A127893 Riordan array (1/(1-x)^3, x/(1-x)^3).

1, 3, 1, 6, 6, 1, 10, 21, 9, 1, 15, 56, 45, 12, 1, 21, 126, 165, 78, 15, 1, 28, 252, 495, 364, 120, 18, 1, 36, 462, 1287, 1365, 680, 171, 21, 1, 45, 792, 3003, 4368, 3060, 1140, 231, 24, 1, 55, 1287, 6435, 12376, 11628, 5985, 1771, 300, 27, 1

Offset: 0

Author

Paul Barry, Feb 04 2007

Keywords

• easy
• nonn
• tabl

Comments

Inverse is A127894.

From Peter Bala, Jul 22 2014: (Start)

Let M denote the unsigned version of the lower unit triangular array A122432 and for k = 0,1,2,... define M(k) to be the lower unit triangular block array

/I_k 0\

\ 0 M/

having the k x k identity matrix I_k as the upper left block; in particular, M(0) = M. Then the present triangle equals the infinite matrix product M(0)*M(1)*M(2)*... (which is clearly well-defined). See the Example section. (End)

Examples

			Triangle begins
   1;
   3,    1;
   6,    6,     1;
  10,   21,     9,     1;
  15,   56,    45,    12,     1;
  21,  126,   165,    78,    15,     1;
  28,  252,   495,   364,   120,    18,     1;
  36,  462,  1287,  1365,   680,   171,    21,    1;
  45,  792,  3003,  4368,  3060,  1140,   231,   24,   1;
  55, 1287,  6435, 12376, 11628,  5985,  1771,  300,  27,  1;
  66, 2002, 12870, 31824, 38760, 26334, 10626, 2600, 378, 30, 1;
  ...
From _Peter Bala_, Jul 22 2014: (Start)
With the arrays M(k) as defined in the Comments section, the infinite product M(0*)M(1)*M(2)*... begins
  / 1         \/1         \/1       \       / 1       \
  | 3  1      ||0  1      ||0 1      |      | 3  1    |
  | 6  3 1    ||0  3 1    ||0 0 1    |... = | 6  6 1  |
  |10  6 3 1  ||0  6 3 1  ||0 0 3 1  |      |10 21 9 1|
  |15 10 6 3 1||0 10 6 3 1||0 0 6 3 1|      |...      |
  |...        ||...       ||...      |      |...      |
(End)
		

Links

• G. C. Greubel, Rows n = 0..99, flattened
• Milan Janjić, Pascal Matrices and Restricted Words, J. Int. Seq., Vol. 21 (2018), Article 18.5.2.

Crossrefs

Cf. A052529, A095263, A122432, A127894.

Programs

GAP

Flat(List([0..10],n->List([0..n],k->Binomial(n+2*k+2,n-k)))); # Muniru A Asiru, Apr 30 2018

Magma

[Binomial(n+2*k+2, n-k): k in [0..n], n in [0..10]]; // G. C. Greubel, Apr 29 2018

Maple

seq(seq(binomial(n+2*k+2,n-k),k=0..n),n=0..10); # Robert Israel, Apr 28 2015

Mathematica

Flatten@ Table[Binomial[n+2k-1, n-k], {n, 10}, {k, n}] (* Michael De Vlieger, Apr 27 2015 *)

PARI

for(n=0,10, for(k=0,n, print1(binomial(n+2*k+2, n-k), ", "))) \\ G. C. Greubel, Apr 29 2018

Sage

flatten([[binomial(n+2*k+2,n-k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Apr 16 2021

Formula

T(n,k) = binomial(n+2*k+2, n-k).

Sum_{k=0..n} T(n, k) = A052529(n+1) (row sums).

Sum_{k=0..floor(n/2)} T(n-k, k) = A095263(n+1) (diagonal sums).

Recurrence: T(n+1, k+1) = Sum_{i = 0..n-k} binomial(i+2, 2)*T(n-i,k). - Peter Bala, Jul 22 2014

G.f.: 1/((1-x)^3-x*y). - Vladimir Kruchinin, Apr 27 2015