A119301 Number triangle binomial(3n-k,n-k).

1, 3, 1, 15, 5, 1, 84, 28, 7, 1, 495, 165, 45, 9, 1, 3003, 1001, 286, 66, 11, 1, 18564, 6188, 1820, 455, 91, 13, 1, 116280, 38760, 11628, 3060, 680, 120, 15, 1, 735471, 245157, 74613, 20349, 4845, 969, 153, 17, 1, 4686825, 1562275, 480700, 134596, 33649, 7315

Offset

0,2

Comments

g(x)=2*sin(arcsin(3*sqrt(3*x)/2)/3)/sqrt(3*x);g(x)=1/(1-xg(x)^2). First column is A005809. Second column is A025174. Row sums are A045721. Inverse is Riordan array (1-3x,x(1-x)^2), A119302.

Links

Table of n, a(n) for n=0..50.

Formula

Riordan array (1/(1-3xg(x)^2),xg(x)^2) where g(x)=1+xg(x)^3.

'Horizontal' recurrence equation: T(n,0) = binomial(3*n,n) and for k >= 1, T(n,k) = sum {i = 1..n+1-k} i*T(n-1,k-2+i). - Peter Bala, Dec 28 2014

Example

Triangle begins
1,
3, 1,
15, 5, 1,
84, 28, 7, 1,
495, 165, 45, 9, 1,
3003, 1001, 286, 66, 11, 1,
18564, 6188, 1820, 455, 91, 13, 1,
116280, 38760, 11628, 3060, 680, 120, 15, 1
...
Horizontal recurrence: T(4,1) = 1*84 + 2*28 + 3*7 + 4*1 = 165. - Peter Bala, Dec 29 2014

Maple

T := proc(n, k) option remember;
`if`(n = 0, 1, add(i*T(n-1, k-2+i), i=1..n+1-k)) end:
for n from 0 to 9 do print(seq(T(n, k), k=0..n)) od; # Peter Luschny, Dec 30 2014

Mathematica

Flatten[Table[Binomial[3n-k, n-k], {n, 0, 10}, {k, 0, n}]] (* Harvey P. Dale, Jul 28 2012 *)

Crossrefs

Sequence in context: A131440 A269950 A190088 * A293157 A121335 A126454

Adjacent sequences: A119298 A119299 A119300 * A119302 A119303 A119304

Keyword

easy,nonn,tabl

Author

Paul Barry, May 13 2006

Status

approved