Let $a(n)$ be the Dirichlet inverse of the Euler totient function:
$$a(n) = \sum\limits_{d|n} d \cdot \mu(d) \tag{1}$$
And let the matrix $T$ be: $$T(n,k)=a(\gcd(n,k)) \tag{2}$$
Compute the ordinary generating functions:
$$q(n)=\sum _{h=0}^{\infty } \left(\sum _{k=1}^n x^{h n+k} a(\gcd (n,k))\right) \tag{3}$$
starting:
$$q(1,2,3,...)=\frac{x}{1-x},\frac{x}{x+1},\frac{2 x^2+x}{x^2+x+1},\frac{x}{x+1},\frac{4 x^4+3 x^3+2 x^2+x}{x^4+x^3+x^2+x+1},\frac{x-2 x^2}{x^2-x+1},\frac{6 x^6+5 x^5+4 x^4+3 x^3+2 x^2+x}{x^6+x^5+x^4+x^3+x^2+x+1},\frac{x}{x+1},\frac{2 x^2+x}{x^2+x+1},\frac{-4 x^4+3 x^3-2 x^2+x}{x^4-x^3+x^2-x+1},...$$
Compute the roots of the denominators of $q(n)$.
Question:
Show that apart from signs, the matrix $T(n,k)=a(\gcd(n,k))$ can be generated from powers of those roots.
In Mathematica the essential command, for which I don't know the latex for, is:
Total[(x /. Solve[denominator[[n]] == 0, x])^k]
where x /. removes the x symbol.
Mathematica program to demonstrate it is true for the first few terms of the matrix:
Clear[s, x, a, t, q]; nn = 7;
a[n_] := Total[Divisors[n]*MoebiusMu[Divisors[n]]];
Monitor[q =
Table[Sum[
Sum[a[GCD[n, k]]*x^(n*h + k), {k, 1, n}], {h, 0, Infinity}], {n,
1, nn}];, n]; denominator = q^-1;
TableForm[Round[Table[Table[a[GCD[n, k]], {k, 1, nn}], {n, 1, nn}]]]
TableForm[
Round[Table[
Table[Total[(x /. Solve[denominator[[n]] == 0, x])^k], {k, 1,
nn}], {n, 1, nn}]]]
Comparison of outputs:
The original infinite matrix starting:
$$\begin{array}{lllllll} 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & -1 & 1 & -1 & 1 & -1 & 1 \\ 1 & 1 & -2 & 1 & 1 & -2 & 1 \\ 1 & -1 & 1 & -1 & 1 & -1 & 1 \\ 1 & 1 & 1 & 1 & -4 & 1 & 1 \\ 1 & -1 & -2 & -1 & 1 & 2 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 & -6 \end{array}$$
compared to the infinite matrix starting:
$$\begin{array}{lllllll} 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ -1 & 1 & -1 & 1 & -1 & 1 & -1 \\ -1 & -1 & 2 & -1 & -1 & 2 & -1 \\ -1 & 1 & -1 & 1 & -1 & 1 & -1 \\ -1 & -1 & -1 & -1 & 4 & -1 & -1 \\ 1 & -1 & -2 & -1 & 1 & 2 & 1 \\ -1 & -1 & -1 & -1 & -1 & -1 & 6 \end{array}$$
where only the signs differ.
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