A002894 -id:A002894 - cp's OEIS frontend

A186414 a(n) = binomial(2n,n)^3/(n+1)^2.

1, 2, 24, 500, 13720, 444528, 16099776, 631628712, 26317863000, 1149330319280, 52120705189696, 2437827529099872, 117006104720150464, 5740095404246000000, 286939169121965760000, 14579498741074214418000

Offset: 0

Emanuele Munarini, Feb 21 2011

Seiichi Manyama, Table of n, a(n) for n = 0..559

Cf. A000108, A002894, A000888, A002897, A186415.

[Binomial(2*n,n)^3/(n+1)^2: n in [0..50]]; // Vincenzo Librandi, Mar 27 2011

Table[Binomial[2n, n]^3/(n + 1)^2, {n, 0, 20}]

makelist(binomial(2*n,n)^3/(n+1)^2,n,0,40);

G.f.: 3F2({1/2, 1/2, 1/2}, {2, 2}, 64x), where 3F2 is a hypergeometric function.

A224734 G.f.: exp( Sum_{n>=1} binomial(2n,n)^2 x^n/n ).

Original entry on oeis.org

1, 4, 26, 216, 2075, 21916, 247326, 2930216, 36028117, 456089076, 5910983050, 78100285784, 1048696065394, 14275198859304, 196610207633100, 2735542102308752, 38400942393884068, 543307627503591440, 7740605626606127512, 110970838624540461472, 1599834676405793089013

Offset: 0

Paul D. Hanna, Apr 16 2013

nonn

The o.g.f. A(x) is the fourth power of the o.g.f. of A158266. - Peter Bala, Jun 04 2015

			G.f.: A(x) = 1 + 4*x + 26*x^2 + 216*x^3 + 2075*x^4 + 21916*x^5 + 247326*x^6 +...
where
log(A(x)) = 2^2*x + 6^2*x^2/2 + 20^2*x^3/3 + 70^2*x^4/4 + 252^2*x^5/5 + 924^2*x^6/6 + 3432^2*x^7/7 + 12870^2*x^8/8 +...+ A000984(n)^2*x^n/n +...

Vaclav Kotesovec, Table of n, a(n) for n = 0..800

Cf. A224732, A224735, A224736, A002894, A000984, A158266.

CoefficientList[Series[Exp[4*x*HypergeometricPFQ[{1, 1, 3/2, 3/2}, {2, 2, 2}, 16*x]], {x, 0, 20}], x] (* Vaclav Kotesovec, Mar 27 2025 *)

{a(n)=polcoeff(exp(sum(k=1,n,binomial(2*k,k)^2*x^k/k)+x*O(x^n)),n)}
for(n=0,20,print1(a(n),", "))

Logarithmic derivative yields A002894.

a(n) ~ c * 16^n / n^2, where c = 0.4942922... - Vaclav Kotesovec, Mar 27 2025

A241530 a(n) = binomial(n,floor(n/2))binomial(n+1,floor(n/2+1/2))(1+floor(n/2))/(1+2*floor(n/2)).

Original entry on oeis.org

1, 2, 4, 12, 36, 120, 400, 1400, 4900, 17640, 63504, 232848, 853776, 3171168, 11778624, 44169840, 165636900, 625739400, 2363904400, 8982836720, 34134779536, 130332794592, 497634306624, 1907598175392, 7312459672336, 28124844893600, 108172480360000

Offset: 0

Peter Luschny, Apr 25 2014

Alois P. Heinz, Table of n, a(n) for n = 0..1000

Cf. A000894, A002894, A005566.

Row n=3 of A275784.

A241530 := n -> binomial(n,iquo(n,2))*binomial(n+1,iquo(n+1,2))
*(1+iquo(n,2))/(1+2*iquo(n,2)); seq(A241530(n), n=0..26);
# second Maple program:
a:= proc(n) option remember; `if`(n<2, 2^n,
     ((8*n-4)*a(n-1)+16*(n-1)*(n-2)*a(n-2))/(n*(n+1)))
    end:
seq(a(n), n=0..30);  # Alois P. Heinz, Aug 10 2016

