A187358 -id:A187358 - cp's OEIS frontend

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

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

Offset: 0

Wolfdieter Lang

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

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

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Paul Barry, Generalized Catalan recurrences, Riordan arrays, elliptic curves, and orthogonal polynomials, arXiv:1910.00875 [math.CO], 2019.
Gi-Sang Cheon, S.-T. Jin, and L. W. Shapiro, A combinatorial equivalence relation for formal power series, Linear Algebra and its Applications, Vol. 491 (2016), pp. 123-137.
Greg Markowsky, A method for deriving hypergeometric and related identities from the H 2 Hardy norm of conformal maps, Integral Transforms and Special Functions, Vol. 24, No. 4 (2013), pp. 302-313; arXiv preprint, arXiv:1205.2458 [math.CV], 2012.
Khodabakhsh Hessami Pilehrood and Tatiana Hessami Pilehrood, Jacobi Polynomials and Congruences Involving Some Higher-Order Catalan Numbers and Binomial Coefficients, J. Int. Seq., Vol. 18 (2015), Article 15.11.7.
Louis Shapiro, Random walks with absorbing points, Discrete Applied Mathematics, Vol. 39 (1992), pp. 57-67.

Cf. A000007, A000108, A024492, A065097, A099250, A187357, A187358, A187359, A228411, A276483.

Cf. A000984, A078531, A182122, A245112, A245113.

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 *)

combinat::dyckWords::count(2*n) $ n = 0..28 // Zerinvary Lajos, Apr 14 2007

/* 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), ", "))

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

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

A024492 Catalan numbers with odd index: a(n) = binomial(4n+2, 2n+1)/(2*n+2).

Original entry on oeis.org

1, 5, 42, 429, 4862, 58786, 742900, 9694845, 129644790, 1767263190, 24466267020, 343059613650, 4861946401452, 69533550916004, 1002242216651368, 14544636039226909, 212336130412243110, 3116285494907301262, 45950804324621742364, 680425371729975800390

Offset: 0

Clark Kimberling

a(n) and Catalan(n) have the same 2-adic valuation (equal to 1 less than the sum of the digits in the binary representation of (n + 1)). In particular, a(n) is odd iff n is of the form 2^m - 1. - Peter Bala, Aug 02 2016

			sqrt((1/2)*(1+sqrt(1-x))) = 1 - (1/8)*x - (5/128)*x^2 - (42/2048)*x^3 - ...

Vincenzo Librandi, Table of n, a(n) for n = 0..100
Elżbieta Liszewska and Wojciech Młotkowski, Some relatives of the Catalan sequence, Advances in Applied Mathematics, Vol. 121 (2020), 102105; arXiv preprint, arXiv:1907.10725 [math.CO], 2019.
Pedro J. Miana and Natalia Romero, Moments of combinatorial and Catalan numbers, Journal of Number Theory, Vol. 130, No. 8 (August 2010), pp. 1876-1887. See Remark 3, p. 1882.

Cf. A048990 (Catalan numbers with even index), A024491, A000108, A000894.

Cf. A000260, A006013, A187357, A187358, A187359, A228329, A276484.

[Factorial(4*n+2)/(Factorial(2*n+1)*Factorial(2*n+2)): n in [0..20]]; // Vincenzo Librandi, Sep 13 2011

with(combstruct):bin := {B=Union(Z,Prod(B,B))}: seq (count([B,bin,unlabeled],size=2*n), n=1..18); # Zerinvary Lajos, Dec 05 2007
a := n -> binomial(4*n+1, 2*n+1)/(n+1):
seq(a(n), n=0..17); # Peter Luschny, May 30 2021

CoefficientList[ Series[1 + (HypergeometricPFQ[{3/4, 1, 5/4}, {3/2, 2}, 16 x] - 1), {x, 0, 17}], x]
CatalanNumber[Range[1,41,2]] (* Harvey P. Dale, Jul 25 2011 *)

a(n):=sum((k+1)^2*binomial(2*(n+1),n-k)^2,k,0,n)/(n+1)^2; /* Vladimir Kruchinin, Oct 14 2014 */

