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A014137 Partial sums of Catalan numbers (A000108). 286
1, 2, 4, 9, 23, 65, 197, 626, 2056, 6918, 23714, 82500, 290512, 1033412, 3707852, 13402697, 48760367, 178405157, 656043857, 2423307047, 8987427467, 33453694487, 124936258127, 467995871777, 1757900019101, 6619846420553, 24987199492705, 94520750408709, 358268702159069 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

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 form 6n + 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

LINKS

T. D. Noe, Table of n, a(n) for n=0..200

W. Chammam, F. Marcellán and R. Sfaxi, Orthogonal polynomials, Catalan numbers, and a general Hankel determinant evaluation, Linear Algebra Appl. (2011), in press

Dennis E. Davenport, Lara K. Pudwell, Louis W. Shapiro, Leon C. Woodson, The Boundary of Ordered Trees, Journal of Integer Sequences, Vol. 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

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

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.

FORMULA

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

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

Sum_{i=1..n} c(i) = Sum_{i=1..n} binomial(2*i-2, i-1)/i = 1/(n-1)! * [n^(n-2) + binomial(n, 2)*n^(n-3) + {8*binomial(n-4, 0) + 19*binomial(n-4, 1) + 24*binomial(n-4, 2) + 14*binomial(n-4, 3) + 3*binomial(n-4, 4)}*n^(n-4) +{18*binomial(n-5, 0) + 82*binomial(n-5, 1) + 229*binomial(n-5, 2) + 323*binomial(n-5, 3) + 244*binomial(n-5, 4) + 95*binomial(n-5, 5) + 15*binomial(n-5, 6)}*n^(n-5) + ... + binomial(n-3, 0)*(n-1)! ] (where c() = Catalan numbers A000108). - André F. Labossière, May 17 2004

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

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

Mathar's conjecture 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

MAPLE

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

MATHEMATICA

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

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

PROG

(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 */

CROSSREFS

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

Cf. A000245. - Gary W. Adamson, May 15 2009

Sequence in context: A000083 A092668 A164039 * A245158 A245159 A245160

Adjacent sequences:  A014134 A014135 A014136 * A014138 A014139 A014140

KEYWORD

nonn,nice

AUTHOR

N. J. A. Sloane

STATUS

approved

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Last modified August 29 08:06 EDT 2015. Contains 261188 sequences.