|
Search: id:A118407
|
|
|
| A118407 |
|
Triangle, read by rows, equal to the matrix square of triangle A118404; also equals the matrix inverse of triangle A118401. |
|
+0
5
|
|
| 1, 0, 1, -2, 0, 1, 2, -2, 0, 1, 0, 2, -2, 0, 1, -2, 0, 2, -2, 0, 1, 4, -2, 0, 2, -2, 0, 1, -6, 4, -2, 0, 2, -2, 0, 1, 4, -6, 4, -2, 0, 2, -2, 0, 1, 6, 4, -6, 4, -2, 0, 2, -2, 0, 1, -20, 6, 4, -6, 4, -2, 0, 2, -2, 0, 1, 26, -20, 6, 4, -6, 4, -2, 0, 2, -2, 0, 1, -12, 26, -20, 6, 4, -6, 4, -2, 0, 2, -2, 0, 1
(list; table; graph; listen)
|
|
|
OFFSET
|
0,4
|
|
|
COMMENT
|
This triangle has an integer matrix square-root (A118404) if the main diagonal of the square-root is allowed to be signed. Even though the columns of this triangle are all the same, the columns of the matrix square-root A118404 are all different.
|
|
FORMULA
|
G.f.: A(x,y) = (1+x)^2/(1+x^2)/(1+2*x+2*x^2)/(1-x*y). Column g.f.: (1+x)^2/(1+x^2)/(1+2*x+2*x^2).
|
|
EXAMPLE
|
Triangle begins:
1;
0, 1;
-2, 0, 1;
2,-2, 0, 1;
0, 2,-2, 0, 1;
-2, 0, 2,-2, 0, 1;
4,-2, 0, 2,-2, 0, 1;
-6, 4,-2, 0, 2,-2, 0, 1;
4,-6, 4,-2, 0, 2,-2, 0, 1;
6, 4,-6, 4,-2, 0, 2,-2, 0, 1;
-20, 6, 4,-6, 4,-2, 0, 2,-2, 0, 1;
26,-20, 6, 4,-6, 4,-2, 0, 2,-2, 0, 1; ...
|
|
PROGRAM
|
(PARI) {T(n, k)=polcoeff(polcoeff((1+x)^2/(1+x^2)/(1+2*x+2*x^2)/(1-x*y+x*O(x^n)), n, x)+y*O(y^k), k, y)}
|
|
CROSSREFS
|
Cf. A118404 (matrix square-root), A118401 (matrix inverse), A118408 (row sums), A118409 (unsigned row sums).
Sequence in context: A094238 A127793 A127771 this_sequence A101663 A062169 A113680
Adjacent sequences: A118404 A118405 A118406 this_sequence A118408 A118409 A118410
|
|
KEYWORD
|
sign,tabl
|
|
AUTHOR
|
Paul D. Hanna (pauldhanna(AT)juno.com), Apr 27 2006
|
|
|
Search completed in 0.002 seconds
|
