A223550 -id:A223550 - cp's OEIS frontend

A067001 Triangle T(n,k) = d(n-k,n), 0 <= k <= n, where d(l,m) = Sum_{k=l..m} 2^k * binomial(2m-2k, m-k) * binomial(m+k, m) * binomial(k, l).

1, 4, 6, 24, 60, 42, 160, 560, 688, 308, 1120, 5040, 8760, 7080, 2310, 8064, 44352, 99456, 114576, 68712, 17556, 59136, 384384, 1055040, 1572480, 1351840, 642824, 134596, 439296, 3294720, 10695168, 19536000, 21778560, 14912064, 5864640, 1038312

Offset: 0

N. J. A. Sloane, Feb 16 2002

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

			Triangle T(n,k) (with rows n >= 0 and columns k = 0..n) starts:
     1;
     4,    6;
    24,   60,   42;
   160,  560,  688,  308;
  1120, 5040, 8760, 7080, 2310;
  ...

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.
V. H. Moll, The evaluation of integrals: a personal story, Notices Amer. Math. Soc., 49 (No. 3, March 2002), 311-317.
V. H. Moll, Combinatorial sequences arising from a rational integral, Onl. J. Anal. Combin. no 2 (2007) #4.

Column k=0 gives A059304.
Row sums give A002458.
Main diagonal gives A004982.
Cf. A223549, A223550.

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:
T:= (n, k)-> d(n-k, n):
seq(seq(T(n, k), k=0..n), n=0..10);

T[n_, k_] := SeriesCoefficient[Sqrt[(1+y)/(1 - 8x (1+y))/(1 + y Sqrt[1 - 8x (1+y)])], {x, 0, n}, {y, 0, k}];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 05 2020 *)

d(l, m) = sum(kk=l, m, 2^kk*binomial(2*m-2*kk,m-kk)*binomial(m+kk,m)*binomial(kk,l));
tabl(nn) = {for (n=0, nn, for (k=0, n, print1(d(n-k, n), ", ");); print(););} \\ Michel Marcus, Jul 18 2015

From Petros Hadjicostas, May 24 2020: (Start)
T(n,k) = 2^(2*n)*A223549(n,n-k)/A223550(n,n-k).
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)

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.

Original entry on oeis.org

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

Jean-François Alcover, Mar 22 2013

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)

			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)

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.

Cf. A067001, A223550 (denominators), A334907.

/* 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

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

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)

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

A334907 Comtet's expansion of the e.g.f. (sqrt(1 + sqrt(8s)) - sqrt(1 - sqrt(8s)))/ sqrt(8s (1 - 8*s)).

Original entry on oeis.org

1, 5, 63, 1287, 36465, 1322685, 58503375, 3053876175, 183771489825, 12525477859125, 953725671273375, 80237355387564375, 7391465178302430225, 739967791738943292525, 79993069900054731795375, 9286937373235386442953375, 1152424501315118408602850625

Offset: 0

Petros Hadjicostas, May 15 2020

nonn

A special case of an integral in Comtet (1967, pp. 85-86) yields

Integral_{t=-oo..oo} dx/(x^2 + t^2)^(2*n) = Pi * a(n-1)/((n-1)! * 2^(3*n - 2) * t^(4*n-1)) for n >= 1 and t > 0. This integral also follows from a theorem in Moll (2002, p. 312, set a=1), but it requires the summation formula for a(n) shown below.

Louis Comtet, Fonctions génératrices et calcul de certaines intégrales, Publikacije Elektrotechnickog faculteta - Serija Matematika i Fizika, No. 181/196 (1967), 77-87; see pp. 81-83.
Petros Hadjicostas, Proof of the claim a(n) = n!*A063079(n+1)/A060818(n), 2020.
V. H. Moll, The evaluation of integrals: a personal story, Notices Amer. Math. Soc., 49 (No. 3, March 2002), 311-317.

