%I #10 Mar 01 2017 11:04:46

%S 1,-4,1,0,-7,1,0,15,-10,1,0,-13,39,-13,1,0,4,-80,72,-16,1,0,0,95,-228,

%T 114,-19,1,0,0,-66,462,-484,165,-22,1,0,0,25,-630,1375,-875,225,-25,1,

%U 0,0,-4,588,-2772,3185,-1428,294,-28,1,0,0,0,-372,4092,-8463,6324,-2170,372

%N Riordan array (1-4x, x(1-x)^3).

%C Inverse of number triangle binomial(4n-k, n-k), A119304. Row sums are A119306.

%H Indranil Ghosh, <a href="/A119305/b119305.txt">Rows 0..101, flattened</a>

%F Number triangle T(n,k) = (C(3k, n-k) + 4*C(3k, n-k-1))(-1)^(n-k).

%e Triangle begins

%e 1;

%e -4, 1;

%e 0, -7, 1;

%e 0, 15, -10, 1;

%e 0, -13, 39, -13, 1;

%e 0, 4, -80, 72, -16, 1;

%e 0, 0, 95, -228, 114, -19, 1;

%t Flatten[Table[(Binomial[3k,n-k]+4Binomial[3k,n-k-1])*(-1)^(n-k),{n,0,11},{k,0,n}]] (* _Indranil Ghosh_, Feb 26 2017 *)

%o (PARI) tabl(nn) = {for (n=0,nn,for (k=0,n,print1((binomial(3*k,n-k)+4*binomial(3*k,n-k-1))*(-1)^(n-k),", "););print(););} \\ _Indranil Ghosh_, Feb 26 2017

%K easy,sign,tabl

%O 0,2

%A _Paul Barry_, May 13 2006