%I #20 Jun 22 2024 07:59:07

%S 1,4,1,28,7,1,220,55,10,1,1820,455,91,13,1,15504,3876,816,136,16,1,

%T 134596,33649,7315,1330,190,19,1,1184040,296010,65780,12650,2024,253,

%U 22,1,10518300,2629575,593775,118755,20475,2925,325,25,1,94143280,23535820

%N Triangle read by rows: T(n,k) = binomial(4n-k,n-k), 0 <= k <= n.

%H Indranil Ghosh, <a href="/A119304/b119304.txt">Rows 0..125, flattened</a>

%F Riordan array (1/(1-4f(x)),f(x)) where f(x)(1-f(x))^3 = x.

%F From _Peter Bala_, Jun 04 2024: (Start)

%F 'Horizontal' recurrence equation: T(n, 0) = binomial(4*n,n) and for k >= 1, T(n, k) = Sum_{i = 1..n+1-k} i*(i+1)/2 * T(n-1, k-2+i).

%F T(n, k) = Sum_{j = 0..n} binomial(n+j-1, j)*binomial(3*n-k-j, 2*n). (End)

%e Triangle begins

%e 1;

%e 4, 1;

%e 28, 7, 1;

%e 220, 55, 10, 1;

%e 1820, 455, 91, 13, 1;

%e 15504, 3876, 816, 136, 16, 1;

%e 134596, 33649, 7315, 1330, 190, 19, 1;

%t Flatten[Table[Binomial[4n-k,n-k],{n,0,9},{k,0,n}]] (* _Indranil Ghosh_, Feb 26 2017 *)

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

%o (Python)

%o from sympy import binomial

%o i=0

%o for n in range(12):

%o for k in range(n+1):

%o print(str(i)+" "+str(binomial(4*n-k,n-k)))

%o i+=1 # _Indranil Ghosh_, Feb 26 2017

%Y Rows sums are A052203. First column is A005810. Inverse of A119305.

%Y Cf. A092392, A119301.

%K easy,nonn,tabl

%O 0,2

%A _Paul Barry_, May 13 2006