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A001791 Binomial coefficients C(2n,n-1).
(Formerly M3500 N1421)
59
0, 1, 4, 15, 56, 210, 792, 3003, 11440, 43758, 167960, 646646, 2496144, 9657700, 37442160, 145422675, 565722720, 2203961430, 8597496600, 33578000610, 131282408400, 513791607420, 2012616400080, 7890371113950, 30957699535776, 121548660036300, 477551179875952 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

COMMENTS

Number of peaks at even level in all Dyck paths of semilength n+1. Example: a(2)=4 because UDUDUD, UDUU*DD, UU*DDUD, UU*DU*DD, UUUDDD, where U=(1,1), D=(1,-1) and the peaks at even level are shown by *. - Emeric Deutsch, Dec 05 2003

Also number of long ascents (i.e., ascents of length at least two) in all Dyck paths of semilength n+1. Example: a(2)=4 because in the five Dyck paths of semilength 3, namely UDUDUD, UD(UU)DD, (UU)DDUD, (UU)DUDD and (UUU)DDD, we have four long ascents (shown between parentheses). Here U=(1,1) and D=(1,-1). Also number of branch nodes (i.e., vertices of outdegree at least two) in all ordered trees with n+1 edges. - Emeric Deutsch, Feb 22 2004

Number of lattice paths from (0,0) to (n,n) with steps E=(1,0) and N=(0,1) which touch or cross the line x-y=1. Example: For n=2 these are the paths EENN, ENEN, ENNE and NEEN. - Herbert Kociemba, May 23 2004

Narayana transform (A001263) of [1, 3, 5, 7, 9,...] = (1, 4, 15, 56, 210,...). Row sums of triangles A136534 and A136536. - Gary W. Adamson, Jan 04 2008

Starting with offset 1 = the Catalan sequence starting (1, 2, 5, 14,...) convolved with A000984: (1, 2, 6, 20,...). - Gary W. Adamson, May 17 2009

Also number of peaks plus number of valleys in all Dyck n-paths. - David Scambler, Oct 08 2012

Apparently counts UDDUD in all Dyck paths of semilength n+2. - David Scambler, Apr 22 2013

Apparently the number of peaks strictly left of the midpoint in all Dyck paths of semilength n+1. - David Scambler, Apr 30 2013

For n>0, a(n) is the number of compositions of n into at most n parts if zeros are allowed as parts (so-called "weak" compositions). - L. Edson Jeffery, Jul 24 2014

Number of paths in the half-plane x>=0, from (0,0) to (2n,2), and consisting of steps U=(1,1) and D=(1,-1). For example, for n=2, we have the 4 paths: UUUD, UUDU, UDUU, DUUU. - José Luis Ramírez Ramírez, Apr 19 2015

REFERENCES

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.

C. Lanczos, Applied Analysis. Prentice-Hall, Englewood Cliffs, NJ, 1956, p. 517.

R. C. Mullin, E. Nemeth and P. J. Schellenberg, The enumeration of almost cubic maps, pp. 281-295 in Proceedings of the Louisiana Conference on Combinatorics, Graph Theory and Computer Science. Vol. 1, edited R. C. Mullin et al., 1970.

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

LINKS

T. D. Noe and Matuszka Tamás, Table of n, a(n) for n = 0..1200 (terms n = 0..200 from T. D. Noe)

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].

Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5.

Paul Barry, On the Hurwitz Transform of Sequences, Journal of Integer Sequences, Vol. 15 (2012), #12.8.7.

Miklós Bóna, Surprising Symmetries in Objects Counted by Catalan Numbers, Electronic J. Combin., 19 (2012), P62, eq. (6).

Guo-Niu Han, Enumeration of Standard Puzzles

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

A. Ivanyi, L. Lucz, T. Matuszka, and S. Pirzada, Parallel enumeration of degree sequences of simple graphs, Acta Univ. Sapientiae, Informatica, 4, 2 (2012) 260-288.

Milan Janjic, Two Enumerative Functions

M. Janjic and B. Petkovic, A Counting Function, arXiv preprint arXiv:1301.4550 [math.CO], 2013.

C. Lanczos, Applied Analysis (Annotated scans of selected pages)

Asamoah Nkwanta and Earl R. Barnes, Two Catalan-type Riordan Arrays and their Connections to the Chebyshev Polynomials of the First Kind, Journal of Integer Sequences, Article 12.3.3, 2012. - From N. J. A. Sloane, Sep 16 2012.

FORMULA

a(n) = n*A000108(n).

