%I #20 May 09 2021 09:51:27
%S 1,1,1,1,1,1,1,1,1,1,1,3,3,3,1,1,10,30,30,10,1,1,31,310,930,310,31,1,
%T 1,94,2914,29140,29140,2914,94,1,1,285,26790,830490,2768300,830490,
%U 26790,285,1,1,865,246525,23173350,239457950,239457950,23173350,246525,865,1
%N Triangle read by rows: T(n,k) = round(c(n)/(c(k)*c(n-k))) where c are partial products of a sequence defined in comments.
%C Let f be the sequence 0, 1, 1, 1, 3, 10, 31, 94, 285, 865, 2626, 7972, 24201.., f(n) = 3*f(n-1)+f(n-4), and c the partial products of f: c(n) = 1, 1, 1, 1, 3, 30, 930, 87420, 24914700, 21551215500, ... . Then T(n,k) = round(c(n)/(c(k)*c(n-k))).
%H G. C. Greubel, <a href="/A172364/b172364.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) = q*f(n-1, 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, 3, 3, 3, 1;
%e 1, 10, 30, 30, 10, 1;
%e 1, 31, 310, 930, 310, 31, 1;
%e 1, 94, 2914, 29140, 29140, 2914, 94, 1;
%e 1, 285, 26790, 830490, 2768300, 830490, 26790, 285, 1;
%e 1, 865, 246525, 23173350, 239457950, 239457950, 23173350, 246525, 865, 1;
%t f[n_, q_]:= f[n, q]= If[n==0,0,If[n<4, 1, q*f[n-1, 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 q*f(n-1, 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): return round(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), this sequence (q=3).
%K nonn,tabl,less
%O 0,12
%A _Roger L. Bagula_, Feb 01 2010
%E Definition corrected to give integral terms, _G. C. Greubel_, May 08 2021