Cf. A001790, A060818, A063079, A223549, A223550.

a(n) = binomial(4*n+2, 2*n+1)*n!/2^(n+1).
a(n) = n!*A063079(n+1)/A060818(n) = n!*A001790(2*n+1)/A060818(n) (see the link for a proof).
a(n) = n!*Sum_{j=0..n} 2^(n-2*j)*binomial(2*n+1,2*j)*binomial(2*j,j).
a(n) = 2^n*n!*Sum_{k=0..n} A223549(n,k)/A223550(n,k).
E.g.f.: 2/(sqrt(1 - 8*s) * (sqrt(1 + sqrt(8*s)) + sqrt(1 - sqrt(8*s)))).
E.g.f.: sqrt(2/(1 + sqrt(1 - 8*s))/(1 - 8*s)).
D-finite with recurrence (2*n+1)*a(n) -(4*n-1)*(4*n+1)*a(n-1)=0. - R. J. Mathar, May 25 2020

A067002 Numerator of Sum_{k=0..n} 2^(k-2n) binomial(2n-2k,n-k) * binomial(n+k,n).

Original entry on oeis.org

1, 3, 21, 77, 1155, 4389, 33649, 129789, 4023459, 15646785, 122044923, 477084699, 7474326951, 29322359577, 230389968105, 906200541213, 57090634096419, 225004263791769, 1775033636579511, 7006711723340175, 110706045228774765

Offset: 0

N. J. A. Sloane, Feb 16 2002

Numerator of e(0,n) (see the Maple line).

The generating function of the full fraction is (1-2*x)^(-3/4). - R. J. Mathar, Nov 06 2011

			1, 3/2, 21/8, 77/16, 1155/128, 4389/256, 33649/1024, 129789/2048, 4023459/32768, ... = A067002/A046161.

V. H. Moll. The evaluation of integrals: a personal story, Notices Amer. Math. Soc., 49 (No. 3, March 2002), 311-317.

Denominators are in A046161.

Cf. A067001, A223549, A223550.

e := proc(l,m) local k; add(2^(k-2*m)*binomial(2*m-2*k,m-k)*binomial(m+k,m)*binomial(k,l),k=l..m); end;

Numerator[Table[Sum[2^(k-2n) Binomial[2n-2k,n-k]Binomial[n+k,n],{k,0,n}],{n,0,30}]] (* Harvey P. Dale, Oct 19 2012 *)

Numerator of 2^n*Gamma(n + 3/4)/(Gamma(3/4)*n!). - R. J. Mathar, Nov 06 2011
Numerator of integral_{x>0} 1/(x^4 + 1)^(n+1) / (Pi*sqrt(2)). - Jean-François Alcover, Apr 29 2013
From Petros Hadjicostas, May 23 2020: (Start)
If fr(n) = A067002(n)/A046161(n), then fr(n) = P_n(0), where P_n(x) is the Boros-Moll polynomial mentioned in A223549 and A223550 (and whose coefficients are the numbers e(l,n) = A067001(n,n-l)/2^(2*n) that are mentioned in the Maple line below with l = 0..n).
Recurrence for fr(n): 4*n*(n - 1)*fr(n) = 8*(2*n - 1)*(n - 1)*fr(n-1) - (16*(n-1)^2 - 1)*fr(n-2) for n >= 2 with fr(0) = 1 and fr(1) = 3/2. (End)

A126936 Coefficients of a polynomial representation of the integral of 1/(x^4 + 2ax^2 + 1)^(n+1) from x = 0 to infinity.

Original entry on oeis.org

1, 6, 4, 42, 60, 24, 308, 688, 560, 160, 2310, 7080, 8760, 5040, 1120, 17556, 68712, 114576, 99456, 44352, 8064, 134596, 642824, 1351840, 1572480, 1055040, 384384, 59136, 1038312, 5864640, 14912064, 21778560, 19536000, 10695168, 3294720

Offset: 0

R. J. Mathar, Mar 17 2007

The integral N(a;n) = Integral_{x=0..infinity} 1/(x^4 + 2*a*x^2 + 1)^(n+1) has a polynomial representation P_n(a) = 2^(n + 3/2) * (a+1)^(n + 1/2) * N(a;n) / Pi (known as the Boros-Moll polynomial). The table contains the coefficients T(n,l) of P_n(a) = 2^(-2*n)*Sum_{l=0..n} T(n,l)*a^l in row n and column l (with n >= 0 and 0 <= l <= n).

