A187357 -id:A187357 - 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

A066802 a(n) = binomial(6n,3n).

Original entry on oeis.org

1, 20, 924, 48620, 2704156, 155117520, 9075135300, 538257874440, 32247603683100, 1946939425648112, 118264581564861424, 7219428434016265740, 442512540276836779204, 27217014869199032015600, 1678910486211891090247320, 103827421287553411369671120, 6435067013866298908421603100

Offset: 0

Benoit Cloitre, Jan 18 2002

For the trisection of a sequence (here A000984) given by its real o.g.f. see a comment and a reference under A187357.

Harry J. Smith, Table of n, a(n) for n = 0..100
R. J. Mathar, The Eggenberger-Polya urn process: Probabilities of revisited ball ratios, vixra:2502.0097 (2025) Table 1

Cf. A001450, A066798, A187364 (binomial(2(3n+1),3n+1)/2), A187365 (binomial(2(3n+2),3n+2)/3!).

[Binomial(6*n, 3*n): n in [0..15]]; // G. C. Greubel, Feb 17 2020

a := n -> hypergeom([-3*n, -3*n], [1], 1):
seq(simplify(a(n)), n=0..13); # Peter Luschny, Mar 19 2018

Table[Binomial[6n, 3n], {n,0,13}] (* Jean-François Alcover, Jun 03 2019 *)

a(n) = { binomial(6*n, 3*n) } \\ Harry J. Smith, Mar 28 2010

[binomial(6*n, 3*n) for n in (0..15)] # G. C. Greubel, Feb 17 2020

a(n) = A000984(3*n).
a(n) = Sum_{i=0..n} Sum_{j=0..n} Sum_{k=0..n} binomial(n, i)*binomial(n, j) *binomial(n, k)*binomial(3n, i+j+k). - Benoit Cloitre, Mar 08 2005
O.g.f. (with a(0):=1): (cb(x^(1/3)) + sqrt(2)*P(x^(1/3))*sqrt(1/P(x^(1/3))+1+2*x^(1/3)))/3, with cb(x):=1/sqrt(1-4*x) (o.g.f. of A000984) and P(x):=P(-1/2,4*x) = 1/sqrt(1+4*x+16*x^2) (o.g.f. of A116091, with P(x,z) the o.g.f. of the Legendre polynomials). - Wolfdieter Lang, Mar 24 2011
D-finite with recurrence n*(3n-1)*(3n-2)*a(n) = 8*(6n-5)*(6n-1)*(2n-1)*a(n-1). - R. J. Mathar, Sep 17 2012
a(n) = GegenbauerC(3*n, -3*n, -1). - Peter Luschny, May 07 2016
a(n) = hypergeom([-3*n, -3*n], [1], 1). - Peter Luschny, Mar 19 2018
a(n) ~ 2^(6*n)/sqrt(3*Pi*n). - Vaclav Kotesovec, Jun 07 2019
From Peter Bala, Feb 16 2020: (Start)
a(m*p^k) == a(m*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers m and k.
a(n) = [(x*y)^(3*n)] (1 + x + y)^(6*n). Cf. A001448. (End)
Conjecture: a(n) = [x^n] G(x)^(2*n), where G(x) = (1 + x)*(1 - 6*x + x^2)/(2*x) + (x^2 - 1)*sqrt(1 - 14*x + x^2)/(2*x) = 1 + 10*x + 81*x^2 + 720*x^3 + .... The algebraic function G(x) satisfies the quadratic equation x*G(x)^2 - (1 - 5*x - 5*x^2 + x^3)*G(x) + (1 + x)^4 = 0. Cf. A001450. - Peter Bala, Oct 27 2022
a(n) = Sum_{k = 0..3*n} binomial(3*n+k-1, k). - Peter Bala, Jun 04 2024
O.g.f: 3F2(1/6,1/2,5/6; 1/3,2/3 ; 64*x). - R. J. Mathar, Jan 11 2025

Extended to a(0)=1 by M. F. Hasler, Oct 06 2014

A187365 Trisection of A000984 (central binomial coefficients): binomial(2(3n+2),3n+2)/3!, n>=0.