CoefficientList[Series[(-EllipticE[16 x^2] + (1 + 4 x) EllipticK[16 x^2])/(2Pi x), {x, 0, 20}], x] (* Benedict W. J. Irwin, Aug 15 2016 *)
Table[Binomial[n, #] Binomial[n + 1, Floor[(n + 1)/2]] (1 + #)/(1 + 2 #) &@ Floor[n/2], {n, 0, 26}] (* Michael De Vlieger, Aug 15 2016 *)

a(n) = ((8*n-4)*a(n-1)+16*(n-1)*(n-2)*a(n-2))/(n*(n+1)) for n>=2, a(n) = 2^n for n<2. - Alois P. Heinz, Apr 25 2014
G.f.: ((1+4*x)*K(4*x) - E(4*x))/(2*Pi*x), where K and E are the complete elliptic integrals of the first and second kind, respectively, with modulus k = 4*x. - Benedict W. J. Irwin, Aug 15 2016
From Wolfdieter Lang, Sep 06 2016 (Start):
The preceding g.f. can be rewritten as ((1+4*x)*F(1/2,1/2;1;(4*x)^2) -
   F(-1/2,1/2;1;(4*x)^2))/(4*x), where F is the hypergyometric function F(a,b;c;z).
This leads to the bisection a(2*k) = ((2*k)!)^2/k!^4 = A002894(k) and a(2*k+1) = 2*(2*k)!*(2*k+1)!/((k+1)*k!^4) = 2*A000894(k), for k >= 0.
(End)

A287317 Number of 5-dimensional cubic lattice walks that start and end at origin after 2n steps, free to pass through origin at intermediate stages.

Original entry on oeis.org

1, 10, 270, 10900, 551950, 32232060, 2070891900, 142317232200, 10277494548750, 770878551371500, 59577647564312020, 4717432065143561400, 381091087190569291900, 31308955091335405435000, 2609450031306515140215000, 220199552765301571338488400

Offset: 0

Peter Luschny, May 23 2017

Case k=5 of A287318.
Cf. A169714, A287316
1-4 dimensional analogs are A000984, A002894, A002896, A039699.

A287317_list := proc(len) series(BesselI(0, 2*sqrt(x))^5, x, len);
seq((2*i)!*coeff(%, x, i), i=0..len-1) end: A287317_list(16);

Table[SeriesCoefficient[BesselI[0, 2 Sqrt[x]]^5, {x, 0, n}] (2 n) !, {n, 0, 15}]
Table[Binomial[2n,n]^2 Sum[(Binomial[n,j]^4/Binomial[2n,2j]) HypergeometricPFQ[{-j,-j,-j}, {1,1/2-j}, 1/4], {j,0,n}], {n,0,15}]
Table[Sum[(2 n)!/(i! j! k! l! (n-i-j-k-l)!)^2, {i,0,n}, {j,0,n-i}, {k,0,n-i-j}, {l,0,n-i-j-k}], {n,0,30}] (* Shel Kaphan, Jan 24 2023 *)

a(n) = (2*n)! [x^n] BesselI(0, 2*sqrt(x))^5.
a(n) = binomial(2*n,n)*A169714(n).
a(n) ~ 2^(2*n) * 5^(2*n + 5/2) / (16 * Pi^(5/2) * n^(5/2)). - Vaclav Kotesovec, Nov 13 2017
a(n) = Sum_{i+j+k+l+m=n, 0<=i,j,k,l,m<=n} multinomial(2n, [i,i,j,j,k,k,l,l,m,m]). - Shel Kaphan, Jan 24 2023

Moved original definition to formula section and reworded definition descriptively similar to sequence A039699, by Dave R.M. Langers, Oct 12 2022

A318417 Scaled g.f. T(u) = Sum_{n>=0} a(n)(3u/48)^n satisfies 3(2u-1)T + d/du(4u(2u-1)(u-1)T') = 0, and a(0)=1; sequence gives a(n).

Original entry on oeis.org

1, 12, 228, 5040, 121380, 3093552, 82047504, 2240162496, 62508328740, 1773580002480, 50988042273168, 1481392081181376, 43413834762798864, 1281498837550545600, 38059165854011995200, 1136249610240102992640, 34076899109906247654180, 1026061759878805529676720

Offset: 0

Bradley Klee, Aug 26 2018

nonn

The defining differential equation determines the period function T(u) along a square-symmetric Hamiltonian family of algebraic plane curves, u=2*H=x^2+y^2-(1/2)*(x^4+y^4). The separatrix with u=1/2 is a product of two congruent ellipses. Transformation u->1-u leaves T(u) invariant.

