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Search: id:A112911
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| A112911 |
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Triangle T, read by rows, such that the matrix inverse satisfies: [T^-1](n,k) = -(k+1)*T(n-1,0) for n>k>=0, with T(n,n)=1 for n>=0. |
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+0
6
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| 1, 1, 1, 3, 2, 1, 14, 8, 3, 1, 85, 44, 15, 4, 1, 621, 298, 96, 24, 5, 1, 5236, 2358, 729, 176, 35, 6, 1, 49680, 21154, 6327, 1492, 290, 48, 7, 1, 521721, 211100, 61380, 14220, 2725, 444, 63, 8, 1, 5994155, 2313030, 655944, 149812, 28425, 4590, 644, 80, 9, 1
(list; table; graph; listen)
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OFFSET
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0,4
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COMMENT
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Matrix inverse square satisfies: [T^-2](3*n+2,n) = 0 for n>=0.
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EXAMPLE
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Triangle T begins:
1;
1,1;
3,2,1;
14,8,3,1;
85,44,15,4,1;
621,298,96,24,5,1;
5236,2358,729,176,35,6,1;
49680,21154,6327,1492,290,48,7,1; ...
Matrix inverse T^-1 begins:
1;
-1,1;
-1,-2*1,1;
-3,-2*1,-3*1,1;
-14,-2*3,-3*1,-4*1,1;
-85,-2*14,-3*3,-4*1,-5*1,1;
-621,-2*85,-3*14,-4*3,-5*1,-6*1,1; ...
where [T^-1](n,k) = -(k+1)*T(n-1,0) for n>k>=0.
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PROGRAM
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(PARI) {T(n, k)=local(A=Mat(1), B); for(m=2, n+1, B=matrix(m, m); for(i=1, m, for(j=1, i, if(j==i, B[i, j]=1, B[i, j]=-j*(A^-1)[i-j, 1] ); )); A=B); return((A^-1)[n+1, k+1])}
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CROSSREFS
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Cf. A088716 (column 0), A112912 (column 1), A112913 (column 2), A112914 (column 3).
Sequence in context: A154921 A127126 A161133 this_sequence A152405 A152400 A111548
Adjacent sequences: A112908 A112909 A112910 this_sequence A112912 A112913 A112914
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KEYWORD
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nonn,tabl
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AUTHOR
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Paul D. Hanna (pauldhanna(AT)juno.com), Oct 06 2005
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