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%I #13 Apr 04 2021 00:50:15
%S 1,1,1,1,8,1,1,30,30,1,1,80,300,80,1,1,175,1750,1750,175,1,1,336,7350,
%T 19600,7350,336,1,1,588,24696,144060,144060,24696,588,1,1,960,70560,
%U 790272,1728720,790272,70560,960,1,1,1485,178200,3492720,14669424,14669424,3492720,178200,1485,1
%N Triangle T(n, k) = ( (k+2)/(2*binomial(k+2, 2)^2) )*binomial(n, k)^2*binomial(n+1, k)*binomial(n+2, k), read by rows.
%C Row sums are 1, 2, 10, 62, 462, 3852, 34974, 338690, 3452306, 36683660, 403472368, ...
%C From _Peter Bala_, May 08 2012: (Start)
%C Define the action of the operator L on a sequence { a(i) }_{0<=i<=n} by L{ a(i) }_{0<=i<=n} = { a(i)^2 - a(i-1)*a(i+1) }_{0<=i<=n} with the conventions a(-1) = a(n+1) = 0. Extend the action of L to a lower triangular array T by letting L act on the rows of T. Then L acting on Pascal's triangle A007318 produces the triangle of Narayana numbers A001263 and L applied to A001263 produces the present triangle.
%C Since the Narayana polynomials are real-rooted it follows by a theorem of Branden that the row polynomials of this array are also real-rooted.
%C (End)
%H G. C. Greubel, <a href="/A142470/b142470.txt">Rows n = 0..50 of the triangle, flattened</a>
%H Petter Brändén, <a href="https://arxiv.org/abs/0909.1927">Iterated sequences and the geometry of zeros</a>, arXiv:0909.1927 [math.CO], 2009-2010; J. Reine Angew. Math. 658 (2011), 115-131.
%F Let f(n, k) = binomial(n, k)*Product_{j=1.2} ( j!*(n+j)!/((k+j)!*(n-k+j)!) ), then T(n, k) = 2^(k-n)*f(n, k)*Sum_{j=k..n} binomial(n, j)*binomial(j, k) = binomial(n, k)*f(n, k).
%F From Peter Bala, May 08 2012: (Start)
%F T(n, k) = C(n, k)^2 * Product {i=1..2} i!*(n+i)!/((k+i)!*(n-k+i)!) = C(n, k)*C(n+2, k)*C(n+2, k+1)*C(n+2, k+2)/(C(n+2, 1)*C(n+2, 2)).
%F T(n, k) = 2/((n+1)*(n+2)*(n+3))*C(n, k)*C(n+1, k)*C(n+2, k+2)*C(n+3, k+1) = C(n, k)*A056939(n, k).
%F (End)
%F T(n, k) = ( (k+2)/(2*binomial(k+2, 2)^2) )*binomial(n, k)^2*binomial(n+1, k)*binomial(n+2, k). - _G. C. Greubel_, Apr 03 2021
%e The triangle begins as:
%e 1;
%e 1, 1;
%e 1, 8, 1;
%e 1, 30, 30, 1;
%e 1, 80, 300, 80, 1;
%e 1, 175, 1750, 1750, 175, 1;
%e 1, 336, 7350, 19600, 7350, 336, 1;
%e 1, 588, 24696, 144060, 144060, 24696, 588, 1;
%e 1, 960, 70560, 790272, 1728720, 790272, 70560, 960, 1;
%e 1, 1485, 178200, 3492720, 14669424, 14669424, 3492720, 178200, 1485, 1;
%t f[n_, k_]:= f[n, k]= Binomial[n, k]*Product[j!*(n+j)!/((k+j)!*(n-k+j)!), {j,1,2}];
%t T[n_, k_]:= Binomial[n, k]*f[n, k];
%t Table[T[n, k], {n, 0, 10}, {k, 0, n}]//Flatten (* modified by _G. C. Greubel_, Apr 03 2021 *)
%o (Magma)
%o A142470:= func< n,k | ( (k+2)/(2*Binomial(k+2, 2)^2) )*Binomial(n, k)^2*Binomial(n+1, k)*Binomial(n+2, k) >;
%o [A142470(n,k): k in [0..n], n in [0..12]]; // _G. C. Greubel_, Apr 03 2021
%o (Sage)
%o def A142470(n, k): return (2/((k+1)^2*(k+2)))*Binomial(n, k)^2*Binomial(n+1, k)*Binomial(n+2, k)
%o flatten([[A142470(n, k) for k in (0..n)] for n in (0..12)]) # _G. C. Greubel_, Apr 03 2021
%Y Cf. A001263, A008459, A056939.
%K nonn,tabl
%O 0,5
%A _Roger L. Bagula_, Sep 20 2008
%E Edited by _G. C. Greubel_, Apr 03 2021
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