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Triangle T(n,k) = binomial(3*n+1, 2*n+k+1), read by rows.
4

%I #21 Sep 08 2022 08:45:44

%S 1,4,1,21,7,1,120,45,10,1,715,286,78,13,1,4368,1820,560,120,16,1,

%T 27132,11628,3876,969,171,19,1,170544,74613,26334,7315,1540,231,22,1,

%U 1081575,480700,177100,53130,12650,2300,300,25,1,6906900,3108105,1184040,376740

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

%C T(n,0) = A045721(n), T(2n,n) = A079590(n).

%H G. C. Greubel, <a href="/A159841/b159841.txt">Rows n=0..100 of triangle, flattened</a>

%H 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 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 Triangle begins:

%e 1;

%e 4, 1;

%e 21, 7, 1;

%e 120, 45, 10, 1;

%e 715, 286, 78, 13, 1;

%e 4368, 1820, 560, 120, 16, 1;

%e ...

%t 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 (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 (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 Cf. A045721, A079590.

%K nonn,tabl

%O 0,2

%A _Philippe Deléham_, Apr 23 2009

%E More terms from _Robert G. Wilson v_, May 31 2009