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A000984 Central binomial coefficients: C(2*n,n) = (2*n)!/(n!)^2.
(Formerly M1645 N0643)
1, 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, 2704156, 10400600, 40116600, 155117520, 601080390, 2333606220, 9075135300, 35345263800, 137846528820, 538257874440, 2104098963720, 8233430727600, 32247603683100 (list; graph; refs; listen; history; text; internal format)



Devadoss refers to these numbers as type B Catalan numbers (cf. A000108).

Equal to the binomial coefficient sum Sum_{k=0..n} binomial(n,k)^2.

Number of possible interleavings of a program with n atomic instructions when executed by two processes - Manuel Carro (mcarro(AT)fi.upm.es), Sep 22 2001

Convolving a(n) with itself yields A000302, the powers of 4. - T. D. Noe, Jun 11 2002

a(n)=Max{ (i+j)!/(i!j!) | 0<=i,j<=n } - Benoit Cloitre, May 30 2002

Number of ordered trees with 2n+1 edges, having root of odd degree and nonroot nodes of outdegree 0 or 2. - Emeric Deutsch, Aug 02 2002

Also number of directed, convex polyominoes having semiperimeter n+2.

Also number of diagonally symmetric, directed, convex polyominoes having semiperimeter 2n+2. - Emeric Deutsch, Aug 03 2002

Also Sum_{k=0..n} binomial(n+k-1,k). - Vladeta Jovovic, Aug 28 2002

The second inverse binomial transform of this sequence is this sequence with interpolated zeros. Its G.f. is (1 - 4*x^2)^(-1/2), with n-th term C(n,n/2)(1+(-1)^n)/2. - Paul Barry, Jul 01 2003

Number of possible values of a 2*n bit binary number for which half the bits are on and half are off. - Gavin Scott (gavin(AT)allegro.com), Aug 09 2003

Ordered partitions of n with zeros to n+1, e.g. for n=4 we consider the ordered partitions of 11110 (5), 11200 (30), 13000 (20), 40000 (5) and 22000 (10), total 70 and a(4)=70. See A001700 (esp. Mambetov Bektur's comment). - Jon Perry, Aug 10 2003

Number of non-decreasing sequences of n integers from 0 to n: a(n) = sum_{i_{1}=0}^{n} sum_{i_{2}=i_{1}}^{n}...sum_{i_{n}=i_{n-1}}^{n}(1). - J. N. Bearden (jnb(AT)eller.arizona.edu), Sep 16 2003

Number of peaks at odd level in all Dyck paths of semilength n+1. Example: a(2)=6 because we have U*DU*DU*D, U*DUUDD, UUDDU*D, UUDUDD, UUU*DDD, where U=(1,1), D=(1,-1) and * indicates a peak at odd level. Number of ascents of length 1 in all Dyck paths of semilength n+1 (an ascent in a Dyck path is a maximal string of up steps). Example: a(2)=6 because we have uDuDuD, uDUUDD, UUDDuD, UUDuDD, UUUDDD, where an ascent of length 1 is indicated by a lower case letter. - Emeric Deutsch, Dec 05 2003

a(n-1)=number of subsets of 2n-1 distinct elements taken n at a time that contain a given element. e.g. n=4 -> a(3)=20 and if we consider the subsets of 7 taken 4 at a time with a 1 we get (1234, 1235, 1236, 1237, 1245, 1246, 1247, 1256, 1257, 1267, 1345, 1346, 1347, 1356, 1357, 1367, 1456, 1457, 1467, 1567) and there are 20 of them. - Jon Perry, Jan 20 2004

The dimension of a particular (necessarily existent) absolutely universal embedding of the unitary dual polar space DSU(2n,q^2) where q>2. - J. Taylor (jt_cpp(AT)yahoo.com), Apr 02 2004.

Number of standard tableaux of shape (n+1, 1^n). - Emeric Deutsch, May 13 2004

Erdos, Graham et al. conjectured that a(n) is never squarefree for sufficiently large n. Sarkozy showed that if s(n) is the square part of a(n), then s(n) is asymptotically (sqrt(2)-2)*(sqrt(n))*(Riemann Zeta Function(1/2)). Granville and Ramare proved that the only squarefree values are a(1)=2, a(2)=6 and a(4)=70. A000984(n)/(n+1) = A000108(n), that is, dividing by (n+1) scales the central binomial coefficients to Catalan numbers also called Segner numbers. - Jonathan Vos Post, Dec 04 2004

p divides a((p-1)/2)-1=A030662(n) for prime p=5, 13, 17, 29, 37, 41, 53, 61, 73, 89, 97, ... = A002144(n) Pythagorean primes: primes of form 4n+1. - Alexander Adamchuk, Jul 04 2006

