OFFSET
1,8
COMMENTS
Matrix inverse of A180051. Where the Excel formula says "randbetween(-9;9)" this table has the values of the Mertens function in the first column. Help with translating the spreadsheet formula would be appreciated.
FORMULA
Contribution from Mats Granvik, Aug 11 2010: (Start)
[from Wouter Meeussen, seqfan]
a(r,c)=0 /; c>r
a(r,c)=1 /; r<=3
a(r,1)=sum(Amu(k),k=1..r)
a(r,c)=a(r,c-1)-sum(a(r-j,c), j=1..c-1)/; c<=3
a(r,c)=sum(a(r-j,c-1), j=1..c-2)-sum(a(r-j,c), j=1..c-1)
(End)
EXAMPLE
Table begins:
1,
1,1,
1,1,1,
-1,-2,-3,1,
-1,1,3,-3,1,
-2,-3,-3,2,-3,1,
-1,2,2,0,2,-3,1,
-2,-4,-3,0,-1,2,-3,1,
-2,2,3,-3,3,-1,2,-3,1,
-2,-4,-4,3,-4,2,-1,2,-3,1,
-1,3,4,-1,0,-1,2,-1,2,-3,1,
-2,-5,-5,1,1,-1,-2,2,-1,2,-3,1,
-2,3,4,-4,3,-1,2,-2,2,-1,2,-3,1,
-3,-6,-5,3,-4,2,-2,1,-2,2,-1,2,-3,1,
MATHEMATICA
Contribution from Mats Granvik, Aug 11 2010: (Start)
[from Wouter Meeussen, seqfan]
Clear[a];
a[r_, c_]:=0 /; c>r;
a[r_, c_]:=1 /; r<=3;
a[r_, 1]:=Sum[MoebiusMu[k], {k, 0, r-1}];
a[r_, c_]:=a[r, c-1]-Sum[a[r-j, c], {j, 1, c-1}]/; c<=3;
a[r_, c_]:=a[r, c]=Sum[a[r-j, c-1], {j, 1, c-2}]-Sum[a[r-j, c], {j, 1, c-1}];
(m=Table[a[i, j], {i, 14}, {j, 14}])//ColumnForm
(End)
PROG
(Excel) Using European dot comma style:
=if(row()>=column(); if(row()<=3; 1; if(column()=1; randbetween(-9; 9); if(or(column()=2; column()=3); sum(indirect(address(row(); column()-1; 4)))-sum(indirect(address(row()-column()+1; column(); 4)&":"&address(row()-1; column(); 4); 4)); sum(indirect(address(row()-column()+2; column()-1; 4)&":"&address(row()-1; column()-1; 4); 4))-sum(indirect(address(row()-column()+1; column(); 4)&":"&address(row()-1; column(); 4); 4))))); 0)
CROSSREFS
KEYWORD
sign,tabl
AUTHOR
Mats Granvik, Aug 08 2010
STATUS
approved