%I #12 Feb 21 2024 08:19:02
%S 1,1,1,1,-1,1,1,0,0,1,1,1,0,1,1,1,-1,1,1,-1,1,1,0,0,-1,0,0,1,1,1,0,-1,
%T -1,0,1,1,1,-1,1,-1,1,-1,1,-1,1,1,0,0,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,
%U 1,1,1,-1,1,0,0,0,0,0,0,1,-1,1,1,0,0,1,0,0,0,0,0,1,0,0,1,1,1,0,1,1,0,0,0,0,1,1,0,1,1,1,-1,1,1,-1,1,0,0,0,1,-1,1,1,-1,1
%N T(n,k)=L(C(n,k)/3) where L(j/p) is the Legendre symbol of j and p.
%C Row sums are A059126. Diagonal sums are A117963. Could be called the Legendre-binomial matrix for p=3.
%C The matrix square equals triangle A117939; the matrix log equals triangle A120854 divided by 2. - _Paul D. Hanna_, Jul 08 2006
%F T(n,k) = balanced ternary digit of C(n,k) mod 3. - _Paul D. Hanna_, Jul 08 2006
%e Triangle begins:
%e 1;
%e 1, 1;
%e 1, -1, 1;
%e 1, 0, 0, 1;
%e 1, 1, 0, 1, 1;
%e ...
%o (PARI) T(n,k)=(binomial(n,k)+1)%3-1 - _Paul D. Hanna_, Jul 08 2006
%Y Cf. A117939 (matrix square), A120854 (2*log).
%K easy,sign,tabl
%O 0,1
%A _Paul Barry_, Apr 05 2006