The number of direct routes from my home to Granny's when Granny lives n blocks south and n blocks east of my home in Grid City. To obtain a direct route, from the 2n blocks, choose n blocks on which one travels south. For example, a(2)=6 because there are 6 direct routes: SSEE, SESE, SEES, EESS, ESES and ESSE. - Dennis P. Walsh, Oct 27 2006

Inverse: With q = -log(log(16)/(pi a(n)^2)), ceiling((q + log(q))/log(16)) = n. - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Feb 26 2007

Number of partitions with Ferrers diagrams that fit in an n X n box (including the empty partition of 0). Example: a(2) = 6 because we have: empty, 1, 2, 11, 21 and 22. - Emeric Deutsch, Oct 02 2007 [So this is the 2-dimensional analogue of A008793. - William Entriken, Aug 06 2013]

The number of walks of length 2n on an infinite linear lattice that begin and end at the origin. - Stefan Hollos (stefan(AT)exstrom.com), Dec 10 2007

The number of lattice paths from (0,0) to (n,n) using steps (1,0) and (0,1). - Joerg Arndt, Jul 01 2011

Integral representation : C(2n,n)=1/Pi Integral [(2x)^(2n)/sqrt(1 - x^2),{x,-1, 1}], i.e. C(2n,n)/4^n is the moment of order 2n of the arcsin distribution on the interval (-1,1). - N-E. Fahssi, Jan 02 2008

Define the array m(1,j)=1 ; m(i,1)=1 ; m(i,j)=m(i,j-1) + m(j,i-1), then a(n) = m(n,n) - philippe lallouet (philip.lallouet(AT)orange.fr), Sep 15 2008

Also the Catalan transform of A000079. - R. J. Mathar, Nov 06 2008

Straub, Amdeberhan and Moll: "... it is conjectured that there are only finitely many indices n such that C_n is not divisible by any of 3, 5, 7 and 11. Finally, we remark that Erdos et al. conjectured that the central binomial coefficients C_n are never squarefree for n > 4 which has been proved by Granville and Ramare." - Jonathan Vos Post, Nov 14 2008

Equals INVERT transform of A081696: (1, 1, 3, 9, 29, 97, 333,...). - Gary W. Adamson, May 15 2009

Also, in sports, the number of ordered ways for a "Best of 2n-1 Series" to progress. For example, a(2) = 6 means there are six ordered ways for a "best of 3" series to progress. If we write A for a win by "team A" and B for a win by "team B" and if we list the played games chronologically from left to right then the six ways are AA, ABA, BAA, BB, BAB, and ABB. (Proof: To generate the a(n) ordered ways: Write down all a(n) ways to designate n of 2n games as won by team A. Remove the maximal suffix of identical letters from each of these.) - Lee A. Newberg, Jun 02 2009

Index the central binomial coefficients with the natural numbers 1,2,3...,n. It appears that dividing the central binomial coefficients by their indexes yields the Catalan numbers (A000108). - Jason Richardson-White, Jun 15 2009

Number of nXn binary arrays with rows, considered as binary numbers, in nondecreasing order, and columns, considered as binary numbers, in nonincreasing order. - R. H. Hardin, Jun 27 2009

Hankel transform is 2^n. - Paul Barry, Aug 05 2009

It appears that a(n) is also the number of quivers in the mutation class of twisted type BC_n for n>=2.

Central terms of Pascal's triangle: a(n) = A007318(2*n,n). - Reinhard Zumkeller, Nov 09 2011

Number of words on {a,b} of length 2n such that no prefix of the word contains more b's than a's. - Jonathan Nilsson, Apr 18 2012

Also a(n) = Product_{k=1..n} (4 - 2/k). - Michel Lagneau, Apr 28 2012

From Pascal's triangle take row(n) with terms in order a1,a2,..a(n) and row(n+1) with terms b1,b2,..b(n), then 2*(a1*b1+a2*b2+...a(n)*b(n)) to get the terms in this sequence. - J. M. Bergot, Oct 07 2012. For example using rows 4 and 5: 2*(1*(1)+4*(5)+6*(10)+4*(10)+1*(5)=252, the sixth term in this sequence.

