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A127793
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Inverse of number triangle A(n,k) = 1/floor((n+2)/2) if k <= n <= 2k, 0 otherwise.
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2
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1, 0, 1, 0, -1, 2, 0, 1, -2, 2, 0, 0, 0, -2, 3, 0, -1, 2, 0, -3, 3, 0, 0, 0, 0, 0, -3, 4, 0, 1, -2, 2, 0, 0, -4, 4, 0, 0, 0, 0, 0, 0, 0, -4, 5, 0, 0, 0, -2, 3, 0, 0, 0, -5, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, -5, 6
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OFFSET
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0,6
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COMMENTS
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It is conjectured that the triangle is an integer triangle. The triangle and its inverse both appear to have row sums equal to the all 1's sequence.
The triangle is equivalent to the lower semi-matrix R = e_{1,1} + Sum_{i>=2} Sum_{p>=0} ( e_{2^p i, i} ceiling(i/2) - e_{2^p (i+1), i} ceiling(i/2) ) , where e_{i,j} is the matrix unit. The conjecture above is true, deduced from the formula of the matrix. - FUNG Cheok Yin, Sep 12 2022
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LINKS
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EXAMPLE
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Triangle begins
1;
0, 1;
0, -1, 2;
0, 1, -2, 2;
0, 0, 0, -2, 3;
0, -1, 2, 0, -3, 3;
0, 0, 0, 0, 0, -3, 4;
0, 1, -2, 2, 0, 0, -4, 4;
0, 0, 0, 0, 0, 0, 0, -4, 5;
0, 0, 0, -2, 3, 0, 0, 0, -5, 5;
0, 0, 0, 0, 0, 0, 0, 0, 0, -5, 6;
0, -1, 2, 0, -3, 3, 0, 0, 0, 0, -6, 6;
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -6, 7;
Inverse of the triangle begins
1;
0, 1;
0, 1/2, 1/2;
0, 0, 1/2, 1/2;
0, 0, 1/3, 1/3, 1/3;
0, 0, 0, 1/3, 1/3, 1/3;
0, 0, 0, 1/4, 1/4, 1/4, 1/4;
0, 0, 0, 0, 1/4, 1/4, 1/4, 1/4;
0, 0, 0, 0, 1/5, 1/5, 1/5, 1/5, 1/5;
0, 0, 0, 0, 0, 1/5, 1/5, 1/5, 1/5, 1/5;
0, 0, 0, 0, 0, 1/6, 1/6, 1/6, 1/6, 1/6, 1/6;
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MATHEMATICA
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rows = 11;
A[n_, k_] := If[k <= n, If[n <= 2 k, 1/Floor[(n+2)/2] , 0], 0];
T = Table[A[n, k], {n, 0, rows-1}, {k, 0, rows-1}] // Inverse;
Table[T[[n, k]], {n, 1, rows}, {k, 1, n}] // Flatten (* Stefano Spezia, Sep 30 2018 *)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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