A067001 -id:A067001 - cp's OEIS frontend

A059304 a(n) = 2^n * (2*n)! / (n!)^2.

1, 4, 24, 160, 1120, 8064, 59136, 439296, 3294720, 24893440, 189190144, 1444724736, 11076222976, 85201715200, 657270374400, 5082890895360, 39392404439040, 305870434467840, 2378992268083200, 18531097667174400

Offset: 0

Henry Bottomley, Jan 25 2001

Number of lattice paths from (0,0) to (n,n) using steps (0,1), and two kinds of steps (1,0). - Joerg Arndt, Jul 01 2011
The convolution square root of this sequence is A004981. - T. D. Noe, Jun 11 2002
Also main diagonal of array: T(i,1)=2^(i-1), T(1,j)=1, T(i,j) = T(i,j-1) + 2*T(i-1,j). - Benoit Cloitre, Feb 26 2003
The Hankel transform (see A001906 for definition) of this sequence with interpolated zeros(1, 0, 4, 0, 24, 0, 160, 0, 1120, ...) = is A036442: 1, 4, 32, 512, 16384, ... . - Philippe Deléham, Jul 03 2005
The Hankel transform of this sequence gives A103488. - Philippe Deléham, Dec 02 2007
Equals the central column of the triangle A038207. - Zerinvary Lajos, Dec 08 2007
Equals number of permutations whose reverse shares the same recording tableau in the Robinson-Schensted correspondence with n=(k-1)/2 for k odd. - Dang-Son Nguyen, Jul 02 2024
Number of ternary strings of length 2*n that have the same number of 0's as the combined number of 1's and 2's. For example, a(2)=24 since the strings of length 4 are the 6 permutations of 0011, the 12 permutations of 0012, and the 6 permutations of 0022. - Enrique Navarrete, Jul 30 2025

Harry J. Smith, Table of n, a(n) for n = 0..200
Paul Barry and Arnauld Mesinga Mwafise, Classical and Semi-Classical Orthogonal Polynomials Defined by Riordan Arrays, and Their Moment Sequences, Journal of Integer Sequences, Vol. 21 (2018), Article 18.1.5.
Hacène Belbachir and Abdelghani Mehdaoui, Recurrence relation associated with the sums of square binomial coefficients, Quaestiones Mathematicae (2021) Vol. 44, Issue 5, 615-624.
Hacène Belbachir, Abdelghani Mehdaoui, and László Szalay, Diagonal Sums in the Pascal Pyramid, II: Applications, J. Int. Seq., Vol. 22 (2019), Article 19.3.5.
H. J. Brothers, Pascal's Prism: Supplementary Material.
Tucker J. Ervin, Blake Jackson, Jay Lane, Kyungyong Lee, Son Dang Nguyen, Jack O'Donohue and Michael Vaughan, Permutations whose Reverse Shares the Same Recording Tableau in the Robinson-Schensted Correspondence, Séminaire Lotharingien de Combinatoire 86 (2022), Article B86a.
Tony D. Noe, On the Divisibility of Generalized Central Trinomial Coefficients, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.7.
Michael Z. Spivey and Laura L. Steil, The k-Binomial Transforms and the Hankel Transform, Journal of Integer Sequences, Vol. 9 (2006), Article 06.1.1.

Diagonal of A013609.
Column k=0 of A067001.
Cf. A000079, A000984, A001906, A036442, A038207, A103488.

[2^n*Binomial(2*n,n): n in [0..25]]; // Vincenzo Librandi, Oct 08 2015

seq(binomial(2*n,n)*2^n,n=0..19); # Zerinvary Lajos, Dec 08 2007

Table[2^n Binomial[2n,n],{n,0,30}] (* Harvey P. Dale, Dec 16 2014 *)

{a(n)=if(n<0, 0, 2^n*(2*n)!/n!^2)} /* Michael Somos, Jan 31 2007 */

{ for (n = 0, 200, write("b059304.txt", n, " ", 2^n * (2*n)! / n!^2); ) } \\ Harry J. Smith, Jun 25 2009