combinat::catalan(2*n+1)$ n = 0..24 // Zerinvary Lajos, Jul 02 2008

combinat::dyckWords::count(2*n+1)$ n = 0..24 // Zerinvary Lajos, Jul 02 2008

a(n)=binomial(4*n+2,2*n+1)/(2*n+2) \\ Charles R Greathouse IV, Sep 13 2011

G.f.: (1/2)*x^(-1)*(1-sqrt((1/2)*(1+sqrt(1-16*x)))).
G.f.: 3F2([3/4, 1, 5/4], [3/2, 2], 16*x). - Olivier Gérard, Feb 16 2011
a(n) = 4^n*binomial(2n+1/2, n)/(n+1). - Paul Barry, May 10 2005
a(n) = binomial(4n+1,2n+1)/(n+1). - Paul Barry, Nov 09 2006
a(n) = (1/(2*Pi))*Integral_{x=-2..2} (2+x)^(2*n)*sqrt((2-x)*(2+x)). - Peter Luschny, Sep 12 2011
D-finite with recurrence (n+1)*(2*n+1)*a(n) -2*(4*n-1)*(4*n+1)*a(n-1)=0. - R. J. Mathar, Nov 26 2012
G.f.: (c(sqrt(x)) - c(-sqrt(x)))/(2*sqrt(x)) = (2-(sqrt(1-4*sqrt(x)) + sqrt(1+4*sqrt(x))))/(4*x), with the g.f. c(x) of the Catalan numbers A000108. - Wolfdieter Lang, Feb 23 2014
a(n) = Sum_{k=0..n} (k+1)^2*binomial(2*(n+1),n-k)^2 /(n+1)^2. - Vladimir Kruchinin, Oct 14 2014
G.f.: A(x) = (1/x)*(inverse series of x - 5*x^2 + 8*x^3 - 4*x^4). - Vladimir Kruchinin, Oct 31 2014
a(n) ~ sqrt(2)*16^n/(sqrt(Pi)*n^(3/2)). - Ilya Gutkovskiy, Aug 02 2016
Sum_{n>=0} 1/a(n) = A276484. - Amiram Eldar, Nov 18 2020
G.f.: A(x) = C(4*x)*C(x*C(4*x)), where C(x) is the g.f. of A000108. - Alexander Burstein, May 01 2021
a(n) = (1/Pi)*16^(n+1)*Integral_{x=0..Pi/2} (cos x)^(4n+2)*(sin x)^2. - Greg Dresden, May 30 2021
Sum_{n>=0} a(n)/4^n = 2 - sqrt(2). - Amiram Eldar, Mar 16 2022
From Peter Bala, Feb 22 2023: (Start)
a(n) = (1/4^n) * Product_{1 <= i <= j <= 2*n} (i + j + 2)/(i + j - 1).
a(n) = Product_{1 <= i <= j <= 2*n} (3*i + j + 2)/(3*i + j - 1). Cf. A000260. (End)

More terms from Wolfdieter Lang

A187357 Catalan trisection: A000108(3*n), n >= 0.

Original entry on oeis.org

1, 5, 132, 4862, 208012, 9694845, 477638700, 24466267020, 1289904147324, 69533550916004, 3814986502092304, 212336130412243110, 11959798385860453492, 680425371729975800390, 39044429911904443959240, 2257117854077248073253720, 131327898242169365477991900, 7684785670514316385230816156, 451959718027953471447609509424

Offset: 0

Wolfdieter Lang, Mar 09 2011

Trisection of a sequence, given by its real o.g.f. G(x), is accomplished by
G(x) = G0(x^3) + x*G1(x^3) + (x^2)*G2(x^3), with the following solutions (using r := exp(2*Pi*i/3) = (-1 + sqrt(3)*i)/2):
G0(x) = (G(x^(1/3) + (G(r*x^(1/3)) + c.c.))/3,
G1(x) = (G(x^(1/3)) + ((1/r)*G(r*x^(1/3)) + c.c.))/(3*x^(1/3)),
G2(x) = (G(x^(1/3)) + (r*G(r*x^(1/3)) + c.c.))/(3*x^(2/3)),
where c.c. denotes the complex conjugate of the preceding expression.
See also the J. Arndt link, sect. 36.1.4,p.688: "Multisection by selecting terms with exponents s mod M", with M=3, where the o.g.f.s for the M-sected sequences with interspersed zeros are given for the general case.

Joerg Arndt, Fxtbook.

Cf. A000108, A024492, A048990, A187358 (C(3*n+1)), A187359 (C(3*n+2)/2), A208745.

Table[CatalanNumber[3*n], {n, 0, 20}] (* Amiram Eldar, Mar 16 2022 *)

a(n) = C(3*n), n >= 0, with C(n):= A000108(n) (Catalan).
O.g.f.: G0(x) = (sqrt(2*sqrt(1 + 4*x^(1/3) + 16*x^(2/3)) - (1 - 4*x^(1/3))) - sqrt(1 - 4*x^(1/3)))/(6*x^(1/3)).
From Ilya Gutkovskiy, Jan 13 2017: (Start)
E.g.f.: 3F3(1/6,1/2,5/6; 2/3,1,4/3; 64*x).
a(n) ~ 64^n/(3*sqrt(3*Pi)*n^(3/2)). (End)
D-finite with recurrence n*(3*n-1)*(3*n+1)*a(n) -8*(6*n-5)*(6*n-1)*(2*n-1)*a(n-1)=0. - R. J. Mathar, Feb 21 2020
Sum_{n>=0} a(n)/4^n = (4/3)^(3/4) (A208745). - Amiram Eldar, Mar 16 2022
a(n) = Product_{1 <= i <= j <= 3*n-1} (3*i + j + 2)/(3*i + j - 1). - Peter Bala, Feb 22 2023

A096304 Numbers k such that 3k does not divide (6k-4)!/((3k-2)!*(3k-1)!).

