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%I #36 Sep 08 2022 08:46:04
%S 1,2,1,8,4,2,16,4,4,2,128,32,32,16,8,256,128,64,8,16,8,1024,512,128,
%T 32,64,32,16,2048,256,256,128,128,32,32,16,32768,4096,4096,2048,2048,
%U 512,512,256,128,65536,32768,8192,2048,4096,2048,1024,64,256,128
%N Triangle T(n,k), read by rows, giving the denominator of the coefficient of x^k in the Boros-Moll polynomial P_n(x) for n >= 0 and 0 <= k <= n.
%C As Chen and Xia (2009) state, the Boros-Moll polynomial P_n(x) can be viewed as a Jacobi polynomial P_n^{a,b}(x) with a = n + (1/2) and b = -(n + (1/2)). For more information about the relation of this polynomial P_n(x) to the theory in Comtet (1967, pp. 81-83 and 85-86), see my comments for A223549. - _Petros Hadjicostas_, May 22 2020
%H Vincenzo Librandi, <a href="/A223550/b223550.txt">Rows n = 0..50, flattened</a>
%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.
%F A223549(n,k)/T(n,k) = 2^(-2*n)*Sum_{j=k..n} 2^j*binomial(2*n - 2*j, n - j)*binomial(n + j, j)*binomial(j, k) = 2^(-2*n)*A067001(n,n-k) for n >= 0 and k = 0..n.
%F P_n(x) = Sum_{k=0..n} (A223549(n,k)/T(n,k))*x^k = ((2*n)!/4^n/(n!)^2)*2F1([-n, n + 1], [1/2 - n], (x + 1)/2).
%F From _Petros Hadjicostas_, May 22 2020: (Start)
%F Recurrence for the polynomial: 4*n*(n - 1)*(x - 1)*P_n(x) = 4*(2*n - 1)*(n - 1)*(x^2 - 2)*P_{n-1}(x) + (16*(n - 1)^2 - 1)*(x + 1)*P_{n-2}(x).
%F P_n(1) = Sum_{k=0..n} A223549(n,k)/T(n,k) = A334907(n)/(2^n*n!). (End)
%e P_3(x) = 77/16 + 43*x/4 + 35*x^2/4 + 5*x^3/2.
%e From _Bruno Berselli_, Mar 22 2013: (Start)
%e Triangle T(n,k) (with rows n >= 0 and columns k=0..n) begins as follows:
%e 1;
%e 2, 1;
%e 8, 4, 2;
%e 16, 4, 4, 2;
%e 128, 32, 32, 16, 8;
%e 256, 128, 64, 8, 16, 8;
%e 1024, 512, 128, 32, 64, 32, 16;
%e 2048, 256, 256, 128, 128, 32, 32, 16;
%e 32768, 4096, 4096, 2048, 2048, 512, 512, 256, 128;
%e 65536, 32768, 8192, 2048, 4096, 2048, 1024, 64, 256, 128;
%e ... (End)
%t t[n_, k_] := 2^(-2*n)*Sum[ 2^j*Binomial[2*n - 2*j, n-j]*Binomial[n+j, j]*Binomial[j, k], {j, k, n}]; Table[t[n, k] // Denominator, {n, 0, 9}, {k, 0, n}] // Flatten
%o (Magma) /* As triangle: */ [[Denominator(2^(-2*n)*&+[2^j*Binomial(2*n-2*j, n-j)*Binomial(n+j, j)*Binomial(j, k): j in [k..n]]): k in [0..n]]: n in [0..10]]; // _Bruno Berselli_, Mar 22 2013
%Y Cf. A067001, A223549 (numerators), A334907.
%K nonn,easy,frac,tabl
%O 0,2
%A _Jean-François Alcover_, Mar 22 2013
%E Name edited by _Petros Hadjicostas_, May 22 2020
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