A142470 - OEIS

%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