Jeffrey Shallit formulated this recurrence for me: $\displaystyle T(n,1)=1, k>1: T(n,k) = \sum\limits_{i=1}^{k-1} T(n-i,k-1)-\sum\limits_{i=1}^{k-1} T(n-i,k)$
which is the lower triangular array equal to 1 if k divides n, 0 otherwise.
By changing the recurrence so that it takes values from either the vertical or horizontal direction:
$\displaystyle T(n,1)=1, T(1,k)=1, n>=k: -\sum\limits_{i=1}^{k-1} T(n-i,k), n<k: -\sum\limits_{i=1}^{n-1} T(k-i,n)$
we get this array starting:
$$\displaystyle T = \left( \begin{array}{ccccccc} +1&+1&+1&+1&+1&+1&+1&\cdots \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{array} \right) $$
Do the series $\displaystyle \sum\limits_{k=1}^{\infty}\frac{T(n,k)}{k}$ $\;$ converge to the Mangoldt function $\Lambda(n)$?
http://mathworld.wolfram.com/MangoldtFunction.html
Edit 14.7.2011, added Mathematica program:
Clear[t];
nn = 100;
mm = 15;
t[n_, 1] = 1;
t[1, k_] = 1;
t[n_, k_] :=
t[n, k] =
If[n < k,
If[And[n > 1, k > 1], Sum[-t[k - i, n], {i, 1, n - 1}], 0],
If[And[n > 1, k > 1], Sum[-t[n - i, k], {i, 1, k - 1}], 0]];
a = Table[Table[t[n, k], {k, 1, mm}], {n, 1, nn}];
b = Range[1, nn];
c = a/b;
MatrixForm[c];
d = N[Table[Total[c[[All, i]]], {i, 1, mm}]]
d[[1]] = 0;
mangoldt = Exp[d]
mangoldtexponentiated = Round[Exp[d]]
that outputs the sequence: $1, 2, 3, 2, 5, 1, 7, 2, 3, 1, 11, 1, 13, 1, 1...$ which is the Mangoldt function exponentiated.
Edit 9.2.2014:
Just for memory:
$$\varphi (n) = n\lim\limits_ {s \rightarrow 1}\zeta (s)\sum\limits_ {d | n}\mu (d) (e^{1/d})^{(s - 1)}$$
$$a(n) = \lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n} \mu(d)(e^{d})^{(s-1)}$$
$$\Lambda(n) =\sum\limits_{k=1}^{\infty} \frac{a(GCD(n,k))}{k}$$
$$\Lambda(n)=\lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}$$
$$\text{Fourier Transform of } \Lambda(n) \sim \sum\limits_{n=1}^{n=k} \frac{1}{n} \zeta(1/2+i \cdot t)\sum\limits_{d|n} \frac{\mu(d)}{d^{(1/2+i \cdot t-1)}}$$
Edit 3.3.2014:
Just for memory: Mathematica:
nn = 12;
mm = nn;
MatrixForm[
Chop[N[Total[
Transpose[
Table[Table[
If[Mod[n1, k1] == 0,
Table[(Table[
Sum[Exp[-a*b/n*2*Pi*I], {b, 1, n}], {a, 1, mm}]), {n, 1,
nn}][[n1/k1]]*MoebiusMu[n1/k1], 0], {k1, 1, nn}], {n1, 1,
nn}]]]]]]
Edit 14.9.2014:
Just for memory: Conjectured formula from Dirichlet characters:
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 *)
Just for memory (18.1.2015) :
A = Table[Table[If[Mod[n, k] == 0, 1, 0], {k, 1, 12}], {n, 1, 12}];
B = Table[
Table[If[Mod[k, n] == 0, MoebiusMu[n]*n, 0], {k, 1, 12}], {n, 1,
12}];
MatrixForm[A.B]
Just for memory (20.1.2015):
nn = 42
Z = Table[Table[If[Mod[n, k] == 0, 1, 0], {k, 1, nn}], {n, 1, nn}];
A = Table[Table[If[Mod[n, k] == 0, k, 0], {k, 1, nn}], {n, 1, nn}];
B = Table[
Table[If[Mod[k, n] == 0, MoebiusMu[n], 0], {k, 1, nn}], {n, 1, nn}];
MatrixForm[T = Z.A.B];
T[[All, 1]] = 0;
a = Table[Total[Total[T[[1 ;; n, 1 ;; n]]]], {n, 1, nn}]
a = Table[Total[Total[T[[1 ;; n, 1 ;; n]]]]/n, {n, 1, nn}]
g1 = ListLinePlot[a];
b = Accumulate[MangoldtLambda[Range[nn]]];
g2 = ListLinePlot[b];
Show[g1, g2]
Relation to square roots:
nn = 32;
A = Table[
Table[If[Mod[n, k] == 0, Sqrt[k], 0], {k, 1, nn}], {n, 1, nn}];
B = Table[
Table[If[Mod[k, n] == 0, MoebiusMu[n]*Sqrt[n], 0], {k, 1, nn}], {n,
1, nn}];
MatrixForm[A.B]