%I #20 May 09 2021 09:52:41
%S 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,4,4,4,4,1,1,7,28,28,28,7,1,1,10,70,
%T 280,280,70,10,1,1,13,130,910,3640,910,130,13,1,1,25,325,3250,22750,
%U 22750,3250,325,25,1,1,46,1150,14950,149500,261625,149500,14950,1150,46,1
%N Triangle read by rows: T(n,k,q) = round(c(n)/(c(k)*c(n-k))) where c are partial products of a sequence defined in comments.
%C Start from A143454 and its partial products c(n) = 1, 1, 1, 1, 1, 4, 28, 280, 3640, 91000, 4186000, ... . Then T(n,k) = round(c(n)/(c(k)*c(n-k))).
%H G. C. Greubel, <a href="/A172369/b172369.txt">Rows n = 0..50 of the triangle, flattened</a>
%F T(n, k, q) = round( c(n,q)/(c(k,q)*c(n-k,q)) ), where c(n, q) = Product_{j=1..n} f(j, q), f(n, q) = f(n-1, q) + q*f(n-4, q), f(0, q) = 0, f(1, q) = f(2, q) = f(3, q) = 1, and q = 3. - _G. C. Greubel_, May 08 2021
%e Triangle begins as:
%e 1;
%e 1, 1;
%e 1, 1, 1;
%e 1, 1, 1, 1;
%e 1, 1, 1, 1, 1;
%e 1, 4, 4, 4, 4, 1;
%e 1, 7, 28, 28, 28, 7, 1;
%e 1, 10, 70, 280, 280, 70, 10, 1;
%e 1, 13, 130, 910, 3640, 910, 130, 13, 1;
%e 1, 25, 325, 3250, 22750, 22750, 3250, 325, 25, 1;
%e 1, 46, 1150, 14950, 149500, 261625, 149500, 14950, 1150, 46, 1;
%t f[n_, q_]:= f[n, q]= If[n==0, 0, If[n<4, 1, f[n-1, q] +q*f[n-4, q]]];
%t c[n_, q_]:= Product[f[j, q], {j,n}];
%t T[n_, k_, q_]:= Round[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 08 2021 *)
%o (Sage)
%o @CachedFunction
%o def f(n,q): return 0 if (n==0) else 1 if (n<4) else f(n-1, q) + q*f(n-4, q)
%o def c(n,q): return product( f(j,q) for j in (1..n) )
%o def T(n,k,q): round(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 08 2021
%Y Cf. A172363 (q=1), A172368 (q=2), this sequence (q=3), A318774.
%K nonn,tabl,less
%O 0,17
%A _Roger L. Bagula_, Feb 01 2010
%E Definition corrected to give integral terms, _G. C. Greubel_, May 08 2021