A228333 -id:A228333 - cp's OEIS frontend

A000515 a(n) = (2n)!(2n+1)!/n!^4, or equally (2n+1)*binomial(2n,n)^2.

1, 12, 180, 2800, 44100, 698544, 11099088, 176679360, 2815827300, 44914183600, 716830370256, 11445589052352, 182811491808400, 2920656969720000, 46670906271240000, 745904795339462400, 11922821963004219300, 190600129650794094000, 3047248986392325330000

Offset: 0

N. J. A. Sloane

a(n) is also the (n,n)-th entry in the inverse of the n-th Hilbert matrix. - Asher Auel, May 20 2001
a(n) is also the ratio of the determinants of the n-th Hilbert matrix to the (n+1)-th Hilbert matrix (see A005249), for n>0. Thus the determinant of the inverse of the n-th Hilbert matrix is the product of a(i) for i from 1 to n. (Claimed by Jud McCranie without proof, Jul 17 2000)
a(n) is the right side of the binomial sum: 2^(4*n) * Sum_{i=0..n} binomial(-1/2, i)*binomial(1/2, i). - Yong Kong (ykong(AT)curagen.com), Dec 26 2000
Right-hand side of Sum_{i=0..n} Sum_{j=0..n} binomial(i+j,j)^2 * binomial(4n-2i-2j,2n-2j).

E. R. Hansen, A Table of Series and Products, Prentice-Hall, Englewood Cliffs, NJ, 1975, p. 96.
A. P. Prudnikov, Yu. A. Brychkov and O. I. Marichev, "Integrals and Series", Volume 1: "Elementary Functions", Chapter 4: "Finite Sums", New York, Gordon and Breach Science Publishers, 1986-1992.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..100
G. E. Andrews and P. Paule, Some questions concerning computer-generated proofs of a binomial double-sum identity, J. Symbolic Computation 11(1994), 1-7.
D. Galakhov, A. Mironov and A. Morozov, Wall Crossing Invariants: from quantum mechanics to knots, arXiv preprint arXiv:1410.8482 [hep-th], 2014. See Eq. (A.15).
R. K. Guy, Letter to N. J. A. Sloane, Sep 1986
J. E. Lauer, Letter to N. J. A. Sloane, Dec 1980
D. H. Lehmer, Review of A. N. Lowan et al., "Table of the zeros of the Legendre polynomials of order 1-16...", in Math. Tables Aids Computation (MTAC), 1 (1943-1945), 52-53.
Pedro J. Miana and Natalia Romero, Moments of combinatorial and Catalan numbers, Journal of Number Theory, Volume 130, Issue 8, August 2010, Pages 1876-1887. See Omega3. Remark 3 p. 1882.
I. Nemes et al., How to do Monthly problems with your computer, Amer. Math. Monthly, 104 (1997), 505-519.
Yidong Sun and Fei Ma, Four transformations on the Catalan triangle, arXiv preprint arXiv:1305.2017 [math.CO], 2013 (see Omega_3).
Yidong Sun and Fei Ma, Some new binomial sums related to the Catalan triangle, Electronic Journal of Combinatorics 21(1) (2014), #P1.33

Cf. A002894, A005249, A002457, A000108, A039598, A024492, A000894, A228329, A000515, A228330, A228331, A228332, A228333.

[(2*n+1)*Binomial(2*n,n)^2: n in [0..25]]; // Vincenzo Librandi, Oct 08 2015

with(linalg): for n from 1 to 24 do print(det(hilbert(n))/det(hilbert(n+1))): od;

A000515[n_] := (2*n + 1)*Binomial[2 n, n]^2 (* Enrique Pérez Herrero, Mar 31 2010 *)
Table[(2 n + 1) (n + 1)^2 CatalanNumber[n]^2, {n, 0, 18}] (* Jan Mangaldan, Sep 23 2021 *)

