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A127793 Inverse of number triangle A(n,k) = 1/floor((n+2)/2) if k <= n <= 2k, 0 otherwise.

%I A127793 #16 Oct 09 2022 21:16:05
%S A127793 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,
%T A127793 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,
%U A127793 0,-5,6
%N A127793 Inverse of number triangle A(n,k) = 1/floor((n+2)/2) if k <= n <= 2k, 0 otherwise.
%C A127793 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.
%C A127793 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
%H A127793 FUNG Cheok Yin, <a href="/A127793/a127793_1.txt">the triangle with the first 61 rows</a>
%e A127793 Triangle begins
%e A127793   1;
%e A127793   0,  1;
%e A127793   0, -1,  2;
%e A127793   0,  1, -2,  2;
%e A127793   0,  0,  0, -2,  3;
%e A127793   0, -1,  2,  0, -3,  3;
%e A127793   0,  0,  0,  0,  0, -3,  4;
%e A127793   0,  1, -2,  2,  0,  0, -4,  4;
%e A127793   0,  0,  0,  0,  0,  0,  0, -4,  5;
%e A127793   0,  0,  0, -2,  3,  0,  0,  0, -5,  5;
%e A127793   0,  0,  0,  0,  0,  0,  0,  0,  0, -5,  6;
%e A127793   0, -1,  2,  0, -3,  3,  0,  0,  0,  0, -6,  6;
%e A127793   0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0, -6,  7;
%e A127793 Inverse of the triangle begins
%e A127793   1;
%e A127793   0,  1;
%e A127793   0, 1/2, 1/2;
%e A127793   0,  0,  1/2, 1/2;
%e A127793   0,  0,  1/3, 1/3, 1/3;
%e A127793   0,  0,   0,  1/3, 1/3, 1/3;
%e A127793   0,  0,   0,  1/4, 1/4, 1/4, 1/4;
%e A127793   0,  0,   0,   0,  1/4, 1/4, 1/4, 1/4;
%e A127793   0,  0,   0,   0,  1/5, 1/5, 1/5, 1/5, 1/5;
%e A127793   0,  0,   0,   0,   0,  1/5, 1/5, 1/5, 1/5, 1/5;
%e A127793   0,  0,   0,   0,   0,  1/6, 1/6, 1/6, 1/6, 1/6, 1/6;
%t A127793 rows = 11;
%t A127793 A[n_, k_] := If[k <= n, If[n <= 2 k, 1/Floor[(n+2)/2] , 0], 0];
%t A127793 T = Table[A[n, k], {n, 0, rows-1}, {k, 0, rows-1}] // Inverse;
%t A127793 Table[T[[n, k]], {n, 1, rows}, {k, 1, n}] // Flatten (* _Stefano Spezia_, Sep 30 2018 *)
%K A127793 sign,tabl
%O A127793 0,6
%A A127793 _Paul Barry_, Jan 29 2007