0,2

This is also the result of applying the transformation on generating functions A -> 1/((1 - x)*(1 - x*A)) to the g.f. for the Catalan numbers.

p divides a(p) - 3 for prime p = 3 and p = {7, 13, 19, 31, 37, 43, ...} = A002476 (Primes of the form 6*n + 1). p^2 divides a(p^2) - 3 for prime p > 3. - Alexander Adamchuk, Jul 11 2006

Prime p divides a(p) for p = {2, 3, 5, 11, 17, 23, 29, 41, 47, 53, 59, 71, 83, 89, 101, ...} = A045309 (Primes congruent to {0, 2} mod 3); and A045309 (Primes p such that x^3 = n (integer) has only one solution mod p). Nonprime numbers n such that n divides a(n) are listed in A128287 = {1, 8, 133, ...}. - Alexander Adamchuk, Feb 23 2007

For p prime >= 5, a(p-1) = 1 or -2 (mod p) according as p = 1 or -1 (mod 3) (see Pan and Sun link). For example, with p=5, a(p-1) = 23 = -2 (mod p). - David Callan, Nov 29 2007

Hankel transform is A010892(n+1). - Paul Barry, Apr 24 2009

Equals INVERTi transform of A000245: (1, 3, 9, 28, ...). - Gary W. Adamson, May 15 2009

The subsequence of prime partial sums of Catalan numbers begins: a(1) = 2, a(4) = 23, a(6) = 197, a(16) = 48760367; see A121852. - Jonathan Vos Post, Feb 10 2010

Number of lattice paths from (0,0) to (n,n) which do not go above the diagonal x=y using steps (1,k), (k,1) with k >= 1 including two kinds of (1,1). - Alois P. Heinz, Oct 14 2015

Binomial transform of A086246(n+1) = [1, 1, 1, 2, 4, 9, ...], or, equivalently, of A001006 (Motzkin numbers) with 1 prepended.

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0 to 200 by T. D. Noe)

G. Alvarez, J. E. Bergner, and R. Lopez, Action Graphs and Catalan Numbers, J. Int. Seq. 18 (2015), 15.7.2.

Maciej Bendkowski and Pierre Lescanne, Combinatorics of explicit substitutions, arXiv:1804.03862 [cs.LO], 2018.

W. Chammam, F. Marcellán and R. Sfaxi, Orthogonal polynomials, Catalan numbers, and a general Hankel determinant evaluation, Linear Algebra Appl. 436(7) (2012), 2105-2116.

Dennis E. Davenport, Lara K. Pudwell, Louis W. Shapiro, and Leon C. Woodson, The Boundary of Ordered Trees, Journal of Integer Sequences, 18 (2015), Article 15.5.8.

Nancy S. S. Gu, Nelson Y. Li, and Toufik Mansour, 2-Binary trees: bijections and related issues, Discr. Math., 308 (2008), 1209-1221.

Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]

Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.

Ângela Mestre and José Agapito, A Family of Riordan Group Automorphisms, J. Int. Seq., Vol. 22 (2019), Article 19.8.5.

I. Pak, Partition identities and geometric bijections, Proc. Amer. Math. Soc. 132 (2004), 3457-3462.

Hao Pan and Zhi-Wei Sun, A combinatorial identity with application to Catalan numbers, arXiv:math/0509648 [math.CO], 2005-2006.

Murray Tannock, Equivalence classes of mesh patterns with a dominating pattern, MSc Thesis, Reykjavik Univ., May 2016. See Appendix B2.

Kevin Topley, Computationally Efficient Bounds for the Sum of Catalan Numbers, arXiv:1601.04223 [math.CO], 2016.

a(n) = A014138(n-1) + 1.