vector(100, n, n--; (2*n+1)*binomial(2*n,n)^2) \\ Altug Alkan, Oct 08 2015

a(n) ~ 2*Pi^-1*2^(4*n). - Joe Keane (jgk(AT)jgk.org), Jun 07 2002
O.g.f.: (2/Pi)*EllipticE(4*sqrt(x))/(1-16*x). - Vladeta Jovovic, Jun 15 2005
E.g.f.: Sum_{n>=0} a(n)*x^(2n)/(2n)! = BesselI(0, 2*x)*(BesselI(0, 2*x) + 4*x*BesselI(1, 2*x)). - Vladeta Jovovic, Jun 15 2005
E.g.f.: Sum_{n>=0} a(n)*x^(2n+1)/(2n+1)! = BesselI(0, 2x)^2*x. - Michael Somos, Jun 22 2005
E.g.f.: x*(BesselI(0, 2*x))^2 = x+(2*x^3)/(U(0)-2*x^2); U(k) = (2*x^2)*(2*k+1) + (k+1)^3 - (2*x^2)*(2*k+3)*((k+1)^3)/U(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 23 2011
n^2*a(n) - 4*(2*n-1)*(2*n+1)*a(n-1) = 0. - R. J. Mathar, Sep 08 2013
O.g.f.: hypergeom([1/2, 3/2], [1], 16*x). - Peter Luschny, Oct 08 2015

A228329 a(n) = Sum_{k=0..n} (k+1)^2*T(n,k)^2 where T(n,k) is the Catalan triangle A039598.

Original entry on oeis.org

1, 8, 98, 1320, 18590, 268736, 3952228, 58837680, 883941750, 13373883600, 203487733020, 3110407163760, 47726453450988, 734694122886080, 11341161925265480, 175489379096245984, 2721169178975361702, 42273090191785999728, 657788911222324942060, 10250564041646388681200

Offset: 0

N. J. A. Sloane, Aug 26 2013

nonn

Let h(m) denote the sequence whose n-th term is Sum_{k=0..n} (k+1)^m*T(n,k)^2, where T(n,k) is the Catalan triangle A039598. This is h(2).

G. C. Greubel, Table of n, a(n) for n = 0..825
Pedro J. Miana and Natalia Romero, Moments of combinatorial and Catalan numbers, Journal of Number Theory, Volume 130, Issue 8, August 2010, Pages 1876-1887. See Remark 3 p. 1882. Omega2(n) = a(n-1).
Yidong Sun and Fei Ma, Four transformations on the Catalan triangle, arXiv:1305.2017 [math.CO], 2013.
Yidong Sun and Fei Ma, Some new binomial sums related to the Catalan triangle, Electronic Journal of Combinatorics 21(1) (2014), #P1.33.

Cf. A039598, A000108, A024492 (h(0)), A000894 (h(1)), A000515 (h(3)), A228330 (h(4)), A228331 (h(5)), A228332 (h(6)), A228333 (h(7)).

Cf. A000142, A007318, A051960.

B:=(n,k)->binomial(2*n, n-k) - binomial(2*n, n-k-2); #A039598
Omega:=(m,n)->add((k+1)^m*B(n,k)^2,k=0..n);
h:=m->[seq(Omega(m,n),n=0..20)];
h(2);
# Second solution:
h := n -> I*HeunG(8/5,0,-1/4,1/4,3/2,1/2,16*x)/sqrt(16*x-1);
seq(coeff(series(h(x),x,n+2),x,n),n=0..19); # Peter Luschny, Nov 26 2013

a[n_] := Binomial[4n, 2n] (3n+1)/(2n+1);
Table[a[n], {n, 0, 19}] (* Jean-François Alcover, Jul 30 2018, after Philippe Deléham *)

@CachedFunction
def A228329(n):
    return A228329(n-1)*(6*n+2)*(4*n-3)*(4*n-1)/(n*(2*n+1)*(3*n-2)) if n>0 else 1
[A228329(n) for n in (0..19)]  # Peter Luschny, Nov 26 2013

Conjecture: n*(2*n+1)*a(n) + 2*(-26*n^2+25*n-11)*a(n-1) + 20*(4*n-5)*(4*n-7)*a(n-2) = 0. - R. J. Mathar, Sep 08 2013
a(n) = ((4n)!*(3n+1))/((2n)!^2*(2n+1)) = binomial(4n,2n)*(3n+1)/(2n+1). - Philippe Deléham, Nov 25 2013
Therefore a(n) = A051960(2*n) / 2. - F. Chapoton, Jun 14 2024
From Peter Luschny, Nov 26 2013: (Start)
a(n) = 16^n*(3*n+1)*gamma(2*n+1/2)/(sqrt(Pi)*gamma(2*n+2)).
a(n) = a(n-1)*(6*n+2)*(4*n-3)*(4*n-1)/(n*(2*n+1)*(3*n-2)) if n > 0 else 1.
a(n) = [x^n] I*HeunG(8/5,0,-1/4,1/4,3/2,1/2,16*x)/sqrt(16*x-1) where [x^n] f(x) is the coefficient of x^n in f(x) and HeunG is the Heun general function. (End)

A228330 Let h(m) denote the sequence whose n-th term is Sum_{k=0..n} (k+1)^m*T(n,k)^2, where T(n,k) is the Catalan triangle A039598. This is h(4).

