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Search: id:A104041
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| A104041 |
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Triangular matrix T, read by rows, such that column k is equal (in absolute value) to row (k-1) in the matrix inverse T^-1 (with extrapolated zeros) for k>0, with T(n,n)=1 (n>=0) and T(n,n-1)=-n (n>=1). |
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+0
3
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| 1, -1, 1, 0, -2, 1, 0, 2, -3, 1, 0, 0, 4, -4, 1, 0, 0, -4, 8, -5, 1, 0, 0, 0, -8, 12, -6, 1, 0, 0, 0, 8, -20, 18, -7, 1, 0, 0, 0, 0, 16, -32, 24, -8, 1, 0, 0, 0, 0, -16, 48, -56, 32, -9, 1, 0, 0, 0, 0, 0, -32, 80, -80, 40, -10, 1, 0, 0, 0, 0, 0, 32, -112, 160, -120, 50, -11, 1, 0, 0, 0, 0, 0, 0, 64, -192, 240, -160, 60, -12, 1
(list; table; graph; listen)
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OFFSET
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0,5
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COMMENT
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Row sums are: {1,0,-1,0, 1,0,-1,0, ...}. Absolute row sums form A038754. Let A(x,y) be the g.f. of T and B(x,y) be the g.f. of T^-1; then B(x,y)=1+x*y*A(-1/y,-x*y^2) and A(x,y)=(B(-x^2*y,-1/x)-1)/(x*y).
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FORMULA
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G.f.: A(x, y) = (1-x+x*y)/(1+2*x^2*y-x^2*y^2).
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EXAMPLE
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Rows of T begin:
1;
-1,1;
0,-2,1;
0,2,-3,1;
0,0,4,-4,1;
0,0,-4,8,-5,1;
0,0,0,-8,12,-6,1;
0,0,0,8,-20,18,-7,1; ...
The matrix inverse T^-1 equals triangle A104040:
1;
1,1;
2,2,1;
4,4,3,1;
8,8,8,4,1;
16,16,20,12,5,1;
32,32,48,32,18,6,1;
64,64,112,80,56,24,7,1; ...
the rows of T^-1 equal columns of T in absolute value.
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PROGRAM
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(PARI) {T(n, k)=local(X=x+x*O(x^n), Y=y+y*O(y^k)); polcoeff(polcoeff((1-X+X*Y)/(1+2*X^2*Y-X^2*Y^2), n, x), k, y)}
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CROSSREFS
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Cf. A104040, A038754.
Sequence in context: A031135 A037181 A051070 this_sequence A104402 A131084 A123949
Adjacent sequences: A104038 A104039 A104040 this_sequence A104042 A104043 A104044
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KEYWORD
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sign,tabl
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AUTHOR
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Paul D. Hanna (pauldhanna(AT)juno.com), Mar 02 2005
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