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Inverse of number triangle A(n,k) = 1/C(n) if k <= n <= 2k, 0 otherwise, where C(n) = A000108(n).
1

%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