Original entry on oeis.org

1, 20, 362, 6504, 114686, 1992536, 34231540, 583027920, 9862508790, 165918037560, 2778642667020, 46358257249200, 770951008563372, 12785838603285104, 211540243555702376, 3492587812271418784, 57557091526140668070, 946970607665938615032, 15557339429900195819164, 255246113991506558429936

Offset: 0

N. J. A. Sloane, Aug 26 2013

nonn

G. C. Greubel, Table of n, a(n) for n = 0..825
Pedro J. Miana and Natalia Romero, Moments of combinatorial and Catalan numbers, Journal of Number Theory, Volume 130, Issue 8, August 2010, Pages 1876-1887. See Remark 3 p. 1882. Omega4(n) = a(n-1).
Yidong Sun and Fei Ma, Four transformations on the Catalan triangle, arXiv:1305.2017 [math.CO], 2013.
Yidong Sun and Fei Ma, Some new binomial sums related to the Catalan triangle, Electronic Journal of Combinatorics 21(1) (2014), #P1.33.

Cf. A000108, A039598, A024492 (h(0)), A000894 (h(1)), A228329 (h(2)), A000515 (h(3)), this sequence (h(4)), A228331 (h(5)), A228332 (h(6)), A228333 (h(7)).

List([0..20], n-> Binomial(4*n,2*n)*(15*n^3+15*n^2+n-1)/((2*n+1)*(4*n-1))); # G. C. Greubel, Mar 02 2019

[Binomial(4*n,2*n)*(15*n^3+15*n^2+n-1)/((2*n+1)*(4*n-1)): n in [0..20]]; // G. C. Greubel, Mar 02 2019

Table[4*Sum[(k+1)^6*(Binomial[2n+1, n-k]/(n+k+2))^2, {k,0,n}], {n,0,20}] (* Vaclav Kotesovec, Dec 08 2013 *)

vector(20, n, n--; binomial(4*n,2*n)*(15*n^3+15*n^2+n-1)/((2*n+1)*(4*n-1))) \\ G. C. Greubel, Mar 02 2019

[binomial(4*n,2*n)*(15*n^3+15*n^2+n-1)/((2*n+1)*(4*n-1)) for n in (0..20)] # G. C. Greubel, Mar 02 2019

Conjecture: n*(2*n+1)*(3467*n-4029)*a(n) + 8*(-36721*n^3 + 109040*n^2 - 137926*n + 69822)*a(n-1) + 4*(4*n-9)*(45706*n-7907)*(4*n-7)*a(n-2) = 0. - R. J. Mathar, Sep 08 2013
Recurrence: n*(2*n+1)*(15*n^3 - 30*n^2 + 16*n - 2)*a(n) = 2*(4*n-5)*(4*n-3)*(15*n^3 + 15*n^2 + n - 1)*a(n-1). - Vaclav Kotesovec, Dec 08 2013
From Vaclav Kotesovec, Dec 08 2013: (Start)
a(n) = binomial(4*n,2*n) * (15*n^3+15*n^2+n-1)/((2*n+1)*(4*n-1)).
a(n) = 4*Sum_{k=0..n} (k+1)^6*(binomial(2*n+1, n-k)/(n+k+2))^2. (End)

A228331 Let h(m) denote the sequence whose n-th term is Sum__{k=0..n} (k+1)^m*T(n,k)^2, where T(n,k) is the Catalan triangle A039598. This is h(5).

Original entry on oeis.org

1, 36, 780, 16240, 321300, 6131664, 114017904, 2079380160, 37356642180, 663144710800, 11657925495216, 203295462691776, 3521108298744400, 60632838691387200, 1038859802556120000, 17721669103065158400, 301147406355880764900, 5099997408534884394000, 86106549929771707182000

Offset: 0

N. J. A. Sloane, Aug 26 2013

nonn

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Pedro J. Miana and Natalia Romero, Moments of combinatorial and Catalan numbers, Journal of Number Theory, Volume 130, Issue 8, August 2010, Pages 1876-1887. See Omega5. Remark 3 p. 1882.
Yidong Sun and Fei Ma, Four transformations on the Catalan triangle, arXiv preprint arXiv:1305.2017, 2013
Yidong Sun and Fei Ma, Some new binomial sums related to the Catalan triangle, Electronic Journal of Combinatorics 21(1) (2014), #P1.33

Cf. A000108, A039598, A024492, A000894, A228329, A000515, A228330, A228332, A228333.