Take from Pascal's triangle row(n) with terms b1, b2,..., b(n+1) and row(n+2) with terms c1, c2,..., c(n+3) and find the sum b1*c2 + b2*c3 + ... + b(n+1)*c(n+2) to get A000984(n+1).  Example using row(3) and row(5) gives sum 1*(5)+3*(10)+3*(10)+1*(5) = 70 = A000984(4). - J. M. Bergot, Oct 31 2012

a(n) == 2 mod n^3 iff n is a prime > 3. (See Mestrovic link, p.4) - Gary Detlefs, Feb 16 2013

Conjecture: For any positive integer n, the polynomial sum_{k=0}^n a(k)x^k is irreducible over the field of rational numbers. In general, for any integer m>1 and n>0, the polynomial f_{m,n}(x) = sum_{k=0}^n(m*k)!/(k!)^m*x^k is irreducible over the field of rational numbers. - Zhi-Wei Sun, Mar 23 2013

This comment generalizes the comment dated Oct 31 2012 and the second of the sequence’s original comments.  For j = 1 to n, a(n) = sum_{k=0..j} C(j,k)* C(2n-j, n-k) = 2*sum_{k=0..j-1} C(j-1,k)*C(2n-j, n-k). Charlie Marion, Jun 07 2013

The differences between consecutive terms of the sequence of the quotients between consecutive terms of this sequence form a sequence containing the reciprocals of the triangular numbers. In other words, a(n+1)/a(n)-a(n)/a(n-1) = 2/(n*(n+1)). - Christian Schulz, Jun 08 2013


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P. Barry, Riordan-Bernstein Polynomials, Hankel Transforms and Somos Sequences, Journal of Integer Sequences, Vol. 15 2012, #12.8.2.

P. Barry, On the Central Coefficients of Riordan Matrices, Journal of Integer Sequences, 16 (2013), #13.5.1.

P. Barry, A Note on a Family of Generalized Pascal Matrices Defined by Riordan Arrays, Journal of Integer Sequences, 16 (2013), #13.5.4.

Paul Barry and Aoife Hennessy, Generalized Narayana Polynomials, Riordan Arrays, and Lattice Paths, Journal of Integer Sequences, Vol. 15, 2012, #12.4.8.

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Oktay Haracci (timetunnel3(AT)hotmail.com), Regular Polygons

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Eric Weisstein's World of Mathematics, Staircase Walk

Eric Weisstein's World of Mathematics, Circle Line Picking

Index entries for "core" sequences


G.f.: A(x) = (1 - 4*x)^(-1/2) = 1 + 2*x + 6*x^2 + 20*x^3 + ...

a(n+1)=2*A001700(n)=A030662(n)+1. a(2*n) = A001448(n), a(2*n+1) = 2*A002458(n).

a(n) = 2^n/n! * product(k=0..n-1, (2*k+1) ).

a(n) = a(n-1)*(4-2/n) = 4*a(n-1)+A002420(n) = A000142(2*n)/(A000142(n)^2) = A001813(n)/A000142(n) = sqrt(A002894(n)) = A010050(n)/A001044(n) = (n+1)*A000108(n) = -A005408(n-1)*A002420(n) - Henry Bottomley, Nov 10 2000

Using Stirling's formula in A000142 it is easy to get the asymptotic expression a(n) ~ 4^n / sqrt(Pi * n) - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 07 2001

Integral representation as n-th moment of a positive function on the interval[0, 4], in Maple notation: a(n)= int(x^n*((x*(4-x))^(-1/2))/Pi, x=0..4), n=0, 1, ... This representation is unique. - Karol A. Penson, Sep 17 2001

sum(n>=1, 1/a(n))=(2*Pi*sqrt(3)+9)/27 [Lehmer, Am. Math. Monthly 92 (1985) 449, eq (15)] - Benoit Cloitre, May 01 2002

E.g.f.: exp(2*x)*I_0(2x), where I_0 is Bessel function. - Michael Somos, Sep 08 2002

E.g.f.: I_0(2*x)=sum a(n)*x^(2*n)/(2*n)!, where I_0 is Bessel function. - Michael Somos, Sep 09, 2002.

a(n) = sum(k=0, n, C(n, k)^2). - Benoit Cloitre, Jan 31 2003

Determinant of n X n matrix M(i, j)=binomial(n+i, j) - Benoit Cloitre, Aug 28 2003