/* as lattice paths: same as in A092566 but use */
steps=[[1, 0], [1, 0], [0, 1]]; /* note the double [1, 0] */
/* Joerg Arndt, Jul 01 2011 */

def A059304(n): return pow(2,n)*binomial(2*n,n)
print([A059304(n) for n in range(41)]) # G. C. Greubel, Jan 18 2025

a(n) = 2^n * C(2*n,n).
D-finite with recurrence a(n) = 4*(2-1/n)*a(n-1).
a(n) = A000079(n)*A000984(n)
G.f.: 1/sqrt(1-8*x) - T. D. Noe, Jun 11 2002
E.g.f.: exp(4*x)*BesselI(0, 4*x). - Vladeta Jovovic, Aug 20 2003
a(n) = A038207(n,n). - Joerg Arndt, Jul 01 2011
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - 4*x*(2*k+1)/(4*x*(2*k+1) + (k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 24 2013
E.g.f.: E(0)/2, where E(k) = 1 + 1/(1 - 4*x/(4*x + (k+1)^2/(2*k+1)/E(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 01 2013
G.f.: Q(0)/(1+2*sqrt(x)), where Q(k) = 1 + 2*sqrt(x)/(1 - 2*sqrt(x)*(2*k+1)/(2*sqrt(x)*(2*k+1) + (k+1)/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 09 2013
O.g.f.: hypergeom([1/2], [], 8*x). - Peter Luschny, Oct 08 2015
a(n) = Sum_{k = 0..2*n} (-1)^(n+k)*binomial(2*n,k)*binomial(3*n-2*k,n)* binomial(n+k,n). - Peter Bala, Aug 04 2016
a(n) ~ 8^n/sqrt(Pi*n). - Ilya Gutkovskiy, Aug 04 2016
From Amiram Eldar, Jul 21 2020: (Start)
Sum_{n>=0} 1/a(n) = 8/7 + 8*sqrt(7)*arcsin(1/sqrt(8))/49.
Sum_{n>=0} (-1)^n/a(n) = (8/27)*(3 - arcsinh(1/sqrt(8))). (End)
a(n) = Sum_{k = n..2*n} binomial(2*n,k)*binomial(k,n). In general, for m >= 1, Sum_{k = n..m*n} binomial(m*n,k)*binomial(k,n) = 2^((m-1)*n)*binomial(m*n,n). - Peter Bala, Mar 25 2023
Conjecture: a(n) = Sum_{0 <= j, k <= n} binomial(n, j)*binomial(n, k)* binomial(k+j, n). - Peter Bala, Jul 16 2024

A002458 a(n) = binomial(4n+1, 2n).

Original entry on oeis.org

1, 10, 126, 1716, 24310, 352716, 5200300, 77558760, 1166803110, 17672631900, 269128937220, 4116715363800, 63205303218876, 973469712824056, 15033633249770520, 232714176627630544, 3609714217008132870, 56093138908331422716, 873065282167813104916

Offset: 0

N. J. A. Sloane

			1 + 10*x + 126*x^2 + 1716*x^3 + 24310*x^4 + 352716*x^5 + 5200300*x^6 + ...

The right-hand side of a binomial coefficient identity in H. W. Gould, Combinatorial Identities, Morgantown, 1982, (3.109), page 35.

T. D. Noe, Table of n, a(n) for n = 0..100

Cf. A000984, A001448, A001700, A024492, A100033, A187364, A187365.

Row sums of A067001.

A002458:=n->binomial(4*n+1,2*n): seq(A002458(n), n=0..30); # Wesley Ivan Hurt, Jan 17 2017

Table[Binomial[4n+1,2n],{n,0,30}] (* Harvey P. Dale, Apr 04 2011 *)
4^Range[0, 22] Simplify[ CoefficientList[ Series[ Sqrt[2]/(((Sqrt[1 - 4 x] + 1)^(1/2))*Sqrt[1 - 4 x]), {x, 0, 22}], x]] (* Robert G. Wilson v, Aug 08 2011 *)

