{j=2}^{m-1} a_{p^j} p^j \right)\\ &=0\\ \end{array} $

If $k=2p$ where $p$ is an odd prime,

$\begin{array}\\ a_{2p} &=\dfrac1{2p}\left(1+2a_2(-1)^p+a_pp(-1)^2\right)\\ &=\dfrac1{2p}\left(1-2\dfrac12+\dfrac{-1}{p}p\right)\\ &=\dfrac1{2p}\left(1-1-1\right)\\ &=\dfrac{-1}{2p}\\ \end{array} $

If $k=pq$ where $p$ and $q$ are distinct odd primes, then $a_{pq} =\dfrac1{pq}\left((-1)^{pq}+a_pp(-1)^p+a_qq(-1)^q\right) =\dfrac1{pq}\left(-1+1+1\right) =\dfrac{1}{pq} $.

I'll leave it at this.