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A118435
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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.
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9
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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; refs; listen; history; internal format)
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OFFSET
| 0,4
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COMMENTS
| 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).
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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.
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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; ...
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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^-1*M)[n+1, k+1]}
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CROSSREFS
| Cf. A118436 (column 0), A118437 (row sums), A118438 (matrix inverse), A118441 (matrix log), A118433 (self-inverse H).
Sequence in context: A123513 A117442 A184182 * A115085 A110616 A059418
Adjacent sequences: A118432 A118433 A118434 * A118436 A118437 A118438
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KEYWORD
| sign,tabl
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AUTHOR
| Paul D. Hanna (pauldhanna(AT)juno.com), Apr 28 2006
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