Original entry on oeis.org

1, 42, 2145, 117572, 6686100, 388934370, 22974421470, 1372238454600, 82653088824684, 5011211083256840, 305437356823765089, 18697712969443807572, 1148770108115543559100, 70797430141465286938140, 4374750896947475198160300, 270950190057528375091435920

Offset: 0

Wolfdieter Lang, Mar 10 2011

See a comment under A187357 concerning trisection.

This appears also in the trisection of A001700: binomial(2*(3*n+1)+1,(3*n+1)+1)/3.

Seiichi Manyama, Table of n, a(n) for n = 0..554

Cf. A066802 binomial(6n,3n), A187364 binomial(2*(3n+1),3n+1)/2, A002458, A100033.

a(n)=binomial(2*(3*n+2),3*n+2)/3!, n>=0.
a(n)=binomial(3*(2*n+1),3*n+2)/3, n>=0.
O.g.f.:(cb(x^(1/3)) - sqrt(2)*P(x^(1/3))*sqrt(1/P(x^(1/3))-(1-4*x^(1/3))/2))/(18*x^(2/3)),
with cb(x):=1/sqrt(1-4*x) (o.g.f. of A000984) and P(x):=P(-1/2,4*x)=1/sqrt(1+4*x+16*x^2) (o.g.f. of A116091, with P(x,z)the o.g.f. of the Legendre polynomials).
From Peter Bala, Mar 19 2023: (Start)
a(n) = (1/6)*Sum_{k = 0..3*n+2} binomial(3*n+2,k)^2.
a(n) = (1/6)*hypergeom([-2 - 3*n, -2 - 3*n], [1], 1).
a(n) = 8*(2*n + 1)*(6*n + 1)*(6*n - 1)/(n*(3*n + 1)*(3*n + 2)) * a(n-1). (End)

A187358 Catalan trisection: A000108(3*n+1), n>=0.

Original entry on oeis.org

1, 14, 429, 16796, 742900, 35357670, 1767263190, 91482563640, 4861946401452, 263747951750360, 14544636039226909, 812944042149730764, 45950804324621742364, 2622127042276492108820, 150853479205085351660700, 8740328711533173390046320, 509552245179617138054608572, 29869166945772625950142417512

Offset: 0

Wolfdieter Lang, Mar 09 2011

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

Cf. A000108, A024492, A048990, A187357 (C(3*n)), A187359 (C(3*n+2)).

Table[CatalanNumber[3*n+1], {n, 0, 20}] (* Amiram Eldar, Mar 16 2022 *)

a(n) = C(3*n+1), n>=0, with C(n) = A000108(n) (Catalan).
O.g.f.: (sqrt(2*sqrt(1+4*x^(1/3)+16*x^(2/3))-(1+8*x^(1/3))) - sqrt(1-4*x^(1/3)))/(6*x^(2/3)).
From Ilya Gutkovskiy, Jan 13 2017: (Start)
E.g.f.: 3F3(1/2,5/6,7/6; 1,4/3,5/3; 64*x).
a(n) ~ 4^(3*n+1)/(3*sqrt(3*Pi)*n^(3/2)). (End)
Sum_{n>=0} a(n)/4^n = 2*sqrt(2*sqrt(3) - 3)/3. - Amiram Eldar, Mar 16 2022
a(n) = Product_{1 <= i <= j <= 3*n} (3*i + j + 2)/(3*i + j - 1). - Peter Bala, Feb 22 2023

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

A235534 a(n) = binomial(6n, 2n) / (4*n + 1).

Original entry on oeis.org

1, 3, 55, 1428, 43263, 1430715, 50067108, 1822766520, 68328754959, 2619631042665, 102240109897695, 4048514844039120, 162250238001816900, 6568517413771094628, 268225186597703313816, 11034966795189838872624, 456949965738717944767791

Offset: 0

Bruno Berselli, Jan 12 2014

This is the case l=4, k=2 of binomial((l+k)*n,k*n)/((l*n+1)/gcd(k,l*n+1)), see Theorem 1.1 in Zhi-Wei Sun's paper.

First bisection of A001764.

Zhi-Wei Sun, On Divisibility Of Binomial Coefficients, Journal of the Australian Mathematical Society 93 (2012), p. 189-201.

