# Is this a recurrence for the Mertens function plus 2?

If we define a symmetric array:

$$T(1,1)=3,\; T(1,2)=2,\; T(2,1)=2$$

$$T(1,k)=\frac{-T(n,k-1)-\sum\limits_{i=2}^{k-1}T(i,k)}{k+1}+T(n,k-1)$$

$$T(n,1)=\frac{-T(n-1,k)-\sum\limits_{i=2}^{n-1}T(n,i)}{n+1}+T(n-1,k)$$

$$n\geq k>1: T(n,k) = -\sum\limits_{i=1}^{k-1}T(n-i,k)$$

$$k>n>1: T(n,k) = -\sum\limits_{i=1}^{n-1}T(k-i,n)$$

starting:

$$T(n,k) = \begin{bmatrix} +3&+2&+1&+1&+0&+1&+0 \\ +2&-2&+2&-2&+2&-2&+2 \\ +1&+2&-3&+1&+2&-3&+1 \\ +1&-2&+1&+0&+1&-2&+1 \\ +0&+2&+2&+1&-5&+0&+2 \\ +1&-2&-3&-2&+0&+6&+1 \\ +0&+2&+1&+1&+2&+1&-7 \end{bmatrix}$$

...and as a Mathematica program:

t[n_, k_] :=
t[n, k] =
If[And[n == 1, k == 1], 3,
If[Or[And[n == 1, k == 2], And[n == 2, k == 1]], 2,
If[n == 1,(-t[n,k-1]-Sum[t[i,k],{i,2,k-1}])/(k+1)+t[n,k-1],
If[k == 1,(-t[n-1,k]-Sum[t[n,i],{i,2,n-1}])/(n+1)+t[n-1,k],
If[n >= k, -Sum[t[n - i, k], {i, 1, k - 1}], -Sum[
t[k - i, n], {i, 1, n - 1}]]]]]];
nn = 81;
MatrixForm[Table[Table[t[n, k], {k, 1, nn}], {n, 1, nn}]];
Table[t[1, k], {k, 1, nn}] - 2


Then we get in the first row and first column a sequence starting:

3, 2, 1, 1, 0, 1, 0, 0, 0, 1, 0, 0, -1, 0, 1, 1, 0, 0, -1, -1,...

By subtracting the sequence with 2, do we then get the Mertens function?

1, 0, -1, -1, -2, -1, -2, -2, -2, -1, -2, -2, -3, -2, -1, -1, -2, -2, -3, -3,...

The Mertens function is the partial sums of the Möbius function.

Edit Nov 6 2011:

Excel spreadsheet formulas for the array:

European version:

When you give the equation for $T(1,k)$, why is there an $n$ on the left hand side? –  Eric Naslund Jul 10 '11 at 23:03
@Eric Naslund: Did you mean the right hand side or is it something that has been edited out? I realize now that the $n$ on the right hand side in the equation for $T(1,k)$ should have been equal to $1$. The same goes for $T(n,1)$ and $k$. $\; k$ is equal to $1$. –  Mats Granvik Jul 11 '11 at 8:02