%I #38 Sep 08 2022 08:45:29
%S 1,3,1,6,6,1,10,21,9,1,15,56,45,12,1,21,126,165,78,15,1,28,252,495,
%T 364,120,18,1,36,462,1287,1365,680,171,21,1,45,792,3003,4368,3060,
%U 1140,231,24,1,55,1287,6435,12376,11628,5985,1771,300,27,1
%N Riordan array (1/(1-x)^3, x/(1-x)^3).
%C Inverse is A127894.
%C From _Peter Bala_, Jul 22 2014: (Start)
%C 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
%C /I_k 0\
%C \ 0 M/
%C 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)
%H G. C. Greubel, <a href="/A127893/b127893.txt">Rows n = 0..99, flattened</a>
%H Milan Janjić, <a href="https://www.emis.de/journals/JIS/VOL21/Janjic2/janjic103.html">Pascal Matrices and Restricted Words</a>, J. Int. Seq., Vol. 21 (2018), Article 18.5.2.
%F T(n,k) = binomial(n+2*k+2, n-k).
%F Sum_{k=0..n} T(n, k) = A052529(n+1) (row sums).
%F Sum_{k=0..floor(n/2)} T(n-k, k) = A095263(n+1) (diagonal sums).
%F Recurrence: T(n+1, k+1) = Sum_{i = 0..n-k} binomial(i+2, 2)*T(n-i,k). - _Peter Bala_, Jul 22 2014
%F G.f.: 1/((1-x)^3-x*y). - _Vladimir Kruchinin_, Apr 27 2015
%e Triangle begins
%e 1;
%e 3, 1;
%e 6, 6, 1;
%e 10, 21, 9, 1;
%e 15, 56, 45, 12, 1;
%e 21, 126, 165, 78, 15, 1;
%e 28, 252, 495, 364, 120, 18, 1;
%e 36, 462, 1287, 1365, 680, 171, 21, 1;
%e 45, 792, 3003, 4368, 3060, 1140, 231, 24, 1;
%e 55, 1287, 6435, 12376, 11628, 5985, 1771, 300, 27, 1;
%e 66, 2002, 12870, 31824, 38760, 26334, 10626, 2600, 378, 30, 1;
%e ...
%e From _Peter Bala_, Jul 22 2014: (Start)
%e With the arrays M(k) as defined in the Comments section, the infinite product M(0*)M(1)*M(2)*... begins
%e / 1 \/1 \/1 \ / 1 \
%e | 3 1 ||0 1 ||0 1 | | 3 1 |
%e | 6 3 1 ||0 3 1 ||0 0 1 |... = | 6 6 1 |
%e |10 6 3 1 ||0 6 3 1 ||0 0 3 1 | |10 21 9 1|
%e |15 10 6 3 1||0 10 6 3 1||0 0 6 3 1| |... |
%e |... ||... ||... | |... |
%e (End)
%p seq(seq(binomial(n+2*k+2,n-k),k=0..n),n=0..10); # _Robert Israel_, Apr 28 2015
%t Flatten@ Table[Binomial[n+2k-1, n-k], {n, 10}, {k, n}] (* _Michael De Vlieger_, Apr 27 2015 *)
%o (PARI) for(n=0,10, for(k=0,n, print1(binomial(n+2*k+2, n-k), ", "))) \\ _G. C. Greubel_, Apr 29 2018
%o (Magma) [Binomial(n+2*k+2, n-k): k in [0..n], n in [0..10]]; // _G. C. Greubel_, Apr 29 2018
%o (GAP) Flat(List([0..10],n->List([0..n],k->Binomial(n+2*k+2,n-k)))); # _Muniru A Asiru_, Apr 30 2018
%o (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
%Y Cf. A052529, A095263, A122432, A127894.
%K easy,nonn,tabl
%O 0,2
%A _Paul Barry_, Feb 04 2007