A118438 Triangle T, read by rows, equal to the matrix product T = H*C*H, where H is the self-inverse triangle A118433 and C is Pascal's triangle.

1, -1, 1, 5, -2, 1, 11, -9, -3, 1, -23, 44, 30, -4, 1, -41, 125, 110, -30, -5, 1, 45, -246, -345, 220, 75, -6, 1, -29, -301, -861, 875, 385, -63, -7, 1, 337, -232, 1260, -2296, -1610, 616, 140, -8, 1, 1199, -3015, -1044, -3612, -5166, 3150, 924, -108, -9, 1

Offset

0,4

Comments

The matrix inverse of H*C*H is H*[C^-1]*H = A118435, where H^2 = I (identity). The matrix log, log(T) = -A118441, is a matrix square root of a triangular matrix with a single diagonal (two rows down from the main diagonal).

Links

Table of n, a(n) for n=0..54.

Formula

Since T + T^-1 = C + C^-1, then [T^-1](n,k) = (1+(-1)^(n-k))*C(n,k) - T(n,k) is a formula for the matrix inverse T^-1 = A118435.

Example

Triangle begins:
1;
-1, 1;
5,-2, 1;
11,-9,-3, 1;
-23, 44, 30,-4, 1;
-41, 125, 110,-30,-5, 1;
45,-246,-345, 220, 75,-6, 1;
-29,-301,-861, 875, 385,-63,-7, 1;
337,-232, 1260,-2296,-1610, 616, 140,-8, 1;
1199,-3015,-1044,-3612,-5166, 3150, 924,-108,-9, 1; ...

Mathematica

nmax = 12;
h[n_, k_] := Binomial[n, k]*(-1)^(Quotient[n+1, 2] - Quotient[k, 2]+n-k);
H = Table[h[n, k], {n, 0, nmax}, {k, 0, nmax}];
Cn = Table[Binomial[n, k], {n, 0, nmax}, {k, 0, nmax}];
Tn = H.Cn.H;
T[n_, k_] := Tn[[n+1, k+1]];
Table[T[n, k], {n, 0, nmax}, {k, 0, n}] // Flatten (* Jean-François Alcover, Apr 08 2024 *)

Prog

(PARI) {T(n, k)=local(M=matrix(n+1, n+1, r, c, if(r>=c, binomial(r-1, c-1)*(-1)^(r\2- (c-1)\2+r-c))), C=matrix(n+1, n+1, r, c, if(r>=c, binomial(r-1, c-1)))); (M*C*M)[n+1, k+1]}

Crossrefs

Cf. A118439 (column 0), A118440 (row sums), A118435 (matrix inverse), A118441 (-matrix log); A118433 (self-inverse H).

Sequence in context: A371848 A011507 A136643 * A336244 A083801 A344557

Adjacent sequences: A118435 A118436 A118437 * A118439 A118440 A118441

Keyword

sign,tabl

Author

Paul D. Hanna, Apr 28 2006

Status

approved