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%I #11 Feb 11 2021 22:57:22
%S 1,1,1,1,6,1,1,12,12,1,1,20,40,20,1,1,30,100,100,30,1,1,42,210,350,
%T 210,42,1,1,56,392,980,980,392,56,1,1,72,672,2352,3528,2352,672,72,1,
%U 1,90,1080,5040,10584,10584,5040,1080,90,1,1,110,1650,9900,27720,38808,27720,9900,1650,110,1
%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 = 1, 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 Michael De Vlieger, <a href="/A174124/b174124.txt">Table of n, a(n) for n = 0..11475</a>
%H Samuele Giraudo, <a href="https://arxiv.org/abs/1903.00677">Tree series and pattern avoidance in syntax trees</a>, arXiv:1903.00677 [math.CO], 2019.
%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 = 1.
%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 = 1.
%F Sum_{k=0..n} T(n, k, 1) = 2*A000108(n+1) - 2 + [n=0]. (End)
%e Triangle begins as:
%e 1;
%e 1, 1;
%e 1, 6, 1;
%e 1, 12, 12, 1;
%e 1, 20, 40, 20, 1;
%e 1, 30, 100, 100, 30, 1;
%e 1, 42, 210, 350, 210, 42, 1;
%e 1, 56, 392, 980, 980, 392, 56, 1;
%e 1, 72, 672, 2352, 3528, 2352, 672, 72, 1;
%e 1, 90, 1080, 5040, 10584, 10584, 5040, 1080, 90, 1;
%e 1, 110, 1650, 9900, 27720, 38808, 27720, 9900, 1650, 110, 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,1], {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,1], {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,1) 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,1): k in [0..n], n in [0..12]]; // _G. C. Greubel_, Feb 11 2021
%Y Cf. this sequence (q=1), A174125 (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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