%I #11 Sep 30 2018 10:45:06
%S 1,0,1,0,-1,2,0,1,-2,5,0,0,0,-5,14,0,-1,2,0,-14,42,0,0,0,0,0,-42,132,
%T 0,1,-2,5,0,0,-132,429,0,0,0,0,0,0,0,-429,1430,0,0,0,-5,14,0,0,0,
%U -1430,4862,0,0,0,0,0,0,0,0,0,-4862,16796
%N Inverse of number triangle A(n,k) = 1/C(n) if k <= n <= 2k, 0 otherwise, where C(n) = A000108(n).
%C It is conjectured that all elements of the triangle are integers.
%e Triangle begins
%e 1;
%e 0, 1;
%e 0, -1, 2;
%e 0, 1, -2, 5;
%e 0, 0, 0, -5, 14;
%e 0, -1, 2, 0, -14, 42;
%e 0, 0, 0, 0, 0, -42, 132;
%e 0, 1, -2, 5, 0, 0, -132, 429;
%e 0, 0, 0, 0, 0, 0, 0, -429, 1430;
%e 0, 0, 0, -5, 14, 0, 0, 0, -1430, 4862;
%e 0, 0, 0, 0, 0, 0, 0, 0, 0, -4862, 16796;
%e Inverse of triangle begins
%e 1;
%e 0, 1;
%e 0, 1/2, 1/2;
%e 0, 0, 1/5, 1/5;
%e 0, 0, 1/14, 1/14, 1/14;
%e 0, 0, 0, 1/42, 1/42, 1/42;
%e 0, 0, 0, 1/132, 1/132, 1/132, 1/132;
%e 0, 0, 0, 0, 1/429, 1/429, 1/429, 1/429;
%e 0, 0, 0, 0, 1/1430, 1/1430, 1/1430, 1/1430, 1/1430;
%e 0, 0, 0, 0, 0, 1/4862, 1/4862, 1/4862, 1/4862, 1/4862;
%e 0, 0, 0, 0, 0, 1/16796, 1/16796, 1/16796, 1/16796, 1/16796, 1/16796;
%o (PARI) row(n) = {n++; my(m = matrix(n, n, i, j, i--; j--; if ((i >= j) && (i <= 2*j), (i+1)/binomial(2*i,i), 0))); m = m^(-1); m[n,];} \\ _Michel Marcus_, Sep 30 2018
%Y Cf. A000108.
%Y Row sums give A127768.
%K sign,tabl
%O 0,6
%A _Paul Barry_, Jan 28 2007