Cf. similar sequences generated by binomial((l+k)*n,k*n)/(l*n+1), where l is divisible by all the factors of k: A000108 (l=1, k=1), A001764 (l=2, k=1), A002293 (l=3, k=1), A002294 (l=4, k=1), A002295 (l=5, k=1), A002296 (l=6, k=1), A007556 (l=7, k=1), A062994 (l=8, k=1), A059968 (l=9, k=1), A230388 (l=10, k=1), A048990 (l=2, k=2), this sequence (l=4, k=2), A235536 (l=6, k=2), A187357 (l=3, k=3), A235535 (l=6, k=3).

l:=4; k:=2; [Binomial((l+k)*n,k*n)/(l*n+1): n in [0..20]]; /* where l is divisible by all the prime factors of k */

Table[Binomial[6 n, 2 n]/(4 n + 1), {n, 0, 20}]

a(n) = A047749(4*n-2) for n>0.
From Ilya Gutkovskiy, Jun 21 2018: (Start)
G.f.: 4F3(1/6,1/3,2/3,5/6; 1/2,3/4,5/4; 729*x/16).
a(n) ~ 3^(6*n+1/2)/(sqrt(Pi)*2^(4*n+7/2)*n^(3/2)). (End)

A235535 a(n) = binomial(9n, 3n) / (6*n + 1).

Original entry on oeis.org

1, 12, 1428, 246675, 50067108, 11124755664, 2619631042665, 642312451217745, 162250238001816900, 41932353590942745504, 11034966795189838872624, 2946924270225408943665279, 796607831560617902288322405, 217550867863011281855594752680

Offset: 0

Bruno Berselli, Jan 12 2014

This is the case l=6, k=3 of binomial((l+k)*n,k*n)/((l*n+1)/gcd(k,l*n+1)), see Theorem 1.1 in Zhi-Wei Sun's paper.

Also, the sequence follows A002296 and A235536, namely binomial(7*n,n)/(6*n+1) and binomial(8*n,2*n)/(6*n+1); naturally, even binomial(10*n,4*n)/(6*n+1) is always integer.

Robert Israel, Table of n, a(n) for n = 0..403
Zhi-Wei Sun, On Divisibility Of Binomial Coefficients, Journal of the Australian Mathematical Society 93 (2012), p. 189-201.

Cf. similar sequences generated by binomial((l+k)*n,k*n)/(l*n+1), where l is divisible by all the factors of k: A000108 (l=1, k=1), A001764 (l=2, k=1), A002293 (l=3, k=1), A002294 (l=4, k=1), A002295 (l=5, k=1), A002296 (l=6, k=1), A007556 (l=7, k=1), A062994 (l=8, k=1), A059968 (l=9, k=1), A230388 (l=10, k=1), A048990 (l=2, k=2), A235534 (l=4, k=2), A235536 (l=6, k=2), A187357 (l=3, k=3), this sequence (l=6, k=3).

l:=6; k:=3; [Binomial((l+k)*n,k*n)/(l*n+1): n in [0..20]]; /* here l is divisible by all the prime factors of k */

seq(binomial(9*n,3*n)/(6*n+1), n=0..30); # Robert Israel, Feb 15 2021

Table[Binomial[9 n, 3 n]/(6 n + 1), {n, 0, 20}]

a(n) = A001764(3*n) = A047749(6*n).
From Ilya Gutkovskiy, Jun 21 2018: (Start)
G.f.: 6F5(1/9,2/9,4/9,5/9,7/9,8/9; 1/3,1/2,2/3,5/6,7/6; 19683*x/64).
a(n) ~ 3^(9*n-1)/(sqrt(Pi)*4^(3*n+1)*n^(3/2)). (End)
D-finite with recurrence 8*(6*n + 5)*(2*n + 1)*(n + 1)*(3*n + 2)*(3*n + 1)*(6*n + 7)*a(n + 1) = 3*(9*n + 8)*(9*n + 7)*(9*n + 5)*(9*n + 4)*(9*n + 2)*(9*n + 1)*a(n). - Robert Israel, Feb 15 2021

A235536 a(n) = binomial(8n, 2n) / (6*n + 1).

