cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-7 of 7 results.

A048990 Catalan numbers with even index (A000108(2*n), n >= 0): a(n) = binomial(4*n, 2*n)/(2*n+1).

Original entry on oeis.org

1, 2, 14, 132, 1430, 16796, 208012, 2674440, 35357670, 477638700, 6564120420, 91482563640, 1289904147324, 18367353072152, 263747951750360, 3814986502092304, 55534064877048198, 812944042149730764, 11959798385860453492, 176733862787006701400
Offset: 0

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Author

Keywords

Comments

With interpolated zeros, this is C(n)*(1+(-1)^n)/2 with g.f. given by 2/(sqrt(1+4x) + sqrt(1-4x)). - Paul Barry, Sep 09 2004
Self-convolution of a(n)/4^n gives Catalan numbers (A000108). - Vladimir Reshetnikov, Oct 10 2016
a(n) is the number of grand Dyck paths from (0,0) to (4n,0) that avoid vertices (2k,0) for all odd k > 0. - Alexander Burstein, May 11 2021
a(n) is the number of lattice paths from (0,0) to (2n,2n) with steps (1,0) and (0,1) that avoid the points (1,1), (3,3), (5,5), ..., (2n-1,2n-1). This is Example 2.5 of the Shapiro reference. - Lucas A. Brown, Jul 24 2025

Examples

			sqrt(2*x^-1*(1-sqrt(1-x))) = 1 + (1/8)*x + (7/128)*x^2 + (33/1024)*x^3 + ...
		

Crossrefs

Programs

  • Mathematica
    a[n_] := CatalanNumber[2n]; Array[a, 18, 0] (* Or *)
    CoefficientList[ Series[ Sqrt[2]/Sqrt[1 + Sqrt[1 - 16 x]], {x, 0, 17}], x] (* Robert G. Wilson v *)
    CatalanNumber[Range[0,40,2]] (* Harvey P. Dale, Mar 19 2015 *)
  • MuPAD
    combinat::dyckWords::count(2*n) $ n = 0..28 // Zerinvary Lajos, Apr 14 2007
    
  • PARI
    /* G.f.: A(x) = exp( x*A(x)^4 + Integral(A(x)^4 dx) ): */
    {a(n)=local(A=1+x); for(i=1, n, A=exp(x*A^4 + intformal(A^4 +x*O(x^n)))); polcoeff(A, n)} \\ Paul D. Hanna, Nov 09 2013
    for(n=0, 30, print1(a(n), ", "))
    
  • Sage
    A048990 = lambda n: hypergeometric([1-2*n,-2*n],[2],1)
    [Integer(A048990(n).n()) for n in range(20)] # Peter Luschny, Sep 22 2014

Formula

a(n) = 2 * A065097(n) - A000007(n).
G.f.: A(x) = sqrt((1/8)*x^(-1)*(1-sqrt(1-16*x))).
G.f.: 2F1( (1/4, 3/4); (3/2))(16*x). - Olivier Gérard Feb 17 2011
D-finite with recurrence n*(2*n+1)*a(n) - 2*(4*n-1)*(4*n-3)*a(n-1) = 0. - R. J. Mathar, Nov 30 2012
E.g.f: 2F2(1/4, 3/4; 1, 3/2; 16*x). - Vladimir Reshetnikov, Apr 24 2013
G.f. A(x) satisfies: A(x) = exp( x*A(x)^4 + Integral(A(x)^4 dx) ). - Paul D. Hanna, Nov 09 2013
G.f. A(x) satisfies: A(x) = sqrt(1 + 4*x*A(x)^4). - Paul D. Hanna, Nov 09 2013
a(n) = hypergeom([1-2*n,-2*n],[2],1). - Peter Luschny, Sep 22 2014
a(n) ~ 2^(4*n-3/2)/(sqrt(Pi)*n^(3/2)). - Ilya Gutkovskiy, Oct 10 2016
From Peter Bala, Feb 27 2020: (Start)
a(n) = (4^n)*binomial(2*n + 1/2, n)/(4*n + 1).
O.g.f.: A(x) = sqrt(c(4*x)), where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. of the Catalan numbers. Cf. A228411. (End)
Sum_{n>=0} 1/a(n) = A276483. - Amiram Eldar, Nov 18 2020
Sum_{n>=0} a(n)/4^n = sqrt(2). - Amiram Eldar, Mar 16 2022
From Peter Bala, Feb 22 2023: (Start)
a(n) = (1/2^(2*n-1)) * Product_{1 <= i <= j <= 2*n-1} (i + j + 2)/(i + j - 1) for n >= 1.
a(n) = Product_{1 <= i <= j <= 2*n-1} (3*i + j + 2)/(3*i + j - 1). Cf. A024492. (End)
a(n) = Sum_{k = 0..2*n-1} (-1)^k * 4^(2*n-k-1)*binomial(2*n-1, k)*Catalan(k+1). - Peter Bala, Apr 29 2024
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^6). - Seiichi Manyama, Jun 20 2025

