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%I #25 Dec 10 2016 17:00:02
%S 1,2,2,1,-2,1,1,2,2,1,0,-2,-3,-2,0,1,2,1,1,2,1,0,-2,2,0,2,-2,0,0,2,-3,
%T 1,1,-3,2,0,0,-2,1,-2,-5,-2,1,-2,0,1,2,2,1,0,0,1,2,2,1,0,-2,-3,0,2,6,
%U 2,0,-3,-2,0,0,2,1,1,2,1,1,2,1,1,2,0,-1,-2,2,-2,1,-2,-7,-2,1,-2,2,-2,-1,0,2,-3,1,-5,-3,0,0,-3,-5,1,-3,2,0,1,-2,1,0,0,-2,2,0,2,-2,0,0,1,-2,1
%N Symmetric square array read by antidiagonals up.
%C The main diagonal is the Mobius function times the natural numbers A055615 (conjecture). For k>1 the first row is the Mertens function + 2 = A002321 + 2 (conjecture). There is one recurrence for n=1 and k=1, and another recurrence for n>1 and k>1.
%H M. Granvik, <a href="http://math.stackexchange.com/questions/50719/">Is this a recurrence for the Mertens function plus 2?</a>
%F T(1,1)=1 or 3, T(1,2)=2, T(2,1)=2, T(1,k)=(-T(n,k-1)-Sum_(i=2)^(k-1) of T(i,k))/(k+1)+T(n,k-1), T(n,1)=(-T(n-1,k)-Sum_(i=2)^(n-1) of T(n,i))/(n+1)+T(n-1,k), n>=k: -Sum_(i=1)^(k-1) of T(n-i,k), n<k: -Sum_(i=1)^(n-1) of T(k-i,n).
%e The array starts:
%e 1..2..1..1..0..1..0..0..0..1...
%e 2.-2..2.-2..2.-2..2.-2..2.-2...
%e 1..2.-3..1..2.-3..1..2.-3..1...
%e 1.-2..1..0..1.-2..1..0..1.-2...
%e 0..2..2..1.-5..0..2..2..1.-5...
%e 1.-2.-3.-2..0..6..1.-2.-3.-2...
%e 0..2..1..1..2..1.-7..0..2..1...
%e 0.-2..2..0..2.-2..0..0..0.-2...
%e 0..2.-3..1..1.-3..2..0..0..0...
%e 1.-2..1.-2.-5.-2..1.-2..0..10...
%t Clear[t]; t[1, 1] = 1; t[2, 1] = t[1, 2] = 2; t[n_Integer, k_Integer] := t[n, k] = Which[n == 1, (-t[n, k - 1] - Sum[t[i, k], {i, 2, k - 1}])/(k + 1) + t[n, k - 1], k == 1, (-t[n - 1, k] - Sum[t[n, i], {i, 2, n - 1}])/(n + 1) + t[n - 1, k], n >= k, -Sum[t[n - i, k], {i, 1, k - 1}], True, -Sum[t[k - i, n], {i, 1, n - 1}]];
%t nn = 12;
%t MatrixForm[Array[t, {nn, nn}]];
%t a = Flatten[Table[Reverse[Range[n]], {n, nn}]];
%t b = Flatten[Table[Range[n], {n, nn}]];
%t Table[t[a[[i]], b[[i]]], {i, 1, nn*(nn + 1)/2}]
%t (* Mats Granvik, _Olivier Gérard_, Jul 10 2011 *)
%t T[ n_, k_] := If[ n < 1 || k < 1, 0, If[ k > n, T[ k, n], T[n, k] = If[ k == 1, If[ n < 3, n, (-T[ n - 1, 1] - Sum[ T[ n, i], {i, 2, n - 1}]) / (n + 1) + T[ n - 1, 1]], If[ n > k, T[ k, Mod[ n, k, 1]], - Sum[ T[ n, i], {i, n - 1}]]]]]; (* _Michael Somos_, Jul 19 2011 *)
%o (Excel cell formula, European version, American version uses "," instead of ";")
%o =if(and(row()=1;column()=1);1;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()))))))))
%o (PARI) {T(n, k) = if( n<1 || k<1, 0, if( k>n, T(k, n), if( k==1, if( n<3, n, (-T(n-1, 1) -sum( i=2, n-1, T(n, i))) / (n+1) + T(n-1, 1)), if( n>k, T(k, (n-1)%k+1), -sum( i=1,n-1, T(n, i))))))}; /* _Michael Somos_, Jul 19 2011 */
%Y Cf. A002321, A055615.
%K sign,tabl
%O 1,2
%A _Mats Granvik_, Jul 09 2011
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