Given m = C(2*n, n), let f be the inverse function, so that f(m) = n. Letting q denote -Log(Log(16)/(m^2*Pi)), we have f(m) = Ceiling( (q + Log(q)) / Log(16) ). - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Oct 30 2003

a(n) = 2*Sum{k= 0...(n-1), a(k)*a(n-k+1)/(k+1)}. - Philippe Deléham, Jan 01 2004

a(n+1)=sum(j=n, n*2+1, binomial(j, n)). E.g. a(4)=C(7, 3)+C(6, 3)+C(5, 3)+C(4, 3)+C(3, 3)=35+20+10+4+1=70 - Jon Perry, Jan 20 2004

a(n) = (-1)^(n)*sum(j=0..(2*n), (-1)^j*binomial(2*n, j)^2) - Helena Verrill (verrill(AT)math.lsu.edu), Jul 12 2004

a(n)=sum{k=0..n, binomial(2n+1, k)*sin((2n-2k+1)*pi/2)}. - Paul Barry, Nov 02 2004

a(n-1)=(1/2)*(-1)^n*sum_{0<=i, j<=n}(-1)^(i+j)*binomial(2n, i+j) - Benoit Cloitre, Jun 18 2005

a(n) = C(2n, n-1) + C(n) = A001791(n) + A000108(n). - Lekraj Beedassy, Aug 02 2005

G.f.: c(x)^2/(2*c(x)-c(x)^2) where c(x) is the g.f. of A000108; - Paul Barry, Feb 03 2006

a(n)=A006480(n)/A005809(n) - Zerinvary Lajos, Jun 28 2007

a(n)=Sum{k, 0<=k<=n}A106566(n,k)*2^k. - Philippe Deléham, Aug 25 2007

a(n)= Sum{k>=0, A039599(n, k)} . a(n)= Sum{k>=0, A050165(n, k)} . a(n)= Sum{k>=0, A059365(n, k)*2^k}, n>0 . a(n+1)= Sum{k>=0, A009766(n, k)*2^(n-k+1)}. - Philippe Deléham, Jan 01 2004

a(n)=4^n*sum{k=0..n, C(n,k)(-4)^(-k)*A000108(n+k)}; - Paul Barry, Oct 18 2007

Row sums of triangle A135091 - Gary W. Adamson, Nov 18 2007

a(n)=Sum_{k, 0<=k<=n}A039598(n,k)*A059841(k). [Philippe Deléham, Nov 12 2008]

A007814(a(n))=A000120(n). [Vladimir Shevelev, Jul 20 2009]

From Paul Barry, Aug 05 2009: (Start)

G.f.: 1/(1-2x-2x^2/(1-2x-x^2/(1-2x-x^2/(1-2x-x^2/(1-... (continued fraction);

G.f.: 1/(1-2x/(1-x/(1-x/(1-x/(1-... (continued fraction). (End)

a(n)=Product(k=1..n)[4-2/k] [David Brown, Sep 19 2009]

If n>=3 is prime, then a(n)==2(mod 2*n). [Vladimir Shevelev, Sep 05 2010]

Let A(x) be the g.f. and B(x)=A(-x), then B(x)=sqrt(1-4*x*B(x)^2) [From Vladimir Kruchinin, Jan 16 2011]

a(n)=(-4)^n*sqrt(Pi)/(gamma((1/2-n))*gamma(1+n)) [Gerry Martens, May 03 2011]

a(n) = upper left term in M^n, M = the infinite square production matrix:

2, 2, 0, 0, 0, 0,...

1, 1, 1, 0, 0, 0,...

1, 1, 1, 1, 0, 0,...

1, 1, 1, 1, 1, 0,...

1, 1, 1, 1, 1, 1,....

- Gary W. Adamson, Jul 14 2011

a(n) = Hypergeometric([-n,-n],[1],1). - Peter Luschny, Nov 01 2011

E.g.f.: exp(2*x)*I_0(2*x)=1+2*x/(Q(0)-2*x); Q(k)=2*x*(2*k+1)+(k+1)^2-2*x*(2*k+3)*(k+1)^2/Q(k+1), where I_0 is Bessel function; (continued fraction). - Sergei N. Gladkovskii, Nov 22 2011

E.g.f.: hypergeometric([1/2],[1],4*x), simplified the e.g.f. given above, also by M. Somos. - Wolfdieter Lang, Jan 13 2012.

a(n) = 2*Sum_{k=0..n-1} (a(k)*A000108(n-k-1)). - Alzhekeyev Ascar M., Mar 09 2012

