

A118435


Triangle T, read by rows, equal to the matrix product T = H*[C^1]*H, where H is the selfinverse triangle A118433 and C is Pascal's triangle.


9



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
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OFFSET

0,4


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).


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)^(nk))*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; ...


PROG

(PARI) {T(n, k)=local(M=matrix(n+1, n+1, r, c, if(r>=c, binomial(r1, c1)*(1)^(r\2 (c1)\2+rc))), C=matrix(n+1, n+1, r, c, if(r>=c, binomial(r1, c1)))); (M*C^1*M)[n+1, k+1]}


CROSSREFS

Cf. A118436 (column 0), A118437 (row sums), A118438 (matrix inverse), A118441 (matrix log), A118433 (selfinverse H).
Sequence in context: A123513 A117442 A184182 * A115085 A110616 A059418
Adjacent sequences: A118432 A118433 A118434 * A118436 A118437 A118438


KEYWORD

sign,tabl


AUTHOR

Paul D. Hanna, Apr 28 2006


STATUS

approved