			Singular Points of u=x^2+y^2-(1/2)*(x^4+y^4).
   u             (x,y)              type
=============================================
   0             (0,0)            circular
  1/2      (0,+/-1) (+/-1,0)     hyperbolic
   1       +/-(1,1) +/-(1,-1)     circular
G.f. = 1 + 12*x + 228*x^2 + 5040*x^3 + 121380*x^4 + 3093552*x^5 + 82047504*x^6 + ... - _Michael Somos_, Aug 27 2018

Bradley Klee, Contour and Period Graphs.

Factors: A000984, A098410. Quartic Periods: A113424, A002894, A318245.

RecurrenceTable[{n^2*a[n]-12*(2*n-1)^2*a[n-1] + 128*(2*n-3)*(2*n-1)*a[n-2] == 0, a[0] == 1, a[1] == 12}, a, {n,0,1000}]
PeriodIntegral[n_]:= CoefficientList[Series[1/Sqrt[1-2*x*((z+y)^4+(z-y)^4)], {x,0,n}], {x, y, z}][[#+1,2*#+1,2*#+1]]&/@Range[0,n];
PeriodIntegral[10]
HadamardProduct[n_]:=Times@@Map[CoefficientList[Normal[Series[#,{x,0,n}]],x,n+1]&, {1/Sqrt[1-12*x+32*x^2], 1/Sqrt[1-4*x]}];
HadamardProduct[10]
a[ n_] := Binomial[2 n, n] SeriesCoefficient[ (1 - 12 x + 32 x^2)^(-1/2), {x, 0, n}]; (* Michael Somos, Aug 27 2018 *)

{a(n) = if( n<0, 0, binomial(2*n, n) * polcoeff( (1 - 12*x + 32*x^2 + x * O(x^n))^(-1/2), n))}; /* Michael Somos, Aug 27 2018 */

Define t(u,z) = 1/sqrt(1-2*u*(cos(z)^4+sin(z)^4)), then certify that:
3*(2*u-1)*t + d/du(4*u*(2*u-1)*(u-1)*dt/du) = d/dz((1/4)*sin(4*z)*t^3).
G.f.: G(x) = T(48*x/3), T(u) = 1/(2*Pi)*Integral_{z=0..2*Pi} t(u)*dz.
n^2*a(n) - 12*(2*n-1)^2*a(n-1) + 128*(2*n-3)*(2*n-1)*a(n-2) = 0.
a(n) = A000984(n)*A098410(n).
0 = +a(n)*(+a(n+1)*(+134217728*a(n+2) -25165824*a(n+3) +819200*a(n+4)) +a(n+2)*(-6291456*a(n+2) +1998848*a(n+3) -78336*a(n+4)) +a(n+3)*(-29184*a(n+3) +1472*a(n+4))) +a(n+1)*(+a(n+1)*(-6291456*a(n+2) +1179648*a(n+3) -38400*a(n+4)) +a(n+2)*(+327680*a(n+2) -110592*a(n+3) +4448*a(n+4)) +a(n+3)*(+1728*a(n+3) -90*a(n+4))) +a(n+2)*(+a(n+2)*(-1536*a(n+2) +800*a(n+3) -36*a(n+4)) +a(n+3)*(-18*a(n+3) +a(n+4))) for all n in Z. - Michael Somos, Aug 27 2018
G.f.: hypergeom([1/2, 1/2],[1],16*x/(32*x-1))/sqrt(1-32*x). - Mark van Hoeij, Dec 13 2024

A329066 Square array T(n,k), n>=0, k>=0, read by antidiagonals, where T(n,k) is the constant term in the expansion of ( (Sum_{j=0..n} x^(2j+1)+1/x^(2j+1)) * (Sum_{j=0..n} y^(2j+1)+1/y^(2j+1)) - (Sum_{j=0..n-1} x^(2j+1)+1/x^(2j+1)) * (Sum_{j=0..n-1} y^(2j+1)+1/y^(2j+1)) )^(2*k).

Original entry on oeis.org

1, 4, 1, 36, 12, 1, 400, 588, 20, 1, 4900, 49440, 2100, 28, 1, 63504, 5187980, 423440, 4956, 36, 1, 853776, 597027312, 117234740, 1751680, 9540, 44, 1, 11778624, 71962945824, 36938855520, 907687900, 5101200, 16236, 52, 1

Offset: 0

Seiichi Manyama, Nov 03 2019

T(n,k) is the number of (2*k)-step closed paths (from origin to origin) in 2-dimensional lattice, using steps (t_1,t_2) (|t_1| + |t_2| = 2*n+1).