Original entry on oeis.org

1, 2, 3, 4, 5, 9, 10, 11, 12, 13, 14, 27, 28, 29, 30, 31, 32, 36, 37, 38, 39, 40, 41, 81, 82, 83, 84, 85, 86, 90, 91, 92, 93, 94, 95, 108, 109, 110, 111, 112, 113, 117, 118, 119, 120, 121, 122, 243, 244, 245, 246, 247, 248, 252, 253, 254, 255, 256, 257, 270, 271

Offset: 1

Ralf Stephan, Aug 03 2004

Equivalently, members of A019469 divisible by 3, divided by 3.
Ralf Stephan's formula is that terms k written in ternary have an arbitrary least significant digit and above that only 0's and 1's (per A340051). - Kevin Ryde, May 22 2021
{3a(n)-1:n>=1} is the set of positive integers k such that the k-th central binomial coefficient is not divisible by (k+1)*(2k-1). Such integers k are characterized by the following property: k is congruent to 2 (mod 3), and at least one of k-1, k+1 has no 2's in its base-3 expansion. - Valerio De Angelis, Aug 08 2022

Harvey P. Dale, Table of n, a(n) for n = 1..600
Kevin Ryde, Proof of Ralf Stephan's formula
Math Stackexchange, Factors of central binomial coefficient
Index entries for 3-automatic sequences.

Cf. A340051 (ternary digits), A005836, A019469, A187358.

Select[Range[300],Mod[(6#-4)!/((3#-2)!(3#-1)!),3#]!=0&] (* Harvey P. Dale, Jun 11 2019 *)

for(n=1,300,if(((6*n-4)!/(3*n-2)!/(3*n-1)!)%(3*n),print1(n",")))

a(n) = my(r);[n,r]=divrem(n,3); fromdigits(concat(binary(n),r), 3); \\ Kevin Ryde, May 22 2021

def A096304(n):
    a, b = divmod(n,3)
    return int(bin(a)[2:],3)*3+b # Chai Wah Wu, Jul 29 2025

a(n) = 9 * A005836(floor(n/6)) + (n mod 6) (conjectured) (confirmed, see links).

G.f.: x*(1+2*x)/(1-x^3) + 3/(1-x) * Sum_{i>=0} 3^i * x^(3*2^i) / (1 + x^(3*2^i)). - Kevin Ryde, May 22 2021

A187359 Catalan trisection: A000108(3*n + 2)/2, n>=0.

Original entry on oeis.org

1, 21, 715, 29393, 1337220, 64822395, 3282060210, 171529806825, 9183676536076, 501121108325684, 27767032438524099, 1558142747453650631, 88366931393503350700, 5056959295818949067010, 291650059796498346544020, 16934386878595523443214745, 989130828878080326811887228, 58078935727891217125276922940, 3426228463922436748774829232156, 202972497563788492865321721683556

Offset: 0

Wolfdieter Lang, Mar 09 2011

See the comment under A187357 for the o.g.f.s of the general trisection of a sequence.

The sequence C(3*n+2) starts as 2, 42, 1430, 58786, 2674440, 129644790, 6564120420, 343059613650, ...

Cf. A000108, A024492, A048990, A187357 (C(3*n)), A187358 (C(3*n+1)).

Table[CatalanNumber[3*n+2]/2, {n, 0, 20}] (* Amiram Eldar, Mar 16 2022 *)

a(n) = C(3*n+2)/2, n>=0, with C(n) = A000108(n).
O.g.f.: (3 - sqrt(1 - 4*x^(1/3)) - sqrt(2)*sqrt(sqrt(1 + 4*x^(1/3) + 16*x^(2/3)) +
  (1 + 2*x^(1/3))))/(12*x).
From Ilya Gutkovskiy, Jan 21 2017: (Start)
E.g.f.: 3F3(5/6,7/6,3/2; 4/3,5/3,2; 64*x).
a(n) ~ 8^(2*n+1)/(3*sqrt(3*Pi)*n^(3/2)). (End)
Sum_{n>=0} a(n)/4^n = 1 - sqrt(3+2*sqrt(3))/3. - Amiram Eldar, Mar 16 2022
a(n) = (1/2)*Product_{1 <= i <= j <= 3*n+1} (3*i + j + 2)/(3*i + j - 1). - Peter Bala, Feb 22 2023

cp's OEIS Frontend

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

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A024492 Catalan numbers with odd index: a(n) = binomial(4*n+2, 2*n+1)/(2*n+2).

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A187357 Catalan trisection: A000108(3*n), n >= 0.

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A096304 Numbers k such that 3k does not divide (6k-4)!/((3k-2)!*(3k-1)!).

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A187359 Catalan trisection: A000108(3*n + 2)/2, n>=0.

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

A024492 Catalan numbers with odd index: a(n) = binomial(4n+2, 2n+1)/(2*n+2).