Table[Sum[(k+1)^5*(Binomial[2n+1, n-k]*2*(k+1)/(n+k+2))^2,{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Dec 08 2013 *)

Conjecture: n^2*a(n) +4*(2*n^2-22*n+11)*a(n-1) +16*(-7*n^2-54*n+166)*a(n-2) -1088*(2*n-3)*(2*n-7)*a(n-3)=0. - R. J. Mathar, Sep 08 2013
Recurrence: n^2*(3*n^2 - 5*n + 1)*a(n) = 4*(2*n-3)*(2*n+1)*(3*n^2 + n - 1)*a(n-1). - Vaclav Kotesovec, Dec 08 2013
a(n) = binomial(2*n,n)^2 * (2*n+1)*(3*n^2+n-1)/(2*n-1). - Vaclav Kotesovec, Dec 08 2013
G.f.: ((28*x+3)*hypergeom([1/2, 5/2],[1],16*x)+20*x*(1-16*x)*hypergeom([3/2, 7/2],[2],16*x))/3. - Mark van Hoeij, Apr 12 2014

A228332 Let h(m) denote the sequence whose n-th term is Sum_{k=0..n} (k+1)^m*T(n,k)^2, where T(n,k) is the Catalan triangle A039598. This is h(6).

Original entry on oeis.org

1, 68, 1778, 43080, 958430, 20119736, 405350788, 7921691280, 151231519350, 2834134359000, 52320693313020, 953960351550960, 17212782834351468, 307826474156801840, 5462948893700675720, 96303960593503261984, 1687752152779483045542, 29424712141610821296408, 510621541414656188646220

Offset: 0

N. J. A. Sloane, Aug 26 2013

nonn

Pedro J. Miana and Natalia Romero, Moments of combinatorial and Catalan numbers, Journal of Number Theory, Volume 130, Issue 8, August 2010, Pages 1876-1887. See Remark 3 p. 1882. Omega6(n) = a(n-1).
Yidong Sun and Fei Ma, Four transformations on the Catalan triangle, arXiv preprint arXiv:1305.2017 [math.CO], 2013.
Yidong Sun and Fei Ma, Some new binomial sums related to the Catalan triangle, Electronic Journal of Combinatorics 21(1) (2014), #P1.33.

Cf. A000108, A039598, A024492, A000894, A228329, A000515, A228330, A228331, A228333.

Table[Sum[(k+1)^6*(Binomial[2n+1, n-k]*2*(k+1)/(n+k+2))^2,{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Dec 08 2013 *)

Recurrence: n*(2*n+1)*(105*n^5 - 420*n^4 + 588*n^3 - 356*n^2 + 96*n - 10)*a(n) = 2*(4*n-7)*(4*n-5)*(105*n^5 + 105*n^4 - 42*n^3 - 62*n^2 - 7*n + 3)*a(n-1). - Vaclav Kotesovec, Dec 08 2013

a(n) = binomial(4*n,2*n) * (105*n^5 + 105*n^4 - 42*n^3 - 62*n^2 - 7*n + 3) / ((2*n+1)*(4*n-3)*(4*n-1)). - Vaclav Kotesovec, Dec 08 2013

cp's OEIS Frontend

A000515 a(n) = (2n)!(2n+1)!/n!^4, or equally (2n+1)*binomial(2n,n)^2.

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A228329 a(n) = Sum_{k=0..n} (k+1)^2*T(n,k)^2 where T(n,k) is the Catalan triangle A039598.

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A228330 Let h(m) denote the sequence whose n-th term is Sum_{k=0..n} (k+1)^m*T(n,k)^2, where T(n,k) is the Catalan triangle A039598. This is h(4).

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A228331 Let h(m) denote the sequence whose n-th term is Sum__{k=0..n} (k+1)^m*T(n,k)^2, where T(n,k) is the Catalan triangle A039598. This is h(5).

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A228332 Let h(m) denote the sequence whose n-th term is Sum_{k=0..n} (k+1)^m*T(n,k)^2, where T(n,k) is the Catalan triangle A039598. This is h(6).

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