T(n,k) is the constant term in the expansion of (Sum_{j=0..2*n+1} (x^j + 1/x^j)*(y^(2*n+1-j) + 1/y^(2*n+1-j)) - x^(2*n+1) - 1/x^(2*n+1) - y^(2*n+1) - 1/y^(2*n+1))^(2*k).

			Square array begins:
   1,  4,   36,     400,       4900, ...
   1, 12,  588,   49440,    5187980, ...
   1, 20, 2100,  423440,  117234740, ...
   1, 28, 4956, 1751680,  907687900, ...
   1, 36, 9540, 5101200, 4190017860, ...

Seiichi Manyama, Antidiagonals n = 0..50, flattened
Wikipedia, Taxicab geometry.

Columns k=0-1 give A000012, A017113.
Rows n=0-2 give A002894, A329024, A329067.
Main diagonal gives A342964.
Cf. A005408, A328718, A329074, A329078.

{T(n, k) = polcoef(polcoef((sum(j=0, 2*n+1, (x^j+1/x^j)*(y^(2*n+1-j)+1/y^(2*n+1-j)))-x^(2*n+1)-1/x^(2*n+1)-y^(2*n+1)-1/y^(2*n+1))^(2*k), 0), 0)}

f(n) = (x^(2*n+2)-1/x^(2*n+2))/(x-1/x);
T(n, k) = sum(j=0, 2*k, (-1)^j*binomial(2*k, j)*polcoef(f(n)^j*f(n-1)^(2*k-j), 0)^2)

See the second code written in PARI.

A078817 Table by antidiagonals giving variants on Catalan sequence: T(n,k)=C(2n,n)C(2k,k)(2k+1)/(n+k+1).

Original entry on oeis.org

1, 3, 1, 10, 4, 2, 35, 15, 9, 5, 126, 56, 36, 24, 14, 462, 210, 140, 100, 70, 42, 1716, 792, 540, 400, 300, 216, 132, 6435, 3003, 2079, 1575, 1225, 945, 693, 429, 24310, 11440, 8008, 6160, 4900, 3920, 3080, 2288, 1430, 92378, 43758, 30888, 24024, 19404

Offset: 0

Henry Bottomley, Dec 07 2002

			Rows start:
     1,     3,    10,    35,   126,   462,  1716,
     1,     4,    15,    56,   210,   792,  3003,
     2,     9,    36,   140,   540,  2079,  8008,
     5,    24,   100,   400,  1575,  6160, 24024,
    14,    70,   300,  1225,  4900, 19404, 76440,
    42,   216,   945,  3920, 15876, 63504,252252,
   132,   693,  3080, 12936, 52920,213444,853776,
etc.

Ira Gessel, Super ballot numbers, J. Symbolic Computation 14 (1992), 179-194.
Isabel Cação, Helmuth R. Malonek, Maria Irene Falcão, and Graça Tomaz, Combinatorial Identities Associated with a Multidimensional Polynomial Sequence, J. Int. Seq., Vol. 21 (2018), Article 18.7.4.
Jovan Mikic, A Note on the Gessel Numbers, arXiv:2203.12931 [math.CO], 2022.

Columns include A000108 (catalan), A038629, A078818 and A078819. Rows include A001700, A001791, A007946 and A078820. Diagonals include A002894 and A060150.

Essentially a reflected version of A033820.

A078817 := proc(n,k)
    binomial(2*n,n)*binomial(2*k,k)*(2*k+1)/(n+k+1) ;
end proc: # R. J. Mathar, Dec 06 2018

T(n, k) = A000984(n)*A002457(k)/(n+k+1) = T(k, n)*(2k+1)/(2n+1).

A254408 a(n) = 2n^2binomial(2*n,n)^2, a closed form for a double binomial sum involving absolute values.

Original entry on oeis.org

0, 8, 288, 7200, 156800, 3175200, 61471872, 1154305152, 21201523200, 382952512800, 6826955907200, 120427502203008, 2105988385632768, 36562298361680000, 630861905459520000, 10827650254927680000, 184984389244186675200, 3147624998233113895200, 53368036302222346320000

Offset: 0

Jean-François Alcover, Jan 30 2015

G. C. Greubel, Table of n, a(n) for n = 0..500
Richard P. Brent, Hideyuki Ohtsuka, Judy-anne H. Osborn, and Helmut Prodinger, Some binomial sums involving absolute values, arXiv:1411.1477 [math.CO], 2014-2016.