A245114 G.f. A(x) satisfies A(x)^3 = 1 + 9*x*A(x)^5.

Original entry on oeis.org

1, 3, 36, 585, 10935, 221697, 4740120, 105225318, 2402040420, 56029889025, 1329627118248, 31998624800220, 779102941714461, 19157195459506230, 475034438632316400, 11865382635213387504, 298265217964573747095, 7539795161286074350785, 191548870595159091038640, 4888023169106780049244275
Offset: 0

Views

Author

Paul D. Hanna, Jul 31 2014

Keywords

Examples

			G.f.: A(x) = 1 + 3*x + 36*x^2 + 585*x^3 + 10935*x^4 + 221697*x^5 +...
where A(x)^3 = 1 + 9*x*A(x)^5:
A(x)^3 = 1 + 9*x + 135*x^2 + 2430*x^3 + 48195*x^4 + 1015740*x^5 +...
A(x)^5 = 1 + 15*x + 270*x^2 + 5355*x^3 + 112860*x^4 + 2480058*x^5 +...
		

Crossrefs

Programs

  • Maple
    rec:= 2*a(n+3)*(n+3)*(n+2)*(n+1)*(2*n+7)=135*a(n)*(5*n+1)*(5*n+4)*(5*n+7)*(5*n+13):
    f:= gfun:-rectoproc({rec,a(0)=1,a(1)=3,a(2)=36},a(n),remember):
    map(f, [$0..30]); # Robert Israel, Jan 30 2018
  • Mathematica
    nmax = 19; sol = {a[0] -> 1};
    Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x]^3 - (1 + 9 x A[x]^5) + O[x]^(n + 1), x] == 0 /. sol; sol = sol ~Join~ Solve[eq][[1]], {n, 1, nmax}];
    sol /. HoldPattern[a[n_] -> k_] :> Set[a[n], k];
    a /@ Range[0, nmax] (* Jean-François Alcover, Nov 01 2019 *)
  • PARI
    /* From A(x)^3 = 1 + 9*x*A(x)^5 : */
    {a(n) = local(A=1+x);for(i=1,n,A=(1 + 9*x*A^5 +x*O(x^n))^(1/3));polcoeff(A,n)}
    for(n=0,20,print1(a(n),", "))
    
  • PARI
    {a(n) = 9^n * binomial((5*n - 2)/3, n) / (2*n+1)}
    for(n=0,20,print1(a(n),", "))

Formula

a(n) = 9^n * binomial((5*n - 2)/3, n) / (2*n + 1).
2*a(n+3)*(n+3)*(n+2)*(n+1)*(2*n+7)=135*a(n)*(5*n+1)*(5*n+4)*(5*n+7)*(5*n+13). - Robert Israel, Jan 30 2018
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^7). - Seiichi Manyama, Jun 20 2025
a(n) ~ 3^n * 5^(5*n/3-1/6) / (sqrt(Pi) * 2^(2*(n+2)/3) * n^(3/2)). - Amiram Eldar, Sep 02 2025

A182122 Expansion of exp( arcsinh( 2*x ) ).

