[seqfan] Re: Mobius function modulo 3.

[Charles Greathouse]

For the first 12 terms.  But if you change nn to 13 you get 427473 = 3
* 142491 as the next term, which suggests mu(13) = 0.

Charles Greathouse
Analyst/Programmer
Case Western Reserve University

On Fri, Jun 10, 2011 at 10:04 AM, Mats Granvik <mgranvik at abo.fi> wrote:
> This Mathematica program produces a sequence 1,-1,2,-6,20,-68... which
> modulo 3 is equal to the mobius function 1,-1,-1,0,-1,1... modulo 3:
> 1,2,2,0,2,1...
>
> Clear[t, n, k, a, b, c, d, e, f, g, h, i, j, k, l, s, u, mat1, mat2,
>  aa, bb, nn, signmatrix];
> nn = 12;
> t[1, 1] = 1;
> t[2, 1] = 1;
> t[3, 1] = 1;
> t[4, 1] = 1;
> t[5, 1] = 1;
> t[6, 1] = 1;
> t[7, 1] = 1;
> t[8, 1] = 1;
> t[9, 1] = 1;
> t[10, 1] = 1;
> t[11, 1] = 1;
> t[12, 1] = 1;
> t[s_, u_] :=
>  t[s, u] =
>   If[And[s > 1, u > 1],
>    Sum[t[s - q, u - 1] + 2*t[s - q, u], {q, 1, u - 1}], 0];
> mat1 = Table[Table[t[s, u], {u, 1, nn}], {s, 1, nn}];
> mat2 = Inverse[mat1];
> MatrixForm[mat2]
> aa = Mod[mat2, 3];
> MatrixForm[aa]
> mat1[[1]][[nn]] = mat1[[nn]][[nn]];
> mat1[[nn]][[nn]] = 0;
> MatrixForm[mat1];
> bb = Det[mat1];
> Mod[bb, 3]
> MatrixForm[(-((-2)^aa) + 1)/3]
>
> Mats Granvik
>
>
>
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