Cf. A000108, A000984, A002894.

[(4*Binomial(n+1,2)*Catalan(n))^2/2: n in [0..30]]; // G. C. Greubel, Mar 31 2021

A254408:= n-> 2*( n*binomial(2*n, n) )^2; seq(A254408(n), n=0..30); # G. C. Greubel, Mar 31 2021

a[n_] := 2*n^2*Binomial[2*n, n]^2; Table[a[n], {n, 0, 20}]

a(n) = 2*n^2*binomial(2*n,n)^2 \\ Charles R Greathouse IV, May 10 2016

[(4*binomial(n+1,2)*catalan_number(n))^2/2 for n in (0..30)] # G. C. Greubel, Mar 31 2021

a(n) = Sum_{k=-n..n} (Sum_{l=-n..n} binomial(2*n, n+k)*binomial(2*n, n+l)*abs(k^2 - l^2)).
From G. C. Greubel, Mar 31 2021: (Start)
a(n) = 8 * binomial(n+1, 2)^2 * C(n)^2, where C(n) = A000108(n) (Catalan numbers).
G.f.: 8*x*Hypergeometric2F1([3/2, 3/2], [1], 16*x) = (16/Pi)*(x/(1-16*x)^2)*( 2*E(16*x) - (1-16*x)*K(16*x) ), where E(x) and K(x) are elliptic functions. (End)
D-finite with recurrence (n-1)^2*a(n) + (n^2-52*n+64)*a(n-1) - 68*(2*n -3)^2*a(n-2) = 0. - R. J. Mathar, Feb 27 2023
a(n) ~ 2^(4*n+1) * n / Pi. - Amiram Eldar, Sep 04 2025

A268148 A double binomial sum involving absolute values.

Original entry on oeis.org

0, 8, 768, 30720, 917504, 23592960, 553648128, 12213813248, 257698037760, 5257039970304, 104453604638720, 2031897488130048, 38843546786070528, 731834939447705600, 13618885273168379904, 250760427251989217280, 4574792530279968800768, 82788987402808467652608

Offset: 0

Richard P. Brent, Jan 27 2016

A fast algorithm follows from Theorem 5 of Brent et al. article.

Colin Barker, Table of n, a(n) for n = 0..800
Richard P. Brent, Hideyuki Ohtsuka, Judy-anne H. Osborn, Helmut Prodinger, Some binomial sums involving absolute values, arXiv:1411.1477v2 [math.CO], 2016.
Index entries for linear recurrences with constant coefficients, signature (48,-768,4096).

Cf. A000984, A002894, A166337, A254408.

a(n) = sum(k=-n,n, sum(l=-n,n, binomial(2*n, n+k)*binomial(2*n, n+l)*abs(k^2 - l^2)^2));

concat(0, Vec(8*x*(1+48*x)/(1-16*x)^3 + O(x^20))) \\ Colin Barker, Feb 11 2016

a(n) = Sum_{k=-n..n} (Sum_{l=-n..n} binomial(2*n, n+k)*binomial(2*n, n+l)*abs(k^2 - l^2)^2).
From Colin Barker, Feb 11 2016: (Start)
a(n) = 2^(4*n-1)*n*(2*n-1).
a(n) = 48*a(n-1)-768*a(n-2)+4096*a(n-3) for n>2.
G.f.: 8*x*(1+48*x) / (1-16*x)^3.
(End)

A329024 Constant term in the expansion of ((x^3 + x + 1/x + 1/x^3)(y^3 + y + 1/y + 1/y^3) - (x + 1/x)(y + 1/y))^(2*n).

Original entry on oeis.org

1, 12, 588, 49440, 5187980, 597027312, 71962945824, 8923789535232, 1128795397492620, 144940851928720848, 18832163401980525168, 2470451402766989534256, 326667449725835512275488, 43485599433527022301377600, 5821983056232777427055717760

Offset: 0

Seiichi Manyama, Nov 02 2019

nonn

Also number of (2*n)-step closed paths (from origin to origin) in 2-dimensional lattice, using steps (t_1,t_2) (|t_1| + |t_2| = 3).
         *
         |
      *-- --*
      |  |  |
   *-- -- -- --*
   |  |  |  |  |
*-- -- --P-- -- --*
   |  |  |  |  |
   *-- -- -- --*
      |  |  |
      *-- --*
         |
         *
Point P move to any position of * in the next step.