Original entry on oeis.org

1, 2, 2, 0, -2, 0, 4, 0, -10, 0, 28, 0, -84, 0, 264, 0, -858, 0, 2860, 0, -9724, 0, 33592, 0, -117572, 0, 416024, 0, -1485800, 0, 5348880, 0, -19389690, 0, 70715340, 0, -259289580, 0, 955277400, 0, -3534526380, 0, 13128240840, 0, -48932534040, 0, 182965127280
Offset: 0

Views

Author

Michael Somos, Apr 13 2012

Keywords

Examples

			G.f. = 1 + 2*x + 2*x^2 - 2*x^4 + 4*x^6 - 10*x^8 + 28*x^10 - 84*x^12 + ...
		

Crossrefs

Programs

  • Magma
    m:=50; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!(Exp(Argsinh(2*x)))); // G. C. Greubel, Aug 12 2018
  • Maple
    s := proc(n) option remember; `if`(n<2, n+1, -4*(n-2)*s(n-2)/(n+1)) end: A127846 := n -> `if`(n<2,n+1,s(n-1)); seq(A127846(n), n=0..47); # Peter Luschny, Sep 23 2014
  • Mathematica
    CoefficientList[Series[Exp[ArcSinh[2x]],{x,0,50}],x] (* Harvey P. Dale, Aug 18 2012 *)
    Table[2 HypergeometricPFQ[{-n+1,2-n},{2},-1],{n,0,46}] (* Peter Luschny, Sep 23 2014 *)
  • PARI
    {a(n) = if( n<2, (n>=0) + (n>0), n = n-2; if( n%2, 0, (-1)^(n/2) * 4 * binomial( n, n/2) / (n + 2)))};
    
  • PARI
    {a(n) = if( n<0, 0, polcoeff( sqrt( 1 + 4*x^2 + x*O(x^n) ) + 2*x, n ) )};
    
  • PARI
    {a(n) = my(A); if( n<0, 0, A = 1 + O(x); for( k=1, n, A = sqrt( 1 + 4*x * A)); polcoeff( A, n))};
    
  • Sage
    def A182122(n):
        if n < 2: return n+1
        if n % 2 == 1: return 0
        return (-1)^(n/2-1)*binomial(n,n/2)/(n-1)
    [A182122(n) for n in range(47)] # Peter Luschny, Sep 23 2014
    

Formula

G.f.: 2*x + sqrt( 1 + 4*x^2 ) = 1 / (1 - 2*x / (1 + x / (1 - x / (1 + x / ... )))).
The g.f. A(x) satisfies: A(x) = sqrt(1 + 4*x * A(x)).
a(n) = (-1)^n * A104624(n). Convolution inverse of A104624.
Conjecture : n*(n+1)*a(n) + (n+2)*(n-1)*a(n-1) +4*(n+1)*(n-3)*a(n-2) +4*(n+2)*(n-4)*a(n-3) = 0.- R. J. Mathar, Jul 24 2012
a(n) = 2*hypergeom([-n+1,2-n],[2],-1). - Peter Luschny, Sep 23 2014
0 = a(n)*(+16*a(n+2) + 10*a(n+4)) + a(n+2)*(-2*a(n+2) + a(n+4)) if n>=0. - Michael Somos, Jan 10 2017
a(n+4) = 2 * a(n+2) * (a(n+2) - 8*a(n)) / (a(n+2) + 10*a(n)) if n>=0 is even. - Michael Somos, Jan 10 2017
G.f. A(x) satisfies A(x) = 1/A(-x). - Seiichi Manyama, Jun 20 2025

A224884 Expansion of x / Series_Reversion(x*sqrt(1 + 4*x)).

