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%I #35 Jun 05 2020 06:04:15
%S 1,4,6,24,60,42,160,560,688,308,1120,5040,8760,7080,2310,8064,44352,
%T 99456,114576,68712,17556,59136,384384,1055040,1572480,1351840,642824,
%U 134596,439296,3294720,10695168,19536000,21778560,14912064,5864640,1038312
%N Triangle T(n,k) = d(n-k,n), 0 <= k <= n, where d(l,m) = Sum_{k=l..m} 2^k * binomial(2*m-2*k, m-k) * binomial(m+k, m) * binomial(k, l).
%C For an explanation on how this triangular array is related to the Boros-Moll polynomial P_n(x) and the theory in Comtet (1967), see my comments in A223549. For example, the bivariate o.g.f. below follows from the theory in Comtet (1967). - _Petros Hadjicostas_, May 24 2020
%H Tewodros Amdeberhan and Victor H. Moll, <a href="http://arxiv.org/abs/0707.2118"> A formula for a quartic integral: a survey of old proofs and some new ones</a>, arXiv:0707.2118 [math.CA], 2007.
%H George Boros and Victor H. Moll, <a href="http://dx.doi.org/10.1016/S0377-0427(99)00081-3">An integral hidden in Gradshteyn and Ryzhik</a>, Journal of Computational and Applied Mathematics, 106(2) (1999), 361-368.
%H William Y. C. Chen and Ernest X. W. Xia, <a href="http://arxiv.org/abs/0806.4333"> The Ratio Monotonicity of the Boros-Moll Polynomials</a>, arXiv:0806.4333 [math.CO], 2009.
%H William Y. C. Chen and Ernest X. W. Xia, <a href="https://doi.org/10.1090/S0025-5718-09-02223-6"> The Ratio Monotonicity of the Boros-Moll Polynomials</a>, Mathematics of Computation, 78(268) (2009), 2269-2282.
%H Louis Comtet, <a href="https://www.jstor.org/stable/43667287">Fonctions génératrices et calcul de certaines intégrales</a>, Publikacije Elektrotechnickog faculteta - Serija Matematika i Fizika, No. 181/196 (1967), 77-87.
%H V. H. Moll, <a href="http://www.ams.org/notices/200203/fea-moll.pdf">The evaluation of integrals: a personal story</a>, Notices Amer. Math. Soc., 49 (No. 3, March 2002), 311-317.
%H V. H. Moll, <a href="https://web.math.rochester.edu/misc/ojac/vol2/Moll_2007.pdf">Combinatorial sequences arising from a rational integral</a>, Onl. J. Anal. Combin. no 2 (2007) #4.
%F From _Petros Hadjicostas_, May 24 2020: (Start)
%F T(n,k) = 2^(2*n)*A223549(n,n-k)/A223550(n,n-k).
%F Bivariate o.g.f.: Sum_{n,k>=0} T(n,k)*x^n*y^k = sqrt((1 + y)/(1 - 8*x*(1 + y))/(1 + y*sqrt(1 - 8*x*(1 + y)))). (End)
%e Triangle T(n,k) (with rows n >= 0 and columns k = 0..n) starts:
%e 1;
%e 4, 6;
%e 24, 60, 42;
%e 160, 560, 688, 308;
%e 1120, 5040, 8760, 7080, 2310;
%e ...
%p d := proc(l,m) local k; add(2^k*binomial(2*m-2*k,m-k)*binomial(m+k,m)*binomial(k,l),k=l..m); end:
%p T:= (n, k)-> d(n-k, n):
%p seq(seq(T(n, k), k=0..n), n=0..10);
%t T[n_, k_] := SeriesCoefficient[Sqrt[(1+y)/(1 - 8x (1+y))/(1 + y Sqrt[1 - 8x (1+y)])], {x, 0, n}, {y, 0, k}];
%t Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Jun 05 2020 *)
%o (PARI) d(l, m) = sum(kk=l, m, 2^kk*binomial(2*m-2*kk,m-kk)*binomial(m+kk,m)*binomial(kk,l));
%o tabl(nn) = {for (n=0, nn, for (k=0, n, print1(d(n-k, n), ", ");); print(););} \\ _Michel Marcus_, Jul 18 2015
%Y Column k=0 gives A059304.
%Y Row sums give A002458.
%Y Main diagonal gives A004982.
%Y Cf. A223549, A223550.
%K nonn,tabl
%O 0,2
%A _N. J. A. Sloane_, Feb 16 2002
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