The way to separately write the terms of the von Mangoldt function $\Lambda$ as Dirichlet character sums seems to be:
$\Lambda (1) = \sum\limits_{n=1}^{\infty } \frac{(e^{\Lambda (1)} \chi _{1,1}(n)-e^{\Lambda (2)}+1)}{n} = \infty$
$\Lambda (2) = \sum\limits_{n=1}^{\infty } \frac{(e^{\Lambda (2)} \chi _{2,1}(n)-e^{\Lambda (2)}+1)}{n}$
$\Lambda (3) = \sum\limits_{n=1}^{\infty } \frac{(e^{\Lambda (3)} \chi _{3,1}(n)-e^{\Lambda (3)}+1)}{n}$
$\Lambda (4) = \sum\limits_{n=1}^{\infty } \frac{(e^{\Lambda (2)} \chi _{4,1}(n)-e^{\Lambda (2)}+1)}{n}$
$\Lambda (5) = \sum\limits_{n=1}^{\infty } \frac{(e^{\Lambda (5)} \chi _{5,1}(n)-e^{\Lambda (5)}+1)}{n}$
$\Lambda (6) = \sum\limits_{n=1}^{\infty } \frac{(e^{\Lambda (2)} \chi _{2,1}(n)-e^{\Lambda (2)}+1) (e^{\Lambda (3)} \chi _{3,1}(n)-e^{\Lambda (3)}+1)}{n}$
$\Lambda (7) = \sum\limits_{n=1}^{\infty } \frac{(e^{\Lambda (7)} \chi _{7,1}(n)-e^{\Lambda (7)}+1)}{n}$
$\Lambda (8) = \sum\limits_{n=1}^{\infty } \frac{(e^{\Lambda (2)} \chi _{8,1}(n)-e^{\Lambda (2)}+1)}{n}$
$\Lambda (9) = \sum\limits_{n=1}^{\infty } \frac{(e^{\Lambda (3)} \chi _{9,1}(n)-e^{\Lambda (3)}+1)}{n}$
$\Lambda (10) = \sum\limits_{n=1}^{\infty } \frac{(e^{\Lambda (2)} \chi _{2,1}(n)-e^{\Lambda (2)}+1) (e^{\Lambda (5)} \chi _{5,1}(n)-e^{\Lambda (5)}+1)}{n}$
$\Lambda (11) = \sum\limits_{n=1}^{\infty } \frac{(e^{\Lambda (11)} \chi _{11,1}(n)-e^{\Lambda (11)}+1)}{n}$
$\Lambda (12) = \sum\limits_{n=1}^{\infty } \frac{(e^{\Lambda (2)} \chi _{2,1}(n)-e^{\Lambda (2)}+1) (e^{\Lambda (3)} \chi _{3,1}(n)-e^{\Lambda (3)}+1)}{n}$
Is there a shorter way to write this?
The special cases above that I don't know how to write are $\Lambda (6)$, $\Lambda (10)$ and $\Lambda (12)$ where a product of the characters defined by prime factors is needed.
I guess that the statements above don't need to be proven because the answer to this question was positive.
The general form appears to be:
$\Lambda(n) = \sum\limits_{k=1}^{\infty} \frac{\prod _{m=1}^{\text{Length}[\text{Divisors}[n]]} \left(\exp ^{-\mu (\text{Divisors}[n])}(\Lambda (\text{Divisors}[n]))[[m]] \chi _{\exp ^{-\mu (\text{Divisors}[n])}(\Lambda (\text{Divisors}[n]))[[m]],j}(k)-\left(\exp ^{-\mu (\text{Divisors}[n])}(\Lambda (\text{Divisors}[n]))[[m]]-1\right)\right)}{k}$
In my browser the expression above splits into three rows but the sum is meant to be the whole expression. See the Mathematica program below for a clearer expression:
j = 1; MatrixForm[ Table[Table[ Product[((Exp[ MangoldtLambda[Divisors[n]]]^-MoebiusMu[Divisors[n]])[[m]]* DirichletCharacter[(Exp[ MangoldtLambda[Divisors[n]]]^-MoebiusMu[ Divisors[n]])[[m]], j, k] - ( (Exp[ MangoldtLambda[Divisors[n]]]^-MoebiusMu[ Divisors[n]])[[m]] - 1)), {m, 1, Length[Divisors[n]]}], {n, 1, 32}], {k, 1, 32}]]
What I am interested in is if: $a(GCD(n,k)) = \prod _{m=1}^{\text{Length}[\text{Divisors}[n]]} \left(\exp ^{-\mu (\text{Divisors}[n])}(\Lambda (\text{Divisors}[n]))[[m]] \chi _{\exp ^{-\mu (\text{Divisors}[n])}(\Lambda (\text{Divisors}[n]))[[m]],j}(k)-\left(\exp ^{-\mu (\text{Divisors}[n])}(\Lambda (\text{Divisors}[n]))[[m]]-1\right)\right)$
where:
$a(n) = \lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n} \mu(d)\exp(d)^{(s-1)}$
which is also known as the Dirichlet inverse of the Euler totient function.
In the expressions above $\exp ^{-\mu (n)}(\Lambda (n))$ is the sequence starting:
1, 2, 3, 1, 5, 1, 7, 1, 1, 1, 11, 1,...
equal to $n$ if $n$ is prime and $1$ otherwise.
nn = 12; b = Table[Exp[MangoldtLambda[Divisors[n]]]^-MoebiusMu[Divisors[n]], {n, 1, nn}]; j = 1; MatrixForm[ Table[Table[ Product[(b[[n]][[m]]* DirichletCharacter[b[[n]][[m]], j, k] - (b[[n]][[m]] - 1)), {m, 1, Length[Divisors[n]]}], {n, 1, nn}], {k, 1, nn}]] (*Conjectured expression as Dirichlet characters.Mats \ Granvik,Nov 23 2013*)
Attempt at Dirichlet L functions:
Table[Limit[ Exp[MangoldtLambda[n]] DirichletL[n, 1, s] - Exp[MangoldtLambda[n]]*Zeta[s] + Zeta[s], s -> 1], {n, 1, 42}]