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A159841 Triangle T(n,k) = binomial(3*n+1, 2*n+k+1), read by rows.

%I A159841 #21 Sep 08 2022 08:45:44
%S A159841 1,4,1,21,7,1,120,45,10,1,715,286,78,13,1,4368,1820,560,120,16,1,
%T A159841 27132,11628,3876,969,171,19,1,170544,74613,26334,7315,1540,231,22,1,
%U A159841 1081575,480700,177100,53130,12650,2300,300,25,1,6906900,3108105,1184040,376740
%N A159841 Triangle T(n,k) = binomial(3*n+1, 2*n+k+1), read by rows.
%C A159841 T(n,0) = A045721(n), T(2n,n) = A079590(n).
%H A159841 G. C. Greubel, <a href="/A159841/b159841.txt">Rows n=0..100 of triangle, flattened</a>
%H A159841 E. H. M. Brietzke, <a href="http://dx.doi.org/10.1016/j.disc.2007.08.050">An identity of Andrews and a new method for the Riordan array proof of combinatorial identities</a>, Discrete Math., 308 (2008), 4246-4262.
%F A159841 T(n,0) = 4*T(n-1,0) + 5*T(n-1,1) + T(n-1,2), T(n+1,k+1) = T(n,k) + 3*T(n,k+1) + 3*T(n,k+2) + T(n,k+3) for k >= 0.
%e A159841 Triangle begins:
%e A159841      1;
%e A159841      4,    1;
%e A159841     21,    7,    1;
%e A159841    120,   45,   10,    1;
%e A159841    715,  286,   78,   13,    1;
%e A159841   4368, 1820,  560,  120,   16,    1;
%e A159841   ...
%t A159841 f[n_,k_]:=Binomial[3n+1,2n+k+1]; Table[ f[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* _Robert G. Wilson v_, May 31 2009 *)
%o A159841 (PARI) for(n=0,10, for(k=0,n, print1(binomial(3*n+1, 2*n+k+1), ", "))) \\ _G. C. Greubel_, May 19 2018
%o A159841 (Magma) /* As triangle */ [[Binomial(3*n+1, 2*n+k+1): k in [0..n]]: n in [0..10]]; // _G. C. Greubel_, May 19 2018
%Y A159841 Cf. A045721, A079590.
%K A159841 nonn,tabl
%O A159841 0,2
%A A159841 _Philippe Deléham_, Apr 23 2009
%E A159841 More terms from _Robert G. Wilson v_, May 31 2009