Original entry on oeis.org

1, 2, -6, 32, -210, 1536, -12012, 98304, -831402, 7208960, -63740820, 572522496, -5209363380, 47915728896, -444799488600, 4161823309824, -39209074920090, 371626340253696, -3541117629057540, 33902753847705600, -325969196485349340, 3146175557067079680, -30471769822097981160
Offset: 0

Views

Author

Paul D. Hanna, Aug 21 2013

Keywords

Comments

Signed version of A206300. - Peter Bala, Mar 05 2020

Examples

			G.f.: A(x) = 1 + 2*x - 6*x^2 + 32*x^3 - 210*x^4 + 1536*x^5 - 12012*x^6 + ..
The coefficients in the powers A(x)^n of the g.f. begin:
n= 1: [1,  2,  -6,   32,  -210,  1536,-12012,  98304, -831402, ...];
n= 2: [1,  4,  -8,   40,  -256,  1848,-14336, 116688, -983040, ...];
n= 3: [1,  6,  -6,   32,  -210,  1536,-12012,  98304, -831402, ...];
n= 4: [1,  8,   0,   16,  -128,  1008, -8192,  68640, -589824, ...];
n= 5: [1, 10,  10,    0,   -50,   512, -4620,  40960, -364650, ...];
n= 6: [1, 12,  24,   -8,     0,   168, -2048,  20592, -196608, ...];
n= 7: [1, 14,  42,    0,    14,     0,  -588,   8192,  -90090, ...];
n= 8: [1, 16,  64,   32,     0,   -32,     0,   2112,  -32768, ...];
n= 9: [1, 18,  90,   96,   -18,     0,    84,      0,   -7722, ...];
n=10: [1, 20, 120,  200,     0,    24,     0,   -240,       0, ...];
n=11: [1, 22, 154,  352,   110,     0,   -44,      0,     726, ...];
n=12: [1, 24, 192,  560,   384,   -48,     0,     96,       0, ...];
n=13: [1, 26, 234,  832,   910,     0,    52,      0,    -234, ...];
n=14: [1, 28, 280, 1176,  1792,   392,     0,    -80,       0, ...];
n=15: [1, 30, 330, 1600,  3150,  1536,  -140,      0,     150, ...];
n=16: [1, 32, 384, 2112,  5120,  4032,     0,    128,       0, ...];
n=17: [1, 34, 442, 2720,  7854,  8704,  1428,      0,    -170, ...];
n=18: [1, 36, 504, 3432, 11520, 16632,  6144,   -432,       0, ...];
n=19: [1, 38, 570, 4256, 16302, 29184, 17556,      0,     342, ...];
n=20: [1, 40, 640, 5200, 22400, 48048, 40960,   5280,       0, ...]; ...
which illustrates the property [x^n] A(x)^(n+2*k) = 0 for k=1..n-1:
[x^2] A(x)^4 = 0;
[x^3] A(x)^5 = 0, [x^3] A(x)^7 = 0;
[x^4] A(x)^6 = 0, [x^4] A(x)^8 = 0, [x^4] A(x)^10 = 0; ...
[x^5] A(x)^7 = 0, [x^5] A(x)^9 = 0, [x^5] A(x)^11 = 0, [x^5] A(x)^13 = 0; ...
Related series:
sqrt(1+4*x) = 1 + 2*x - 2*x^2 + 4*x^3 - 10*x^4 + 28*x^5 - 84*x^6 + 264*x^7 - 858*x^8 + ... + (-1)^(n-1)*2*A000108(n-1)*x^n + ...
		

