A055138 Matrix inverse of Losanitsch's triangle A034851.

1, -1, 1, 0, -1, 1, 1, 0, -2, 1, -1, 2, 0, -2, 1, -1, -3, 6, 0, -3, 1, 4, -3, -7, 8, 0, -3, 1, -1, 18, -18, -13, 17, 0, -4, 1, -19, -4, 56, -28, -22, 20, 0, -4, 1, 31, -127, 60, 136, -98, -34, 36, 0, -5, 1, 120, 163, -511, 80, 288, -126, -50, 40, 0, -5, 1

Offset

0,9

Comments

Also the adjoint matrix of Losanitsch's triangle, since the determinants of n X n subarrays, rooted at (0,0), of Losanitsch's triangle are 1 and since for a matrix A, A^(-1) = 1/det(A) * (adjoint of A). - Gerald McGarvey, Oct 30 2007

Links

Michel Marcus, Rows n=0..50 of triangle, flattened

Example

Triangle begins:
   1;
  -1, 1;
   0,-1, 1;
   1, 0,-2, 1;
  -1, 2, 0,-2, 1;
  ...

Maple

b:= proc(n, k) (binomial(irem(n, 2, 'i'), irem(k, 2, 'j'))*
                binomial(i, j)+binomial(n, k))/2 end:
T:= n-> (M-> seq(M[n+1, j], j=1..n+1))(Matrix(n+1,
             (i, j)-> b(i-1, j-1))^(-1)):
seq(T(n), n=0..10);  # Alois P. Heinz, Mar 01 2022

Mathematica

nmax = 10;
b[n_?EvenQ, k_?OddQ] := Binomial[n, k]/2;
b[n_, k_] := (Binomial[n, k] + Binomial[Quotient[n, 2], Quotient[k, 2]])/2;
M = Table[b[n, k], {n, 0, nmax}, {k, 0, nmax}] // Inverse;
T[n_, k_] := M[[n+1, k+1]];
Table[T[n, k], {n, 0, nmax}, {k, 0, n}] // Flatten (* Jean-François Alcover, Oct 13 2022 *)

Prog

(PARI) T(n, k) = (1/2)*(binomial(n, k) + binomial(n%2, k%2) * binomial(n\2, k\2)); \\ A034851
row(n) = my(m=matrix(n+1, n+1, i, j, i--; j--; T(i, j))); vector(n+1, i, (1/m)[n+1, i]); \\ Michel Marcus, Mar 01 2022

Crossrefs

Cf. A034851.

Sequence in context: A357332 A347367 A194529 * A177717 A155997 A326171

Adjacent sequences: A055135 A055136 A055137 * A055139 A055140 A055141

Keyword

sign,tabl

Author

Christian G. Bower, May 09 2000

Extensions

Typo in definition corrected by Georg Fischer, Mar 01 2022

Status

approved