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%I #7 Feb 11 2021 22:57:29
%S 1,1,1,1,8,1,1,15,15,1,1,24,45,24,1,1,35,105,105,35,1,1,48,210,336,
%T 210,48,1,1,63,378,882,882,378,63,1,1,80,630,2016,2940,2016,630,80,1,
%U 1,99,990,4158,8316,8316,4158,990,99,1,1,120,1485,7920,20790,28512,20790,7920,1485,120,1
%N Triangle T(n, k, q) = (q+1)*binomial(n, k)*(Pochhammer(q+1, n)/(Pochhammer(q+1, k)*Pochhammer(q+1, n-k))), with T(n, 0) = T(n, n) = 1, and q = 2, read by rows.
%C Triangles of this class, depending upon q, are of the form T(n, k, q) = (q+1)*binomial(n, k)*(Pochhammer(q+1, n)/(Pochhammer(q+1, k)*Pochhammer(q+1, n-k))), with T(n, 0) = T(n, n) = 1, and have the row sums Sum_{k=0..n} T(n, k, q) = q*(q+1)*C_{n+q}/binomial(n+2*q, q-1) - 2*q + q*[n=0], where C_{n} are the Catalan numbers (A000108) and [] is the Iverson bracket. - _G. C. Greubel_, Feb 11 2021
%H G. C. Greubel, <a href="/A174125/b174125.txt">Rows n = 0..100 of the triangle, flattened</a>
%F Let c(n, q) = Product_{j=2..n} j*(j+q) for n > 2, otherwise 1, then the number triangle is given by T(n, k, q) = c(n, q)/(c(k, q)*c(n-k, q)) for q = 2.
%F From _G. C. Greubel_, Feb 11 2021: (Start)
%F T(n, k, q) = (q+1)*binomial(n, k)*(Pochhammer(q+1, n)/(Pochhammer(q+1, k)*Pochhammer(q+1, n-k))), with T(n, 0) = T(n, n) = 1, and q = 2.
%F Sum_{k=0..n} T(n, k, 1) = 6*A000108(n+2)/(n+4) - 4 + 2*[n=0]. (End)
%e Triangle begins as:
%e 1;
%e 1, 1;
%e 1, 8, 1;
%e 1, 15, 15, 1;
%e 1, 24, 45, 24, 1;
%e 1, 35, 105, 105, 35, 1;
%e 1, 48, 210, 336, 210, 48, 1;
%e 1, 63, 378, 882, 882, 378, 63, 1;
%e 1, 80, 630, 2016, 2940, 2016, 630, 80, 1;
%e 1, 99, 990, 4158, 8316, 8316, 4158, 990, 99, 1;
%e 1, 120, 1485, 7920, 20790, 28512, 20790, 7920, 1485, 120, 1;
%t (* First program *)
%t c[n_, q_]:= If[n<2, 1, Product[i*(i+q), {i,2,n}]];
%t T[n_, m_, q_]:= c[n, q]/(c[k, q]*c[n-k, q]);
%t Table[T[n,k,2], {n,0,12}, {k,0,n}]//Flatten
%t (* Second program *)
%t T[n_, k_, q_]:= If[k==0 || k==n, 1, (q+1)*Binomial[n, k]*(Pochhammer[q+1, n]/(Pochhammer[q+1, k]*Pochhammer[q+1, n-k]))];
%t Table[T[n,k,2], {n,0,12}, {k,0,n}]//Flatten (* _G. C. Greubel_, Feb 11 2021 *)
%o (Sage)
%o def T(n,k,q): return 1 if (k==0 or k==n) else (q+1)*binomial(n, k)*(rising_factorial(q+1, n)/(rising_factorial(q+1, k)*rising_factorial(q+1, n-k)))
%o flatten([[T(n,k,2) for k in (0..n)] for n in (0..12)]) # _G. C. Greubel_, Feb 11 2021
%o (Magma)
%o c:= func< n,q | n lt 2 select 1 else (&*[j*(j+q): j in [2..n]]) >;
%o T:= func< n,k,q | c(n, q)/(c(k, q)*c(n-k, q)) >;
%o [T(n,k,2): k in [0..n], n in [0..12]]; // _G. C. Greubel_, Feb 11 2021
%Y Cf. A174124 (q=1), this sequence (q=2).
%Y Cf. A000108, A174116, A174117, A174119.
%K nonn,tabl
%O 0,5
%A _Roger L. Bagula_, Mar 09 2010
%E Edited by _G. C. Greubel_, Feb 11 2021
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