G.f.: 1/sqrt(1-4*x)= 1 + 2*x/(U(0)-2*x) where U(k)= 2*(2*k+1)*x + (k+1) - 2*(k+1)*(2*k+3)*x/U(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Jun 28 2012

a(n) = sum_{k=0..n} binomial(n,k)^2*H(k))/(2*H(n)-H(2*n), n>0, where H(n) is the n-th harmonic number. - Gary Detlefs, Mar 19 2013

G.f.: Q(0)*(1-4*x), where Q(k)= 1 + 4*(2*k+1)*x/( 1 - 1/(1 + 2*(k+1)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 11 2013

G.f.: G(0)/2, where G(k)= 1 + 1/(1 - 2*x*(2*k+1)/(2*x*(2*k+1) + (k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 24 2013

E.g.f.: E(0)/2, where E(k)= 1 + 1/(1 - 2*x/(2*x + (k+1)^2/(2*k+1)/E(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 01 2013

Special values of Jacobi polynomials, in Maple notation: a(n) = 4^n*JacobiP(n,0,-1/2-n,-1). - Karol A. Penson, Jul 27 2013.

a(n) = 2^(4*n)/((2*n+1)*sum(k=0..n, (-1)^k*C(2*n+1,n-k)/(2*k+1))). - Mircea Merca, Nov 12 2013

a(n) = C(2*n-1,n-1)*C(4*n^2,2)/(3*n*C(2*n+1,3)), n>0. - Gary Detlefs, Jan 02 2014

Sum(n>=0, a(n)/n!) = A234846. - Richard R. Forberg, Feb 10 2014


A000984 := n-> binomial(2*n, n);

with(combstruct); [seq(count([S, {S=Prod(Set(Z, card=i), Set(Z, card=i))}, labeled], size=(2*i)), i =0..20)];

with(combstruct); [seq(count([S, {S=Sequence(Union(Arch, Arch)), Arch=Prod(Epsilon, Sequence(Arch), Z)}, unlabeled], size=i), i=0..25)];

Z:=(1-sqrt(1-z))*4^n/sqrt(1-z): Zser:=series(Z, z=0, 32): seq(coeff(Zser, z, n), n=0..24); # Zerinvary Lajos, Jan 01 2007

with(combstruct):bin := {B=Union(Z, Prod(B, B))}: seq (count([B, bin, unlabeled], size=n)*n, n=1..25); # Zerinvary Lajos, Dec 05 2007


Table[Binomial[2n, n], {n, 0, 24}] (* Alonso del Arte, Nov 10 2005 *)

CoefficientList[Series[1/Sqrt[1-4x], {x, 0, 25}], x]  (* Harvey P. Dale, Mar 14 2011 *)


(MAGMA) a:= func< n | Binomial(2*n, n) >; [ a(n) : n in [0..10]];

(PARI) A000984(n)=binomial(2*n, n) \\ much more efficient than 2n!/n!^2. - M. F. Hasler, Feb 26 2014

(PARI) fv(n, p)=my(s); while(n\=p, s+=n); s

a(n)=prodeuler(p=2, 2*n, p^(fv(2*n, p)-2*fv(n, p))) \\ Charles R Greathouse IV, Aug 21 2013

(PARI) fv(n, p)=my(s); while(n\=p, s+=n); s

a(n)=my(s=1); forprime(p=2, 2*n, s*=p^(fv(2*n, p)-2*fv(n, p))); s \\ Charles R Greathouse IV, Aug 21 2013


a000984 n = a007318_row (2*n) !! n  -- Reinhard Zumkeller, Nov 09 2011

(Maxima) A000984(n):=(2*n)!/(n!)^2$ makelist(A000984(n), n, 0, 30); [Martin Ettl, Oct 22 2012]


Cf. A000108, A002420, A002457, A030662, A002144, A135091, A152229, A158815, A081696, A205946, A182400. Differs from A071976 at 10-th term.

Bisection of A001405 and of A226302. See also A025565, the same ordered partitions but without all in which are two successive zeros: 11110 (5), 11200 (18), 13000 (2), 40000 (0) and 22000 (1), total 26 and A025565(4)=26.

Cf. A226078, A051924 (first differences).

Row sums of A059481, A008459, A152229, A158815, A205946.

Sequence in context: A056616 A065346 A071976 * A087433 A119373 A151284

Adjacent sequences:  A000981 A000982 A000983 * A000985 A000986 A000987




N. J. A. Sloane.



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Last modified April 20 09:30 EDT 2014. Contains 240779 sequences.