A192763 Symmetric square array read by antidiagonals up.

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, 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, 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

Offset

1,2

Comments

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.

Links

Table of n, a(n) for n=1..120.

M. Granvik, Is this a recurrence for the Mertens function plus 2?

Formula

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).

Example

The array starts:
1..2..1..1..0..1..0..0..0..1...
2.-2..2.-2..2.-2..2.-2..2.-2...
1..2.-3..1..2.-3..1..2.-3..1...
1.-2..1..0..1.-2..1..0..1.-2...
0..2..2..1.-5..0..2..2..1.-5...
1.-2.-3.-2..0..6..1.-2.-3.-2...
0..2..1..1..2..1.-7..0..2..1...
0.-2..2..0..2.-2..0..0..0.-2...
0..2.-3..1..1.-3..2..0..0..0...
1.-2..1.-2.-5.-2..1.-2..0..10...

Mathematica

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}]];
nn = 12;
MatrixForm[Array[t, {nn, nn}]];
a = Flatten[Table[Reverse[Range[n]], {n, nn}]];
b = Flatten[Table[Range[n], {n, nn}]];
Table[t[a[[i]], b[[i]]], {i, 1, nn*(nn + 1)/2}]
(* Mats Granvik, Olivier Gérard, Jul 10 2011 *)
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 *)

Prog

(Excel cell formula, European version, American version uses ", " instead of "; ")
=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()))))))))
(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 */

Crossrefs

Cf. A002321, A055615.

Sequence in context: A071292 A088569 A246144 * A001285 A088424 A317960

Adjacent sequences: A192760 A192761 A192762 * A192764 A192765 A192766

Keyword

sign,tabl

Author

Mats Granvik, Jul 09 2011

Status

approved