G.f.: (1 - (1 - 4*x)^(1/2))/(2*x*(1 - x)).

a(n) = Sum_{k=0..n} (2k)!/(k!)^2/(k+1). - Alexander Adamchuk, Jul 11 2006

D-finite with recurrence: (n+1)*a(n) + (1-5*n)*a(n-1) + 2*(2*n-1)*a(n-2) = 0. - R. J. Mathar, Dec 14 2011

Mathar's formula reduces to 2*(2*n-1)*C(n-1) = (n+1)*C(n), which is a known recurrence of the Catalan numbers, so the conjecture is true. - Peter J. Taylor, Mar 23 2015

Let C(n+1) = binomial(2*n+2,n+1)/(n+2) and H(n) = hypergeometric([1,n+3/2],[n+3],4) then A014137(n) = -(-1)^(2/3) - C(n+1)*H(n) and A014138(n) = -I^(2/3) - C(n+1)*H(n). - Peter Luschny, Aug 09 2012

G.f. (conjecture): Q(0)/(1-x), where Q(k)= 1 + (4*k + 1)*x/(k + 1 - 2*x*(k + 1)*(4*k + 3)/(2*x*(4*k + 3) + (2*k + 3)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 14 2013

a(n) ~ 2^(2*n + 2)/(3*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Dec 10 2013

0 =  a(n)*(16*a(n+1) - 26*a(n+2) + 10*a(n+3)) + a(n+1)*(-14*a(n+1) + 23*a(n+2) - 11*a(n+3)) + a(n+2)*(a(n+2) + a(n+3)) if n >= 0. - Michael Somos, Oct 24 2015

a(n) = (1 + A000108(n)*(3*(n+1)*hypergeom([1,-n], [1/2-n], 1/4) - 4*n - 2))/2. - Vladimir Reshetnikov, Oct 03 2016

G.f. A(x) satisfies: A(x) = 1 / (1 - x) + x * (1 - x) * A(x)^2. - Ilya Gutkovskiy, Jul 25 2021

G.f. = 1 + 2*x + 4*x^2 + 9*x^3 + 23*x^4 + 65*x^5 + 197*x^6 + 626*x^7 + 2056*x^8 + ...

a:= proc(n) option remember; `if`(n<2, n+1,

      ((5*n-1)*a(n-1)-(4*n-2)*a(n-2))/(n+1))

    end:

seq(a(n), n=0..30);  # Alois P. Heinz, May 18 2013

A014137List := proc(m) local A, P, n; A := [1]; P := [1];

for n from 1 to m - 2 do P := ListTools:-PartialSums([op(P), P[-n]]);

A := [op(A), P[-1]] od; A end: A014137List(30); # Peter Luschny, Mar 26 2022

Table[Sum[(2k)!/(k!)^2/(k+1), {k, 0, n}], {n, 0, 30}] (* Alexander Adamchuk, Jul 11 2006 *)

Accumulate[CatalanNumber[Range[0, 30]]] (* Harvey P. Dale, May 08 2012 *)

a[ n_] := SeriesCoefficient[ (1 - (1 - 4 x)^(1/2)) / (2 x (1 - x)), {x, 0, n}]; (* Michael Somos, Oct 24 2015 *)

Table[(1 + CatalanNumber[n] (3 (n + 1) Hypergeometric2F1[1, -n, 1/2 - n, 1/4] - 4 n - 2))/2, {n, 0, 20}] (* Vladimir Reshetnikov, Oct 03 2016 *)

(PARI) Vec((1-(1-4*x)^(1/2))/(2*x*(1-x))+O(x^99)) \\ Charles R Greathouse IV, Feb 11 2011

(PARI)

sm(v)={my(s=vector(#v)); s[1]=v[1]; for(n=2, #v, s[n]=v[n]+s[n-1]); s; }

C(n)=binomial(2*n, n)/(n+1);

sm(vector(66, n, C(n-1)))

/* Joerg Arndt, May 04 2013 */

(Python)

from __future__ import division

A014137_list, b, s = [], 1, 0

for n in range(10**2):

    s += b

    A014137_list.append(s)

    b = b*(4*n+2)//(n+2) # Chai Wah Wu, Jan 28 2016

(Sage)

def A014137():

    f, c, n = 1, 1, 1

    while True:

        yield f

        n += 1

        c = c * (4*n - 6) // n

        f = c + f

a = A014137()

print([next(a) for _ in range(29)]) # Peter Luschny, Nov 30 2016

Cf. A000108, A000245, A000984, A001246, A002476, A002897, A006134, A033536, A045309, A079727, A082894, A094638, A094639, A128287.

Sequence in context: A000083 A092668 A164039 * A245158 A245159 A245160

Adjacent sequences:  A014134 A014135 A014136 * A014138 A014139 A014140

nonn,nice

N. J. A. Sloane

approved