

A000984


Central binomial coefficients: binomial(2*n,n) = (2*n)!/(n!)^2.
(Formerly M1645 N0643)


779



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, 126410606437752, 495918532948104, 1946939425648112
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OFFSET

0,2


COMMENTS

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+k1,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 nth term C(n,n/2)(1+(1)^n)/2.  Paul Barry, Jul 01 2003
Number of possible values of a 2nbit 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 nondecreasing 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_{n1}..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(n1) = number of subsets of 2n1 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
Erdős, Graham et al. conjectured that a(n) is never squarefree for sufficiently large n (cf. Graham, Knuth, Patashnik, Concrete Math., 2nd ed., Exercise 112). Sárközy 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.  Jonathan Vos Post, Dec 04 2004 [For more about this conjecture, see A261009.  N. J. A. Sloane, Oct 25 2015]
The MathOverflow link contains the following comment (slightly edited): The Erdős squarefree conjecture (that a(n) is never squarefree for n>4) was proved in 1980 by Sárközy, A. (On divisors of binomial coefficients. I. J. Number Theory 20 (1985), no. 1, 7080.) who showed that the conjecture holds for all sufficiently large values of n, and by A. Granville and O. Ramaré (Explicit bounds on exponential sums and the scarcity of squarefree binomial coefficients. Mathematika 43 (1996), no. 1, 73107) who showed that it holds for all n>4.  Fedor Petrov, Nov 13 2010. [From N. J. A. Sloane, Oct 29 2015]
A000984(n)/(n+1) = A000108(n), the Catalan numbers.
p divides a((p1)/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 2dimensional analog of A008793.  William Entriken, Aug 06 2013
The number of walks of length 2n on an infinite linear lattice that begins and ends 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).  NE. Fahssi, Jan 02 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 Erdős 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 2n1 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
Number of n X n 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
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.  ZhiWei 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(2nj, nk) = 2*Sum_{k=0..j1} C(j1,k)*C(2nj, nk).  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(n1) = 2/(n*(n+1)).  Christian Schulz, Jun 08 2013
Number of distinct strings of length 2n using n letters A and n letters B.  Hans Havermann, May 07 2014
From Fung Lam, May 19 2014: (Start)
Expansion of G.f. A(x) = 1/(1+q*x*c(x)), where parameter q is positive or negative (except q=1), and c(x) is the g.f. of A000108 for Catalan numbers. The case of q=1 recovers the g.f. of A000108 as xA^2A+1=0. The present sequence A000984 refers to q=2. Recurrence: (1+q)*(n+2)*a(n+2) + ((q*q4*q4)*n + 2*(q*qq1))*a(n+1)  2*q*q*(2*n+1)*a(n) = 0, a(0)=1, a(1)=q. Asymptotics: a(n) ~ ((q+2)/(q+1)*(q^2/(q1))^n, q<=3, a(n) ~ (1)^n*((q+2)/(q+1))*(q^2/(q+1))^n, q>=5, and a(n) ~ Kq*2^(2*n)/sqrt(Pi*n^3), where the multiplicative constant Kq is given by K1=1/9 (q=1), K2=1/8 (q=2), K3=3/25 (q=3), K4=1/9 (q=4). These formulas apply to existing sequences A126983 (q=1), A126984 (q=2), A126982 (q=3), A126986 (q=4), A126987 (q=5), A127017 (q=6), A127016 (q=7), A126985 (q=8), A127053 (q=9), and to A007854 (q=3), A076035 (q=4), A076036 (q=5), A127628 (q=6), A126694 (q=7), A115970 (q=8). (End)
a(n)*(2^n)^(j2) equals S(n), where S(n) is the nth number in the selfconvolved sequence which yields the powers of 2^j for all integers j, n>=0. For example, when n=5 and j=4, a(5)=252; 252*(2^5)^(42) = 252*1024 = 258048. The selfconvolved sequence which yields powers of 16 is {1, 8, 96, 1280, 17920, 258048, ...}; i.e., S(5) = 258048. Note that the convolved sequences will be composed of numbers decreasing from 1 to 0, when j<2 (exception being j=1, where the first two numbers in the sequence are 1 and all others decreasing).  Bob Selcoe, Jul 16 2014
The variance of the nth difference of a sequence of pairwise uncorrelated random variables each with variance 1.  Liam Patrick Roche, Jun 04 2015
Number of ordered trees with n edges where vertices at level 1 can be of 2 colors. Indeed, the standard decomposition of ordered trees leading to the equation C = 1 + zC^2 (C is the Catalan function), yields this time G = 1 + 2zCG, from where G = 1/sqrt(14z).  Emeric Deutsch, Jun 17 2015
Number of monomials of degree at most n in n variables.  Ran Pan, Sep 26 2015
Let V(n, r) denote the volume of an ndimensional sphere with radius r, then V(n, 2^n) / Pi = V(n1, 2^n) * a(n/2) for all even n.  Peter Luschny, Oct 12 2015
a(n) is the number of sets {i1,...,in} of length n such that n >= i1 >= i2 >= ... >= in >= 0. For instance, a(2) = 6 as there are only 6 such sets: (2,2) (2,1) (2,0) (1,1) (1,0) (0,0).  Anton Zakharov, Jul 04 2016
From Ralf Steiner, Apr 07 2017: (Start)
By analytic continuation to the entire complex plane there exist regularized values for divergent sums such as:
Sum_{k>=0} a(k)/(2)^k = 1/sqrt(3).
Sum_{k>=0} a(k)/(1)^k = 1/sqrt(5).
Sum_{k>=0} a(k)/(1/2)^k = 1/3.
Sum_{k>=0} a(k)/(1/2)^k = 1/sqrt(7)i.
Sum_{k>=0} a(k)/(1)^k = 1/sqrt(3)i.
Sum_{k>=0} a(k)/2^k = i. (End)
Number of sequences (e(1), ..., e(n+1)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) > e(j). [Martinez and Savage, 2.18]  Eric M. Schmidt, Jul 17 2017
The o.g.f. for the sequence equals the diagonal of any of the following the rational functions: 1/(1  (x + y)), 1/(1  (x + y*z)), 1/(1  (x + x*y + y*z)) or 1/(1  (x + y + y*z)).  Peter Bala, Jan 30 2018
From Colin Defant, Sep 16 2018: (Start)
Let s denote West's stacksorting map. a(n) is the number of permutations pi of [n+1] such that s(pi) avoids the patterns 132, 231, and 321. a(n) is also the number of permutations pi of [n+1] such that s(pi) avoids the patterns 132, 312, and 321.
a(n) is the number of permutations of [n+1] that avoid the patterns 1342, 3142, 3412, and 3421. (End)
All binary selfdual codes of length 4n, for n>0, must contain at least a(n) codewords of weight 2n. More to the point, there will always be at least one, perhaps unique, binary selfdual code of length 4n that will contain exactly a(n) codewords that have a hamming weight equal to half the length of the code (2n). This code can be constructed by direct summing the unique binary selfdual code of length 2 (up to permutation equivalence) to itself an even number of times. A permutation equivalent code can be constructed by augmenting two identity matrices of length 2n together.  Nathan J. Russell, Nov 25 2018
From Isaac Saffold, Dec 28 2018: (Start)
Let [b/p] denote the Legendre symbol and 1/b denote the inverse of b mod p. Then, for m and n, where n is not divisible by p,
[(m+n)/p] == [n/p]*Sum_{k=0..(p1)/2} (m/(4*n))^k * a(k) (mod p).
Evaluating this identity for m = 1 and n = 1 demonstrates that, for all odd primes p, Sum_{k=0..(p1)/2} (1/4)^k * a(k) is divisible by p. (End)
Number of vertices of the subgraph of the (2n1)dimensional hypercube induced by all bitstrings with n1 or n many 1s. The middle levels conjecture asserts that this graph has a Hamilton cycle.  Torsten Muetze, Feb 11 2019