			The table T(n,l) (with rows n >= 0 and columns l = 0..n) starts:
      1;
      6,     4;
     42,    60,     24;
    308,   688,    560,   160;
   2310,  7080,   8760,  5040,  1120;
  17556, 68712, 114576, 99456, 44352, 8064;
  ...
For n = 2, N(a;2) = Integral_{x=0..oo} dx/(x^4 + 2*a*x + 1)^3 = 2^(-2*2)*(Sum_{l=0..2} T(2,l)*a^l) * Pi/(2^(2 + 3/2) * (a + 1)^(2 + 1/2) = (42 + 60*a + 24*a^2) * Pi/(32 * (2*(a+1))^(5/2)) for a > -1. - _Petros Hadjicostas_, May 25 2020

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.
Victor H. Moll, The evaluation of integrals: a personal story, Notices Amer. Math. Soc., 49 (No. 3, March 2002), 311-317.
Victor H. Moll, Combinatorial sequences arising from a rational integral, Onl. J. Anal. Combin., no 2 (2007), #4.

Cf. A002458 (row sums), A004982 (column l=0), A059304 (main diagonal), A067001 (rows reversed), A223549, A223550, A334907.

A126936 := proc(m, l)
    add(2^k*binomial(2*m-2*k, m-k)*binomial(m+k, m)*binomial(k, l), k=l..m):
end:
seq(seq(A126936(m,l), l=0..m), m=0..12); # R. J. Mathar, May 25 2020

t[m_, l_] := Sum[2^k*Binomial[2*m-2*k, m-k]*Binomial[m+k, m]*Binomial[k, l], {k, l, m}]; Table[t[m, l], {m, 0, 11}, {l, 0, m}] // Flatten (* Jean-François Alcover, Jan 09 2014, after Maple, adapted May 2020 *)

From Petros Hadjicostas, May 25 2020: (Start)
T(n,l) = A067001(n, n-l) = 2^(2*n) * A223549(n,l)/A223550(n,l).
Sum_{l=0..n} T(n,l) = A002458(n) = A334907(n)*2^n/n!.
Bivariate o.g.f.: Sum_{n,l >= 0} T(n,l)*x^n*y^l = sqrt((1 + y)/(1 - 8*x*(1 + y))/(y + sqrt(1 - 8*x*(1 + y)))). (End)

Corrected by Petros Hadjicostas, May 23 2020

cp's OEIS Frontend

A067001 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).

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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.

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A334907 Comtet's expansion of the e.g.f. (sqrt(1 + sqrt(8*s)) - sqrt(1 - sqrt(8*s)))/ sqrt(8*s * (1 - 8*s)).

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A067002 Numerator of Sum_{k=0..n} 2^(k-2*n) * binomial(2*n-2*k,n-k) * binomial(n+k,n).

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A126936 Coefficients of a polynomial representation of the integral of 1/(x^4 + 2*a*x^2 + 1)^(n+1) from x = 0 to infinity.

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A067001 Triangle T(n,k) = d(n-k,n), 0 <= k <= n, where d(l,m) = Sum_{k=l..m} 2^k * binomial(2m-2k, m-k) * binomial(m+k, m) * binomial(k, l).

A334907 Comtet's expansion of the e.g.f. (sqrt(1 + sqrt(8s)) - sqrt(1 - sqrt(8s)))/ sqrt(8s (1 - 8*s)).

A067002 Numerator of Sum_{k=0..n} 2^(k-2n) binomial(2n-2k,n-k) * binomial(n+k,n).

A126936 Coefficients of a polynomial representation of the integral of 1/(x^4 + 2ax^2 + 1)^(n+1) from x = 0 to infinity.