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

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