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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:

=if(and(row()=1; column()=1); 3; if(or(and(row()=1; column()=2); and(row()=2; column()=1)); 2; if(row()=1; (-indirect(address(row(); column()-1))-sum(indirect(address(2; column())&":"&address(column()-1; column()))))/(column()+1)+indirect(address(row(); column()-1)); if(column()=1; (-indirect(address(row()-1; column()))-sum(indirect(address(row(); 2)&":"&address(row(); row()-1))))/(row()+1)+indirect(address(row()-1; column())); if(row()>=column(); -sum(indirect(address(row()-column()+1; column())&":"&address(row()-1; column()))); -sum(indirect(address(column()-row()+1; row())&":"&address(column()-1; row()))))))))

American version:

=if(and(row()=1, column()=1), 3, if(or(and(row()=1, column()=2), and(row()=2, column()=1)), 2, if(row()=1, (-indirect(address(row(), column()-1))-sum(indirect(address(2, column())&":"&address(column()-1, column()))))/(column()+1)+indirect(address(row(), column()-1)), if(column()=1, (-indirect(address(row()-1, column()))-sum(indirect(address(row(), 2)&":"&address(row(), row()-1))))/(row()+1)+indirect(address(row()-1, column())), if(row()>=column(), -sum(indirect(address(row()-column()+1, column())&":"&address(row()-1, column()))), -sum(indirect(address(column()-row()+1, row())&":"&address(column()-1, row()))))))))

share|improve this question
    
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

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