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%I #11 May 10 2021 03:45:57
%S 1,1,1,1,1,1,1,1,1,1,1,3,3,3,1,1,3,9,9,3,1,1,5,15,45,15,5,1,1,9,45,
%T 135,135,45,9,1,1,11,99,495,495,495,99,11,1,1,19,209,1881,3135,3135,
%U 1881,209,19,1,1,29,551,6061,18183,30305,18183,6061,551,29,1
%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 Start from the sequence A159284 and its partial products c(n) = 1, 1, 1, 1, 3, 9, 45, 405, 4455, 84645, 2454705, ... . Then T(n,k) = round( c(n)/(c(k)*c(n-k)) ).
%H G. C. Greubel, <a href="/A172358/b172358.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-2, q) + q*f(n-3, q), f(0,q)=0, f(1,q) = f(2,q) = 1, and q = 2. - _G. C. Greubel_, May 09 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, 3, 9, 9, 3, 1;
%e 1, 5, 15, 45, 15, 5, 1;
%e 1, 9, 45, 135, 135, 45, 9, 1;
%e 1, 11, 99, 495, 495, 495, 99, 11, 1;
%e 1, 19, 209, 1881, 3135, 3135, 1881, 209, 19, 1;
%e 1, 29, 551, 6061, 18183, 30305, 18183, 6061, 551, 29, 1;
%t f[n_, q_]:= f[n, q]= If[n<3, Fibonacci[n], f[n-2, q] + q*f[n-3, 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, 2], {n,0,12}, {k,0,n}]//Flatten (* modified by _G. C. Greubel_, May 09 2021 *)
%o (Sage)
%o @CachedFunction
%o def f(n,q): return fibonacci(n) if (n<3) else f(n-2, q) + q*f(n-3, 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,2) for k in (0..n)] for n in (0..12)]) # _G. C. Greubel_, May 09 2021
%Y Cf. A172353 (q=1), this sequence (q=2), A172359 (q=4), A172360 (q=5).
%K nonn,tabl,less
%O 0,12
%A _Roger L. Bagula_, Feb 01 2010
%E Definition corrected to give integral terms by _G. C. Greubel_, May 09 2021
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