G.f.: x*diff(c(x), x) where c(x) = Catalan g.f. - Wolfdieter Lang

Convolution of A001700( central binomial of odd order) and A000108 (Catalan): a(n+1)=sum(C(k)*binomial(2*(n-k)+1, n-k), k=0..n), C(k): Catalan. - Wolfdieter Lang

E.g.f.: exp(2x) I_1(2x), where I_1 is Bessel function. - Michael Somos, Sep 08 2002

a(n) = sum{k=0..n, C(n, k)*C(n, k+1) }. - Paul Barry, May 15 2003

a(n) = sum(binomial(i+n-1, n), i=1..n).

G.f.: [1-2x-sqrt(1-4x)]/[2xsqrt(1-4x)]. - Emeric Deutsch, Dec 05 2003

A092956/(n!). - Amarnath Murthy, Jun 16 2004

a(n) = binomial(2n,n) - A000108(n). - Paul Barry, Apr 21 2005.

a(n) = (1/(2*pi))*Int(x^n*(x-2)/sqrt(x(4-x)),x,0,4) is the moment sequence representation. - Paul Barry, Jan 11 2007

Row sums of triangle A132812 starting (1, 4, 15, 56, 210,...). - Gary W. Adamson, Sep 01 2007

Starting (1, 4, 15, 56, 210,...) gives the binomial transform of A025566 starting (1, 3, 8, 22, 61, 171,...). - Gary W. Adamson, Sep 01 2007

For n>=1,a(2^n) = 2^(n+1)*A001795(2^(n-1)). - Vladimir Shevelev, Sep 05 2010

(n-1)*(n+1)*a(n) = 2*n*(2n-1)*a(n-1). - R. J. Mathar, Dec 17 2011

From Sergei N. Gladkovskii, Jul 07 2012: (Start)

G.f.: -1/(2*x) - G(0) where G(k)= 1 - 1/(2*x - 8*x^3*(2*k+1)/(4*x^2*(2*k+1)- (k+1)/G(k+1))); (continued fraction, 3rd kind, 3-step);

E.g.f.: BesselI(1,2*x)*exp(2*x)= x* G(0) where G(k)= 1 + 2*x*(4*k+3)/((2*k+1)*(2*k+3) - x*(2*k+1)*(2*k+3)*(4*k+5)/(x*(4*k+5) + 2*(k+1)*(k+2)/G(k+1))); (continued fraction, 3rd kind, 3-step).

(End)

G.f.: c(x)^3/(2-c(x)) where c(x) is the g.f. for A000108. - Cheyne Homberger, May 05 2014

G.f.: z*C(z)^2/(1-2*z*C(z)), where C(z) is the g.f. of Catalan numbers. - José Luis Ramírez Ramírez, Apr 19 2015

G.f.: x*2F1(3/2,2;3;4x). - R. J. Mathar, Aug 09 2015

a(n) = Sum_{i=1..n}((binomial(2*i-2,i-1)*binomial(2*(n-i+1),n-i+2))/(n-i+1)). - Vladimir Kruchinin, Sep 07 2015

MAPLE

[seq(binomial(2*n, n)/(n+1)*n, n=0..30)]; # Zerinvary Lajos, Jun 25 2006

MATHEMATICA

CoefficientList[ Series[4/(((Sqrt[1 - 4 x] + 1)^2)*Sqrt[1 - 4 x]), {x, 0, 22}], x] (* Robert G. Wilson v, Aug 08 2011 *)

Table[Binomial[2n, n-1], {n, 0, 30}] (* Harvey P. Dale, Jul 12 2012 *)

PROG

(PARI) a(n)=if(n<1, 0, (2*n)!/(n+1)!/(n-1)!)

(Mupad) combinat::catalan(n) *binomial(n, 1) $ n = 0..24 /* Zerinvary Lajos, Feb 15 2007 */

(Maxima) A001791(n):=binomial(2*n, n-1)$

makelist(A001791(n), n, 0, 30); /* Martin Ettl, Nov 05 2012 */

(MAGMA) [Binomial(2*n, n-1): n in [0..30]]; // Vincenzo Librandi, Apr 20 2015

CROSSREFS

Cf. A000984.

Diagonal 3 of triangle A100257.

First differences are in A076540.

Cf. A000108, A000984, A002378.

Cf. A025566, A132812.

Sequence in context: A026030 A047038 A158500 * A047128 A087438 A131497

Adjacent sequences:  A001788 A001789 A001790 * A001792 A001793 A001794

KEYWORD

nonn,easy,nice

AUTHOR

N. J. A. Sloane

STATUS

approved

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Last modified December 3 03:54 EST 2016. Contains 278698 sequences.