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[x/InverseSeries[Series[x*Sqrt[1+4*x],{x,0,20}],x],{x,0,20}],x] (* Vaclav Kotesovec, Aug 22 2013 *)
  • PARI
    {a(n)=polcoeff(x/serreverse(x*sqrt(1+4*x +x^2*O(x^n))),n)}
    for(n=0,25,print1(a(n),", "))

Formula

G.f. A(x) satisfies:
(1) A(x) = A(x)^3 - 4*x.
(2) A(x) = sqrt(1 + 4*x/A(x)).
(3) A(x*sqrt(1+4*x)) = sqrt(1+4*x).
(4) [x^n] A(x)^(n+2*k) = 0 for k=1..n-1, for n >= 2.
From Vaclav Kotesovec, Aug 22 2013: (Start)
a(n) = (-1)^(n+1) * 3^(3*n/2-1) * 4^(n-1) * GAMMA(n/2 - 1/6) * GAMMA(n/2 + 1/6)/(Pi*n!).
|a(n)| ~ 6^(n-1)*3^(n/2)/(sqrt(Pi/2)*n^(3/2)).
D-finite with recurrence: (n-1)*n*a(n) = 12*(3*n-7)*(3*n-5)*a(n-2). (End)
G.f.: (2/sqrt(3))*cosh(1/3*arccosh(sqrt(108)*x)). - Vladimir Kruchinin, Oct 11 2022
G.f. A(x) satisfies A(x) = 1/A(-x/A(x)^4). - Seiichi Manyama, Jun 20 2025

A245113 G.f. A(x) satisfies A(x)^2 = 1 + 4*x*A(x)^6.

Original entry on oeis.org

1, 2, 22, 340, 6118, 120060, 2492028, 53798888, 1195684230, 27175425004, 628705751828, 14756641134872, 350529497005532, 8410852483002200, 203561027031883320, 4963404936414528720, 121810229481173225670, 3006555636255509030220, 74585744314812449403300, 1858695101618327423328312
Offset: 0

Views

Author

Paul D. Hanna, Jul 31 2014

Keywords

Comments

Radius of convergence of g.f. A(x) is r = 1/27 where A(r) = sqrt(3/2).

Examples

			G.f.: A(x) = 1 + 2*x + 22*x^2 + 340*x^3 + 6118*x^4 + 120060*x^5 + ...
where A(x)^2 = 1 + 4*x*A(x)^6:
A(x)^2 = 1 + 4*x + 48*x^2 + 768*x^3 + 14080*x^4 + 279552*x^5 + ...
A(x)^6 = 1 + 12*x + 192*x^2 + 3520*x^3 + 69888*x^4 + 1462272*x^5 + ...
Related series:
A(x)^5 = 1 + 10*x + 150*x^2 + 2660*x^3 + 51750*x^4 + 1068012*x^5 + ...
A(x)^10 = 1 + 20*x + 400*x^2 + 8320*x^3 + 179200*x^4 + 3969024*x^5 + ...
where A(x) = sqrt(1 + 4*x^2*A(x)^10) + 2*x*A(x)^5.
		

Crossrefs

Programs

  • Maple
    A245113:=n->4^n*binomial((6*n-1)/2,n)/(4*n+1): seq(A245113(n), n=0..30); # Wesley Ivan Hurt, Aug 11 2015
  • Mathematica
    Table[4^n*Binomial[(6 n - 1)/2, n]/(4 n + 1), {n, 0, 20}] (* Wesley Ivan Hurt, Aug 11 2015 *)
  • PARI
    /* From A(x)^2 = 1 + 4*x*A(x)^6 : */
    {a(n) = local(A=1+x);for(i=1,n,A=sqrt(1 + 4*x*A^6 +x*O(x^n)));polcoeff(A,n)}
    for(n=0,20,print1(a(n),", "))
    
  • PARI
    {a(n) = 4^n * binomial((6*n - 1)/2, n) / (4*n + 1)}
    for(n=0,20,print1(a(n),", "))
    
