A024491 -id:A024491 - cp's OEIS frontend

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

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

A241477 Triangle read by rows, number of orbitals classified with respect to the first zero crossing, n>=1, 1<=k<=n.

Original entry on oeis.org

1, 0, 2, 2, 2, 2, 0, 4, 0, 2, 6, 12, 4, 2, 6, 0, 12, 0, 4, 0, 4, 20, 60, 12, 12, 12, 4, 20, 0, 40, 0, 12, 0, 8, 0, 10, 70, 280, 40, 60, 36, 24, 40, 10, 70, 0, 140, 0, 40, 0, 24, 0, 20, 0, 28, 252, 1260, 140, 280, 120, 120, 120, 60, 140, 28, 252, 0, 504, 0

Offset: 1

Peter Luschny, Apr 23 2014

For the combinatorial definitions see A232500. An orbital w over n sectors has its first zero crossing at k if k is the smallest j such that the partial sum(1<=i<=j, w(i))) = 0, where w(i) are the jumps of the orbital represented by -1, 0, 1.

			[1], [ 1]
[2], [ 0,  2]
[3], [ 2,  2,  2]
[4], [ 0,  4,  0,  2]
[5], [ 6, 12,  4,  2,  6]
[6], [ 0, 12,  0,  4,  0, 4]
[7], [20, 60, 12, 12, 12, 4, 20]

Row sums: A056040.

Cf. A232500.

A241477 := proc(n, k)
  if n = 0 then 1
elif k = 0 then 0
elif irem(n, 2) = 0 and irem(k, 2) = 1 then 0
elif k = 1 then (n-1)!/iquo(n-1,2)!^2
else 2*(n-k)!*(k-2)!/iquo(k,2)/(iquo(k-2,2)!*iquo(n-k,2)!)^2
  fi end:
for n from 1 to 9 do seq(A241477(n, k), k=1..n) od;

T[n_, k_] := Which[n == 0, 1, k == 0, 0, Mod[n, 2] == 0 && Mod[k, 2] == 1,  0, k == 1, (n-1)!/Quotient[n-1, 2]!^2, True, 2*(n-k)!*(k-2)!/Quotient[k, 2]/(Quotient[k-2, 2]!*Quotient[n-k, 2]!)^2];
Table[T[n, k], {n, 1, 12}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jun 20 2018, from Maple *)

def A241477_row(n):
    if n == 0: return [1]
    Z = [0]*n; T = [0] if is_odd(n) else []
    for i in (1..n//2): T.append(-1); T.append(1)
    for p in Permutations(T):
        i = 0; s = p[0]
        while s != 0: i += 1; s += p[i];
        Z[i] += 1
    return Z
for n in (1..9): A241477_row(n)

If n is even and k is odd then T(n, k) = 0 else if k = 1 then T(n, 1) = A056040(n-1) else T(n, k) = 2*A057977(k-2)*A056040(n-k).
T(n, n) = A241543(n).
T(n+1, 1) = A126869(n).
T(2*n, 2*n) = |A002420(n)|.
T(2*n+1, 1) = A000984(n).
T(2*n+1, n+1) = A241530(n).
T(2*n+2, 2) = A028329(n).
T(4*n, 2*n) = |A010370(n)|.
T(4*n, 4*n) = |A024491(n)|.
T(4*n+1, 1) = A001448(n).
T(4*n+1, 2*n+1) = A002894(n).

A245112 G.f. A(x) satisfies A(x)^2 = 1 + 4xA(x)^5.

Original entry on oeis.org

1, 2, 18, 224, 3230, 50688, 840420, 14483456, 256856886, 4656988160, 85929839996, 1608379269120, 30463651429484, 582796191989760, 11245047027447240, 218581150665277440, 4276257634911525670, 84135742205488791552, 1663738200769421021580, 33047906167191995678720

Offset: 0

Paul D. Hanna, Jul 31 2014

Radius of convergence of g.f. A(x) is r = (3/5)^(5/2) / 6 where A(r) = sqrt(5/3).

			G.f.: A(x) =  = 1 + 2*x + 18*x^2 + 224*x^3 + 3230*x^4 + 50688*x^5 +...
where A(x)^2 = 1 + 4*x*A(x)^5:
A(x)^2 = 1 + 4*x + 40*x^2 + 520*x^3 + 7680*x^4 + 122360*x^5 +...
A(x)^5 = 1 + 10*x + 130*x^2 + 1920*x^3 + 30590*x^4 + 512512*x^5 +...
Related series:
A(x)^4 = 1 + 8*x + 96*x^2 + 1360*x^3 + 21120*x^4 + 347760*x^5 +...
A(x)^8 = 1 + 16*x + 256*x^2 + 4256*x^3 + 73216*x^4 + 1294560*x^5 +...
where A(x) = sqrt(1 + 4*x^2*A(x)^8) + 2*x*A(x)^4.

Gi-Sang Cheon, Sung-Tae Jin, and Louis W. Shapiro, A combinatorial equivalence relation for formal power series, Linear Algebra and its Applications, Volume 491, 15 February 2016, Pages 123-137.

Cf. A000108, A024491, A214553, A245113.

a[n_] := 4^n * Binomial[(5*n - 1)/2, n] / (3*n + 1); Array[a, 20, 0] (* Amiram Eldar, Sep 02 2025 *)

/* From A(x)^2 = 1 + 4*x*A(x)^5 : */
{a(n) = local(A=1+x);for(i=1,n,A=sqrt(1 + 4*x*A^5 +x*O(x^n)));polcoeff(A,n)}
for(n=0,20,print1(a(n),", "))

{a(n) = 4^n * binomial((5*n - 1)/2, n) / (3*n + 1)}
for(n=0,20,print1(a(n),", "))

/* From A(x) = sqrt(1 + 4*x^2*A(x)^8) + 2*x*A(x)^4 : */
{a(n) = local(A=1+x);for(i=1,n,A = sqrt(1 + 4*x^2*A^8 +x*O(x^n)) + 2*x*A^4);polcoeff(A,n)}
for(n=0,20,print1(a(n),", "))

a(n) = 4^n * binomial((5*n - 1)/2, n) / (3*n + 1).
G.f. A(x) satisfies: A(x) = sqrt(1 + 4*x^2*A(x)^8) + 2*x*A(x)^4.
Self convolution yields A214553.
G.f. A(x) = 1/x * series reversion of x*sqrt(1 - 4*x*C(4*x)), where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the generating function of the Catalan numbers A000108. See A024491. - Peter Bala, Mar 27 2023
D-finite with recurrence 3*n*(n-1)*(3*n-1)*(3*n+1)*a(n) - 20*(5*n-9)*(5*n-3)*(5*n-7)*(5*n-1)*a(n-2) = 0. - R. J. Mathar, Nov 22 2024
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^8). - Seiichi Manyama, Jun 20 2025
a(n) ~ 2^(n-1/2) * 5^(5*n/2) / (3^(3*n/2+1) * n^(3/2) * sqrt(Pi)). - Amiram Eldar, Sep 02 2025

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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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A241477 Triangle read by rows, number of orbitals classified with respect to the first zero crossing, n>=1, 1<=k<=n.

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A245112 G.f. A(x) satisfies A(x)^2 = 1 + 4*x*A(x)^5.

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

A245112 G.f. A(x) satisfies A(x)^2 = 1 + 4xA(x)^5.