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A127793
Inverse of number triangle A(n,k) = 1/floor((n+2)/2) if k <= n <= 2k, 0 otherwise.
2
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
OFFSET
0,6
COMMENTS
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
EXAMPLE
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;
MATHEMATICA
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 *)
CROSSREFS
Sequence in context: A036862 A285124 A094238 * A127771 A248806 A118407
KEYWORD
sign,tabl
AUTHOR
Paul Barry, Jan 29 2007
STATUS
approved