PARI
```
a(n) = binomial( 4*n + 1, 2*n)
```

a(n) = Sum_{k=0..n} 4^k * binomial( n + k, n) * binomial( 2*n - 2*k, n - k). - Michael Somos, Feb 25 2012
a(n) = A001700(2*n) = (n+1)*A000108(2*n+1).
G.f.: (4 - (1+4*y)*c(y) - (1-4*y)*c(-y))/(2*(1 - (4*y)^2)) with y^2 = x, c(y) = g.f. for A000108 (Catalan). - Wolfdieter Lang, Dec 13 2001
a(n) ~ 2^(1/2)*Pi^(-1/2)*n^(-1/2)*2^(4*n)*{1 - 5/16*n^-1 + ...}. - Joe Keane (jgk(AT)jgk.org), Jun 11 2002
a(n) = A024492(n)*(n+1). - R. J. Mathar, Aug 10 2015
G.f.: 2F1(3/4,5/4; 3/2; 16*x). - R. J. Mathar, Aug 10 2015
D-finite with recurrence n*(2*n + 1)*a(n) - 2*(4*n - 1)*(4*n + 1)*a(n-1) = 0. - R. J. Mathar, Aug 10 2015
From Peter Bala, Nov 04 2015: (Start)
a(n) = 4^n*binomial(2*n + 1/2, n).
O.g.f.: sqrt(c(4*x)/(1 - 16*x)) = sqrt(2/(1 - 16*x)/(1 + sqrt(1 - 16*x))), where
c(y) = g.f. for A000108 (Catalan). In general, c(x)^k/sqrt(1 - 4*x) is the o.g.f. for the sequence binomial(2*n + k, n). (End) [Edited by Petros Hadjicostas, May 25 2020]
From Ilya Gutkovskiy, Jan 17 2017: (Start)
E.g.f.: 2F2(3/4,5/4; 1,3/2; 16*x).
Sum_{n>=0} 1/a(n) = 3F2(1,1,3/2; 3/4,5/4; 1/16) = 1.108563435104316693... (End)
From Peter Bala, Mar 16 2018: (Start)
The right-hand side of the binomial coefficient identity Sum_{k = 0..n} 4^(n-k) * C(2*n+1, 2*k) * C(2*k, k) = a(n).
a(n) = 4^n*hypergeom([-n, -n-1/2], [1], 1). (End)
From Peter Bala, Mar 20 2023: (Start)
a(n) = Sum_{k = 0..n} binomial(2*n+1,k)^2.
a(n) = (1/2)*hypergeom([-1 - 2*n, -1 - 2*n], [1], 1). (End)

A004982 a(n) = (2^n/n!) * Product_{k=0..n-1} (4*k + 3).

Original entry on oeis.org

1, 6, 42, 308, 2310, 17556, 134596, 1038312, 8046918, 62587140, 488179692, 3816677592, 29897307804, 234578876616, 1843119744840, 14499208659408, 114181268192838, 900017055167076, 7100134546318044, 56053693786721400, 442824180915099060, 3500419715805068760, 27685137752276452920

Offset: 0

Joe Keane (jgk(AT)jgk.org)

Conjecture: a(p*n) = a(n) (mod p^2) for prime p == 1 (mod 4) and all positive integers n. Cf. A004981. - Peter Bala, Dec 22 2019

Alois P. Heinz, Table of n, a(n) for n = 0..400

Main diagonal of A067001. Cf. A004981.

List([0..25], n-> 2^n*Product([0..n-1], k-> 4*k+3)/Factorial(n) ); # G. C. Greubel, Aug 22 2019

[1] cat [2^n*&*[4*k+3: k in [0..n-1]]/Factorial(n): n in [1..25]]; // G. C. Greubel, Aug 22 2019

A004982 := n -> (-8)^n*binomial(-3/4, n):
seq(A004982(n), n=0..25); # Peter Luschny, Oct 23 2018

Table[2^n/n! Product[4k+3,{k,0,n-1}],{n,0,30}] (* Harvey P. Dale, Oct 03 2011 *)
Table[Sum[2^k*Binomial[2*n-2*k,n-k]*Binomial[n+k,n],{k,0,n}],{n,0,25}] (* Vaclav Kotesovec, Sep 15 2013 *)
FullSimplify[Table[8^n*Gamma[n+3/4]/(n!*Gamma[3/4]), {n, 0, 25}]] (* Vaclav Kotesovec, Sep 15 2013 *)
max = 30; s = Hypergeometric1F1[3/4, 1, 8x] + O[x]^(max+1);
CoefficientList[s, x]*(Range[0, max]!) (* Jean-François Alcover, Dec 19 2015, after Karol A. Penson *)

a(n)=2^n/n!*prod(k=0,n-1,4*k+3)
for(n=0,25,print(a(n)))

x='x+O('x^66); Vec((1-8*x)^(-3/4)) \\ Joerg Arndt, Apr 20 2013

[8^n*rising_factorial(3/4, n)/factorial(n) for n in (0..25)] # G. C. Greubel, Aug 22 2019

G.f.: (1 - 8*x)^(-3/4).
a(n) ~ Gamma(3/4)^-1*n^(-1/4)*2^(3*n)*{1 - 3/32*n^-1 + ...}
a(n) = 8^n*Gamma(n+3/4)/(n!*Gamma(3/4)). - Vaclav Kotesovec, Sep 15 2013
From Karol A. Penson, Dec 19 2015: (Start)
a(n) = (-8)^n*binomial(-3/4,n).
E.g.f.: is the hypergeometric function of type 1F1, in Maple notation hypergeom([3/4], [1], 8*x).
Representation as n-th moment of a positive function on (0, 8): a(n) = Integral_{x=0..8} ( x^n*2^(1/4)/(8*Pi*x^(1/4)*(1-x/8)^(3/4)) ) dx, n >= 0. This function is the solution of the Hausdorff moment problem on (0, 8) with moments equal to a(n). As a consequence this representation is unique. (End)
D-finite with recurrence: n*a(n) +2*(-4*n+1)*a(n-1)=0. - R. J. Mathar, Jan 16 2020

More terms from Rick L. Shepherd, Mar 03 2002

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

A223550 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.

