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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*)
share|improve this question
    
What is $\chi_{2,1}(n)$? By the way, I find it difficult to believe that, say, log(2) would have an expression like above. –  user27126 Nov 4 '12 at 8:48
    
I don't know. But in mathematica this is written with the command: "DirichletCharacter[2, 1, n]". –  Mats Granvik Nov 4 '12 at 8:50
    
OK. For example in $\Lambda(2)$. I think your RHS is (1 - 2 + 1 - 2 + ...) Which I don't see why it's the same as LHS. –  user27126 Nov 4 '12 at 8:55
    
Yes you are right. I forgot to divide by "n". I will correct that now. –  Mats Granvik Nov 4 '12 at 8:57
    
In any case, I find it very strange that one may want to write Mangoldt function in terms of Dirichlet characters. Can you explain your motivation? –  user27126 Nov 4 '12 at 9:00

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