A223549 Triangle T(n,k), read by rows, giving the numerator of the coefficient of x^k in the Boros-Moll polynomial P_n(x) for n >= 0 and 0 <= k <=n.

1, 3, 1, 21, 15, 3, 77, 43, 35, 5, 1155, 885, 1095, 315, 35, 4389, 8589, 7161, 777, 693, 63, 33649, 80353, 42245, 12285, 16485, 3003, 231, 129789, 91635, 233001, 170145, 152625, 20889, 6435, 429, 4023459, 3283533, 9804465, 8625375, 9695565, 1772199, 819819, 109395, 6435, 15646785, 58019335, 49782755, 25638305, 69324255, 31726695, 9794785, 245245, 230945, 12155

Offset: 0

Author

Jean-François Alcover, Mar 22 2013

Keywords

• nonn
• easy
• frac
• tabl

Comments

From Petros Hadjicostas, May 21 2020: (Start)

Let P_n(x) = Sum_{k=0..n} (T(n,k)/A223550(n,k))*x^k be the Boros-Moll polynomial. It follows from the theory in Comtet (1967, pp. 81-83 and 85-86) that the polynomial Q_n(x) = 2^n*n!*P_n(x) has integer coefficients and satisfies the recurrence (x-1)*Q_n(x) = 2*(2*n - 1)*(x^2 - 2)*Q_{n-1}(x) + (16*(n-1)^2 - 1)*(x + 1)*Q_{n-2}(x).

We have integral_{y = 0..infinity} dy/(y^4 + 2*x*y + 1)^(n + 1) = Pi * P_n(x)/(2^(n + (3/2)) * (x + 1)^(n + (1/2))) = Pi * Q_n(x)/(2^(2*n + (3/2)) * n! * (x + 1)^(n + (1/2))) for x > -1 and n integer >= 0.

It also follows from the theory in Comtet (1967, pp. 81-83) that g(t) = (sqrt(x + sqrt(x^2 - 1 + t)) - sqrt(x - sqrt(x^2 - 1 + t))) / sqrt((1 - t) * (x^2 - 1 + t)) = Sum_{n >= 0} t^n * P_n(x)/(2^(n - (1/2)) * (x + 1)^(n + (1/2))) for x >= 1 and 0 <= t < 1.

From Comtet's result, we get g(t)^2 = 2*(x - sqrt(1-t))/((1-t) * (x^2 - 1 + t)) = 2/((1-t) * (x + sqrt(1-t))) = Sum_{n >= 0} (Sum_{k=0..n} P_k(x) * P_{n-k}(x)) / (2^(n-1) * (x+1)^(n+1)) * t^n for 0 <= t < 1 and x > 1. (End)

Examples

			P_3(x) = 77/16 + 43*x/4 + 35*x^2/4 + 5*x^3/2.
As a result, integral_{y = 0..infinity} dy/(y^4 + 2*x*y + 1)^4 = Pi * P_3(x)/(2^(3 + (3/2)) * (x + 1)^(3 + (1/2))) = Pi * (40*x^3 + 140*x^2 + 172*x + 77)/(32 * sqrt(2*(x + 1)^7)) for x > -1. - _Petros Hadjicostas_, May 22 2020
From _Bruno Berselli_, Mar 22 2013: (Start)
Triangle T(n,k) (with rows n >= 0 and columns k = 0..n) begins as follows:
       1;
       3,     1;
      21,    15,      3;
      77,    43,     35,      5;
    1155,   885,   1095,    315,     35;
    4389,  8589,   7161,    777,    693,    63;
   33649, 80353,  42245,  12285,  16485,  3003,  231;
  129789, 91635, 233001, 170145, 152625, 20889, 6435, 429;
  ... (End)
		

Links

• Vincenzo Librandi, Rows n = 0..50, flattened
• Tewodros Amdeberhan and Victor H. Moll, A formula for a quartic integral: a survey of old proofs and some new ones, arXiv:0707.2118 [math.CA], 2007.
• George Boros and Victor H. Moll, An integral hidden in Gradshteyn and Ryzhik, Journal of Computational and Applied Mathematics, 106(2) (1999), 361-368.
• William Y. C. Chen and Ernest X. W. Xia, The Ratio Monotonicity of the Boros-Moll Polynomials, arXiv:0806.4333 [math.CO], 2009.
• William Y. C. Chen and Ernest X. W. Xia, The Ratio Monotonicity of the Boros-Moll Polynomials, Mathematics of Computation, 78(268) (2009), 2269-2282.
• Louis Comtet, Fonctions génératrices et calcul de certaines intégrales, Publikacije Elektrotechnickog faculteta - Serija Matematika i Fizika, No. 181/196 (1967), 77-87.

Crossrefs

Cf. A067001, A223550 (denominators), A334907.

Programs

Magma

/* As triangle: */ [[Numerator(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

Mathematica

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] // Numerator, {n, 0, 9}, {k, 0, n}] // Flatten

Formula

T(n,k)/A223550(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.

P_n(x) = Sum_{k=0..n} (T(n, k)/A223550(n,k))*x^k = ((2*n)!/4^n/(n!)^2)*2F1([-n, n + 1], [1/2 - n], (x + 1)/2).

From Petros Hadjicostas, May 22 2020: (Start)

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).

O.g.f. for P_n(x): sqrt((x + 1)/(1 - 2*(x + 1)*w)/(x + sqrt(1 - 2*(x + 1)*w))). [It follows from Comtet's theory and my comments.]

P_n(1) = Sum_{k=0..n} T(n,k)/A223550(n,k) = A334907(n)/(2^n*n!). (End)

Extensions

Various sections and name edited by Petros Hadjicostas, May 22 2020