Original entry on oeis.org

1, 2, 1, 8, 4, 2, 16, 4, 4, 2, 128, 32, 32, 16, 8, 256, 128, 64, 8, 16, 8, 1024, 512, 128, 32, 64, 32, 16, 2048, 256, 256, 128, 128, 32, 32, 16, 32768, 4096, 4096, 2048, 2048, 512, 512, 256, 128, 65536, 32768, 8192, 2048, 4096, 2048, 1024, 64, 256, 128

Offset: 0

Jean-François Alcover, Mar 22 2013

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

			P_3(x) = 77/16 + 43*x/4 + 35*x^2/4 + 5*x^3/2.
From _Bruno Berselli_, Mar 22 2013: (Start)
Triangle T(n,k) (with rows n >= 0 and columns k=0..n) begins as follows:
      1;
      2,     1;
      8,     4,    2;
     16,     4,    4,    2;
    128,    32,   32,   16,    8;
    256,   128,   64,    8,   16,    8;
   1024,   512,  128,   32,   64,   32,   16;
   2048,   256,  256,  128,  128,   32,   32,  16;
  32768,  4096, 4096, 2048, 2048,  512,  512, 256, 128;
  65536, 32768, 8192, 2048, 4096, 2048, 1024,  64, 256, 128;
  ... (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, A223549 (numerators), A334907.

/* 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

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

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.
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).
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).
P_n(1) = Sum_{k=0..n} A223549(n,k)/T(n,k) = A334907(n)/(2^n*n!). (End)

Name edited by Petros Hadjicostas, May 22 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

A335183 T(n,k) = Sum_{j=1..n} 2^jbinomial(2n-2j, n-j)binomial(n+j, n)*binomial(j, k), triangle read by rows (n >= 0 and 0 <= k <= n).

Original entry on oeis.org

0, 4, 4, 36, 60, 24, 288, 688, 560, 160, 2240, 7080, 8760, 5040, 1120, 17304, 68712, 114576, 99456, 44352, 8064, 133672, 642824, 1351840, 1572480, 1055040, 384384, 59136, 1034880, 5864640, 14912064, 21778560, 19536000, 10695168, 3294720, 439296

Offset: 0

Petros Hadjicostas, May 25 2020

This was the original version of A126936.

			Table T(n,k) (with rows n >= 0 and columns k = 0..n) begins as follows:
       0;
       4,      4;
      36,     60,      24;
     288,    688,     560,     160;
    2240,   7080,    8760,    5040,    1120;
   17304,  68712,  114576,   99456,   44352,   8064;
  133672, 642824, 1351840, 1572480, 1055040, 384384, 59136;
  ...

Cf. A000984, A067001, A069722 (main diagonal), A126936.

t[l_, m_] := Sum[2^k*Binomial[2*m-2*k, m-k]*Binomial[m+k, m]*Binomial[k, l], {k, 1, m}]; Table[t[l, m], {m, 0, 7}, {l, 0, m}] // Flatten (* Jean-François Alcover, Jan 09 2014 from the original version of A126936 *)

T(n,k) = sum(j=1, n, 2^j*binomial(2*n-2*j, n-j)*binomial(n+j, n)*binomial(j, k));
tabl(nn) = {for (n=0, nn, for (k=0, n, print1(T(n,k), ", "); ); print(); ); }

T(n,n) = A069722(n+1) for n >= 0.
T(n,k) = A126936(n,k) = A067001(n,n-k) for n >= k >= 1.
T(n,0) = A126936(n,0) - binomial(2*n, n) = A067001(n,n) - A000984(n) for n >= 0.
Bivariate o.g.f.: Sum_{n,k >= 0} T(n,k)*x^n*y^k = -1/sqrt(1 - 4*x) + sqrt((1 + y)/(1 - 8*x*(1 + y))/(y + sqrt(1 - 8*x*(1 + y)))).

cp's OEIS Frontend

A059304 a(n) = 2^n * (2*n)! / (n!)^2.

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A002458 a(n) = binomial(4*n+1, 2*n).

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A004982 a(n) = (2^n/n!) * Product_{k=0..n-1} (4*k + 3).

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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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A223550 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.

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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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A002458 a(n) = binomial(4n+1, 2n).

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.

A335183 T(n,k) = Sum_{j=1..n} 2^jbinomial(2n-2j, n-j)binomial(n+j, n)*binomial(j, k), triangle read by rows (n >= 0 and 0 <= k <= n).