Seiichi Manyama, Table of n, a(n) for n = 0..400 (terms 0..185 from Vaclav Kotesovec)
Vaclav Kotesovec, Recurrence of order 4 (conjectured)

Row n=1 of A329066.

Cf. A002894, A094061, A254129.

{a(n) = polcoef(polcoef(((x^3+x+1/x+1/x^3)*(y^3+y+1/y+1/y^3)-(x+1/x)*(y+1/y))^(2*n), 0), 0)}

{a(n) = polcoef(polcoef((sum(k=0, 3, (x^k+1/x^k)*(y^(3-k)+1/y^(3-k)))-x^3-1/x^3-y^3-1/y^3)^(2*n), 0), 0)}

f(n) = (x^(2*n+2)-1/x^(2*n+2))/(x-1/x);
a(n) = sum(k=0, 2*n, (-1)^k*binomial(2*n, k)*polcoef(f(1)^k*f(0)^(2*n-k), 0)^2)

Conjecture: a(n) ~ 3 * 144^n / (19*Pi*n). - Vaclav Kotesovec, Nov 04 2019

cp's OEIS Frontend

A186414 a(n) = binomial(2n,n)^3/(n+1)^2.

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Mathematica

Maxima

Formula

A224734 G.f.: exp( Sum_{n>=1} binomial(2*n,n)^2 * x^n/n ).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A241530 a(n) = binomial(n,floor(n/2))*binomial(n+1,floor(n/2+1/2))*(1+floor(n/2))/(1+2*floor(n/2)).

Views

Author

Keywords

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A287317 Number of 5-dimensional cubic lattice walks that start and end at origin after 2n steps, free to pass through origin at intermediate stages.

Views

Author

Keywords

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions

A318417 Scaled g.f. T(u) = Sum_{n>=0} a(n)*(3*u/48)^n satisfies 3*(2*u-1)*T + d/du(4*u*(2*u-1)*(u-1)*T') = 0, and a(0)=1; sequence gives a(n).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A329066 Square array T(n,k), n>=0, k>=0, read by antidiagonals, where T(n,k) is the constant term in the expansion of ( (Sum_{j=0..n} x^(2*j+1)+1/x^(2*j+1)) * (Sum_{j=0..n} y^(2*j+1)+1/y^(2*j+1)) - (Sum_{j=0..n-1} x^(2*j+1)+1/x^(2*j+1)) * (Sum_{j=0..n-1} y^(2*j+1)+1/y^(2*j+1)) )^(2*k).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

PARI

Formula

A078817 Table by antidiagonals giving variants on Catalan sequence: T(n,k)=C(2n,n)*C(2k,k)*(2k+1)/(n+k+1).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

Formula

A254408 a(n) = 2*n^2*binomial(2*n,n)^2, a closed form for a double binomial sum involving absolute values.

Views

A224734 G.f.: exp( Sum_{n>=1} binomial(2n,n)^2 x^n/n ).

A241530 a(n) = binomial(n,floor(n/2))binomial(n+1,floor(n/2+1/2))(1+floor(n/2))/(1+2*floor(n/2)).

A318417 Scaled g.f. T(u) = Sum_{n>=0} a(n)(3u/48)^n satisfies 3(2u-1)T + d/du(4u(2u-1)(u-1)T') = 0, and a(0)=1; sequence gives a(n).

A329066 Square array T(n,k), n>=0, k>=0, read by antidiagonals, where T(n,k) is the constant term in the expansion of ( (Sum_{j=0..n} x^(2j+1)+1/x^(2j+1)) * (Sum_{j=0..n} y^(2j+1)+1/y^(2j+1)) - (Sum_{j=0..n-1} x^(2j+1)+1/x^(2j+1)) * (Sum_{j=0..n-1} y^(2j+1)+1/y^(2j+1)) )^(2*k).

A078817 Table by antidiagonals giving variants on Catalan sequence: T(n,k)=C(2n,n)C(2k,k)(2k+1)/(n+k+1).

A254408 a(n) = 2n^2binomial(2*n,n)^2, a closed form for a double binomial sum involving absolute values.

A329024 Constant term in the expansion of ((x^3 + x + 1/x + 1/x^3)(y^3 + y + 1/y + 1/y^3) - (x + 1/x)(y + 1/y))^(2*n).