  • PARI
    /* From A(x) = sqrt(1 + 4*x^2*A(x)^10) + 2*x*A(x)^5 : */
    {a(n) = local(A=1+x);for(i=1,n,A = sqrt(1 + 4*x^2*A^10 +x*O(x^n)) + 2*x*A^5);polcoeff(A,n)}
    for(n=0,20,print1(a(n),", "))

Formula

a(n) = 4^n * binomial((6*n - 1)/2, n) / (4*n + 1).
G.f. A(x) satisfies A(x) = sqrt(1 + 4*x^2*A(x)^10) + 2*x*A(x)^5.
G.f.: A(x) = sqrt(D(4*x)), where D(x) is the g.f. of A001764. - Werner Schulte, Aug 10 2015
From Karol A. Penson, Mar 19 2024: (Start)
a(n) = 4^n*binomial(3*n+1/2,n)/(6*n+1).
G.f.: 3F2([1/6, 1/2, 5/6], [3/4, 5/4], 27*x).
G.f.: sqrt(2)*sqrt((-(sqrt(1 - 27*x) + 3*i*sqrt(3)*sqrt(x))^(1/3) + (sqrt(1 - 27*x) - 3*i*sqrt(3)*sqrt(x))^(1/3))*i)*3^(3/4)/(6*x^(1/4)), where i is the imaginary unit.
a(n) = Integral_{x=0...27} x^n*W(x), where W(x) = h1(x) + h2(x) + h3(x) and
h1(x) = 2^(2/3)*3F2([-1/12, 1/6, 5/12], [1/3, 2/3], x/27)/(4*Pi*x^(5/6));
h2(x) = -3F2([1/4, 1/2, 3/4], [2/3, 4/3], x/27)/(12*Pi*sqrt(x));
h3(x) = -2^(1/3)*3F2([7/12, 5/6, 13/12], [4/3, 5/3], x/27)/(576*Pi*x^(1/6)).
This integral representation is unique as W(x) is the solution of the Hausdorff power moment problem on x = (0, 27). Using only the definition of a(n), W(x) can be proven to be positive. W(x) is singular at x = 0 and for x > 0 is monotonically decreasing to zero at x = 27. For x -> 27, W'(x) tends to -infinity. (End)
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^10). - Seiichi Manyama, Jun 20 2025
a(n) ~ 27^n / (4 * n^(3/2) * sqrt(2*Pi)). - Amiram Eldar, Sep 04 2025

A214553 G.f. A(x) satisfies A(x) = 1 + 4*x*A(x)^(5/2).

Original entry on oeis.org

1, 4, 40, 520, 7680, 122360, 2050048, 35600400, 635043840, 11566760920, 214221455360, 4021962900592, 76374500966400, 1464312851075760, 28307243610931200, 551140224522544160, 10797908842864705536, 212721273248318069400, 4211238736846158561280
Offset: 0

Views

Author

Paul D. Hanna, Jul 20 2012

Keywords

Comments

Radius of convergence of g.f. A(x) is r = (3/5)^(5/2) / 6 where A(r) = 5/3.
The two-parameter Fuss-Catalan sequence is A_n(p,r) := r*binomial(n*p + r, n)/(n*p + r). This sequence is 4^n*A_n(5/2,1). - Peter Bala, Oct 15 2015

Examples

			G.f.: A(x) = 1 + 4*x + 40*x^2 + 520*x^3 + 7680*x^4 + 122360*x^5 + 2050048*x^6 +... where A(x) = 1 + 4*x*A(x)^(5/2).
Radius of convergence: r = (3/5)^(5/2)/6 = 0.046475800...
Related expansions:
A(x)^(5/2) = 1 + 10*x + 130*x^2 + 1920*x^3 + 30590*x^4 + 512512*x^5 +...
A(x)^(1/2) = 1 + 2*x + 18*x^2 + 224*x^3 + 3230*x^4 + 50688*x^5 + 840420*x^6 + 14483456*x^7 + 256856886*x^8 +...
		

