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Number triangle T(n,k)=sum{j=0..n, C(floor((n+k+j)/3),k)C(k,floor((n+k+j)/3))}.
1

%I #2 Mar 30 2012 18:59:13

%S 1,2,1,1,3,1,0,2,3,1,0,1,3,3,1,0,0,2,3,3,1,0,0,1,3,3,3,1,0,0,0,2,3,3,

%T 3,1,0,0,0,1,3,3,3,3,1,0,0,0,0,2,3,3,3,3,1,0,0,0,0,1,3,3,3,3,3,1,0,0,

%U 0,0,0,2,3,3,3,3,3,1,0,0,0,0,0,1,3,3,3,3,3,3,1,0,0,0,0,0,0,2,3,3,3,3,3,3,1

%N Number triangle T(n,k)=sum{j=0..n, C(floor((n+k+j)/3),k)C(k,floor((n+k+j)/3))}.

%C Row sums are A114119. Diagonal sums are A008619(n+1).

%e Triangle begins

%e 1...........................= 1 = 1 mod 3

%e 2,1.........................= 3 = 0 mod 3

%e 1,3,1.......................= 5 = 2 mod 3

%e 0,2,3,1.....................= 6 = 0 mod 3

%e 0,1,3,3,1...................= 8 = 2 mod 3

%e 0,0,2,3,3,1.................= 9 = 0 mod 3

%e 0,0,1,3,3,3,1...............= 11 = 2 mod 3

%e 0,0,0,2,3,3,3,1.............= 12 = 0 mod 3

%e 0,0,0,1,3,3,3,3,1...........= 14 = 2 mod 3

%e 0,0,0,0,2,3,3,3,3,1.........= 15 = 0 mod 3

%e 0,0,0,0,1,3,3,3,3,3,1.......= 17 = 2 mod 3

%e 0,0,0,0,0,2,3,3,3,3,3,1.....= 18 = 0 mod 3

%K easy,nonn,tabl

%O 0,2

%A _Paul Barry_, Nov 13 2005