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A118435 Triangle T, read by rows, equal to the matrix product T = H*[C^-1]*H, where H is the self-inverse triangle A118433 and C is Pascal's triangle. +0
10
1, 1, 1, -3, 2, 1, -11, 15, 3, 1, 25, -44, -18, 4, 1, 41, -115, -110, 50, 5, 1, -43, 246, 375, -220, -45, 6, 1, 29, 315, 861, -805, -385, 105, 7, 1, -335, 232, -1204, 2296, 1750, -616, -84, 8, 1, -1199, 3033, 1044, 3780, 5166, -2898, -924, 180, 9, 1 (list; table; graph; listen)
OFFSET

0,4

COMMENT

The matrix inverse of H*[C^-1]*H is H*C*H = A118438, 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).

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 = A118438.

EXAMPLE

Triangle begins:

1;

1, 1;

-3, 2, 1;

-11, 15, 3, 1;

25,-44,-18, 4, 1;

41,-115,-110, 50, 5, 1;

-43, 246, 375,-220,-45, 6, 1;

29, 315, 861,-805,-385, 105, 7, 1;

-335, 232,-1204, 2296, 1750,-616,-84, 8, 1;

-1199, 3033, 1044, 3780, 5166,-2898,-924, 180, 9, 1; ...

The matrix log, LOG(T) = A118441, starts:

0;

1, 0;

-4, 2, 0;

-12, 12, 3, 0;

32,-48,-24, 4, 0;

80,-160,-120, 40, 5, 0; ...

where matrix square, LOG(T)^2, is a single diagonal:

0;

0,0;

2,0,0;

0,6,0,0;

0,0,12,0,0;

0,0,0,20,0,0; ...

PROGRAM

(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^-1*M)[n+1, k+1]}

CROSSREFS

Cf. A118436 (column 0), A118437 (row sums), A118438 (matrix inverse), A118441 (matrix log), A118433 (self-inverse H).

Sequence in context: A104219 A123513 A117442 this_sequence A115085 A110616 A059418

Adjacent sequences: A118432 A118433 A118434 this_sequence A118436 A118437 A118438

KEYWORD

sign,tabl

AUTHOR

Paul D. Hanna (pauldhanna(AT)juno.com), Apr 28 2006

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Last modified November 18 20:14 EST 2008. Contains 147244 sequences.


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