Crossrefs

Programs

  • Maple
    seq(4^n*binomial(5*n/2,n)/(3*n/2+1),n=0..50); # Robert Israel, Oct 18 2015
  • Mathematica
    m = 19; A[_] = 0;
    Do[A[x_] = 1 + 4*x*A[x]^(5/2) + O[x]^m, {m}];
    CoefficientList[A[x], x] (* Jean-François Alcover, Oct 20 2019 *)
  • PARI
    {a(n)=4^n*binomial(5*n/2,n)/(3*n/2+1)}
    
  • PARI
    {a(n)=local(A=1+x);for(i=1,n,A =1+4*x*(A+x*O(x^n))^(5/2));polcoeff(A,n)}
    for(n=0,30,print1(a(n),", "))

Formula

a(n) = 4^n * binomial(5*n/2, n) / (3*n/2 + 1).
From Peter Bala, Oct 13 2015: (Start)
O.g.f. A(x) satisfies A(x) = C(4*x*sqrt(A)), where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108.
Sqrt(A(x)) = 1/x * series reversion( x/sqrt(C(4*x)) ) is the o.g.f. for A245112. (End)
D-finite with recurrence 3*n*(n-1)*(3*n+2)*(3*n-2)*a(n) - 20*(5*n-4)*(5*n-8)*(5*n-2)*(5*n-6)*a(n-2) = 0. - R. J. Mathar, Nov 22 2024
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^4). - Seiichi Manyama, Jun 17 2025
a(n) ~ 2^(n+1/2) * 5^((5*n+1)/2) / (3^(3*(n+1)/2) * n^(3/2) * sqrt(Pi)). - Amiram Eldar, Sep 02 2025

A024491 a(n) = (1/(4n-1))*C(4n,2n).

Original entry on oeis.org

-1, 2, 10, 84, 858, 9724, 117572, 1485800, 19389690, 259289580, 3534526380, 48932534040, 686119227300, 9723892802904, 139067101832008, 2004484433302736, 29089272078453818, 424672260824486220, 6232570989814602524, 91901608649243484728, 1360850743459951600780
Offset: 0

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Examples

			sqrt(1/2*(1+sqrt(1-x))) = 1 - 1/8*x - 5/128*x^2 - 21/1024*x^3 - ...
		

Crossrefs

Programs

  • Magma
    [(1/(4*n-1))*Binomial(4*n,2*n) : n in [0..20]]; // Wesley Ivan Hurt, Jan 06 2024
  • Mathematica
    Table[1/(4n-1) Binomial[4n,2n],{n,0,20}] (* or *) With[{c=4Sqrt[x]}, CoefficientList[ Series[(-Sqrt[1-c]-Sqrt[1+c])/2,{x,0,30}],x]] (* Harvey P. Dale, Mar 10 2013 *)

Formula

G.f.: A(x) = -sqrt((1/2)*(1+sqrt(1-16*x))).
With interpolated zeros, this has g.f. -(sqrt(1-4x)+sqrt(1+4x))/2. - Paul Barry, Dec 23 2006
D-finite with recurrence n*(2*n-1)*a(n) - 2*(4*n-3)*(4*n-5)*a(n-1) = 0. - R. J. Mathar, Nov 13 2012
a(n) = A001448(n)/(4*n-1). - R. J. Mathar, Apr 27 2020
From Peter Bala, Apr 02 2023: (Start)
O.g.f. A(x) = - sqrt(1 - 4*x*C(4*x)), where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. of the Catalan numbers A000108.
The series reversion of -x*A(x) is equal to x * the o.g.f. of A245112. (End)
a(n) ~ 2^(4*n-5/2) / (n^(3/2) * sqrt(Pi)). - Amiram Eldar, Sep 04 2025

Extensions

More terms from Harvey P. Dale, Mar 10 2013
Showing 1-7 of 7 results.