Original entry on oeis.org

1, 4, 140, 7084, 420732, 27343888, 1882933364, 134993766600, 9969937491420, 753310723010608, 57956002331347120, 4524678117939182220, 357557785658996609700, 28545588568201512137904, 2298872717007844035521848, 186533392975795702301759056

Offset: 0

Bruno Berselli, Jan 12 2014

This is the case l=6, k=2 of binomial((l+k)*n,k*n)/((l*n+1)/gcd(k,l*n+1)), see Theorem 1.1 in Zhi-Wei Sun's paper.
First bisection of A002293.
Also, the sequence is between A002296 and A235535.

Zhi-Wei Sun, On Divisibility Of Binomial Coefficients, Journal of the Australian Mathematical Society 93 (2012), p. 189-201.

Cf. similar sequences generated by binomial((l+k)*n,k*n)/(l*n+1), where l is divisible by all the factors of k: A000108 (l=1, k=1), A001764 (l=2, k=1), A002293 (l=3, k=1), A002294 (l=4, k=1), A002295 (l=5, k=1), A002296 (l=6, k=1), A007556 (l=7, k=1), A062994 (l=8, k=1), A059968 (l=9, k=1), A230388 (l=10, k=1), A048990 (l=2, k=2), A235534 (l=4, k=2), this sequence (l=6, k=2), A187357 (l=3, k=3), A235535 (l=6, k=3).

l:=6; k:=2; [Binomial((l+k)*n,k*n)/(l*n+1): n in [0..20]]; /* where l is divisible by all the prime factors of k */

Table[Binomial[8 n, 2 n]/(6 n + 1), {n, 0, 20}]

a(n) = A124753(6*n).
From Ilya Gutkovskiy, Jun 21 2018: (Start)
G.f.: 6F5(1/8,1/4,3/8,5/8,3/4,7/8; 1/3,1/2,2/3,5/6,7/6; 65536*x/729).
a(n) ~ 2^(16*n-1)/(sqrt(Pi)*3^(6*n+3/2)*n^(3/2)). (End)

A187361 Pell trisection: Pell(3*n+1), n >= 0.

Original entry on oeis.org

1, 12, 169, 2378, 33461, 470832, 6625109, 93222358, 1311738121, 18457556052, 259717522849, 3654502875938, 51422757785981, 723573111879672, 10181446324101389, 143263821649299118, 2015874949414289041, 28365513113449345692, 399133058537705128729, 5616228332641321147898

Offset: 0

Wolfdieter Lang, Mar 09 2011

For the general computation of the o.g.f.s for the trisection of a sequence, given by its real o.g.f., see a Wolfdieter Lang comment under A187357.

Colin Barker, Table of n, a(n) for n = 0..850
Index entries for linear recurrences with constant coefficients, signature (14,1).

Cf. A142588 (Pell(3*n)), A187362 (Pell(3*n+2)).

Table[Fibonacci[3n + 1, 2], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 16 2016 *)
LinearRecurrence[{14,1},{1,12},20] (* Harvey P. Dale, Jul 06 2023 *)

Vec((1-2*x)/(1-14*x-x^2) + O(x^20)) \\ Colin Barker, Jan 25 2016

a(n) = Pell(3*n+1), n >= 0, with Pell(n):=A000129(n).
O.g.f.: (1-2*x)/(1-14*x-x^2).
a(n) = 14*a(n-1) + a(n-2), a(0)= 1, a(1)=12.
a(n) = (((7-5*sqrt(2))^n*(-1+sqrt(2))+(1+sqrt(2))*(7+5*sqrt(2))^n))/(2*sqrt(2)). - Colin Barker, Jan 25 2016

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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A066802 a(n) = binomial(6*n,3*n).

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A187365 Trisection of A000984 (central binomial coefficients): binomial(2(3n+2),3n+2)/3!, n>=0.

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A187358 Catalan trisection: A000108(3*n+1), n>=0.

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

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A235534 a(n) = binomial(6*n, 2*n) / (4*n + 1).

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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).

A066802 a(n) = binomial(6n,3n).

A235534 a(n) = binomial(6n, 2n) / (4*n + 1).

A235535 a(n) = binomial(9n, 3n) / (6*n + 1).

A235536 a(n) = binomial(8n, 2n) / (6*n + 1).