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Triangle T(n, k, q) = c(n,q)/( c(k,q)*c(n-k,q) ), where c(n, q) = Product_{j=1..n} f(n, q), f(n, q) = ( (1-q^n)*(1+(-1)^n) + n!*(1-(-1)^n) )/2, and q = 3, read by rows.
3

%I #6 May 08 2021 01:41:27

%S 1,1,1,1,-8,1,1,6,6,1,1,-80,60,-80,1,1,120,1200,1200,120,1,1,-728,

%T 10920,-145600,10920,-728,1,1,5040,458640,9172800,9172800,458640,5040,

%U 1,1,-6560,4132800,-501446400,752169600,-501446400,4132800,-6560,1,1,362880,297561600,249951744000,2274560870400,2274560870400,249951744000,297561600,362880,1

%N Triangle T(n, k, q) = c(n,q)/( c(k,q)*c(n-k,q) ), where c(n, q) = Product_{j=1..n} f(n, q), f(n, q) = ( (1-q^n)*(1+(-1)^n) + n!*(1-(-1)^n) )/2, and q = 3, read by rows.

%H G. C. Greubel, <a href="/A172428/b172428.txt">Rows n = 0..30 of the triangle, flattened</a>

%F T(n, k, q) = c(n,q)/( c(k,q)*c(n-k,q) ), where c(n, q) = Product_{j=1..n} f(n, q), f(n, q) = ( (1-q^n)*(1+(-1)^n) + n!*(1-(-1)^n) )/2, and q = 3.

%e Triangle begins as:

%e 1;

%e 1, 1;

%e 1, -8, 1;

%e 1, 6, 6, 1;

%e 1, -80, 60, -80, 1;

%e 1, 120, 1200, 1200, 120, 1;

%e 1, -728, 10920, -145600, 10920, -728, 1;

%e 1, 5040, 458640, 9172800, 9172800, 458640, 5040, 1;

%e 1, -6560, 4132800, -501446400, 752169600, -501446400, 4132800, -6560, 1;

%t f[n_, q_]:= ((1-q^n)*(1+(-1)^n) + n!*(1-(-1)^n))/2;

%t c[n_, q_]:= Product[f[j, q], {j, n}];

%t T[n_, k_, q_]:= c[n, q]/(c[k, q]*c[n-k, q]);

%t Table[T[n, k, 3], {n,0,12}, {k,0,n}]//Flatten (* modified by _G. C. Greubel_, May 07 2021 *)

%o (Sage)

%o @CachedFunction

%o def f(n,q): return ((1-q^n)*(1+(-1)^n) + factorial(n)*(1-(-1)^n))/2

%o def c(n,q): return product( f(j,q) for j in (1..n) )

%o def T(n,k,q): return c(n,q)/(c(k,q)*c(n-k,q))

%o flatten([[T(n,k,3) for k in (0..n)] for n in (0..12)]) # _G. C. Greubel_, May 07 2021

%Y Cf. A172427 (q=3), this sequence (q=4), A172429 (q=5).

%K sign,tabl

%O 0,5

%A _Roger L. Bagula_, Feb 02 2010

%E Edited by _G. C. Greubel_, May 07 2021