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.
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 160.
Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 575, line 3, with a=b=n.
E. Deutsch and L. Shapiro, Seventeen Catalan identities, Bulletin of the Institute of Combinatorics and its Applications, 31, 3138, 2001.
H. W. Gould, Combinatorial Identities, Morgantown, 1972, (3.66), page 30.
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. AddisonWesley, Reading, MA, Second Ed., see Exercise 112.
M. Griffiths, The Backbone of Pascal's Triangle, United Kingdom Mathematics Trust (2008), 3124.
Leonard Lipshitz and A. van der Poorten. "Rational functions, diagonals, automata and arithmetic." In Number Theory, Richard A. Mollin, ed., Walter de Gruyter, Berlin (1990): 339358.
J. C. P. Miller, editor, Table of Binomial Coefficients. Royal Society Mathematical Tables, Vol. 3, Cambridge Univ. Press, 1954.
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

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Anders Hyllengren, Four integer sequences, Oct 04 1985. Observes essentially that A000984 and A002426 are inverse binomial transforms of each other, as are A000108 and A001006.
Milan Janjic, Two Enumerative Functions
I. Jensen, Series expansions for selfavoiding polygons
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Index entries for "core" sequences


FORMULA

G.f.: A(x) = (1  4*x)^(1/2) = 1F0(1/2;;4x).
a(n+1) = 2*A001700(n) = A030662(n) + 1. a(2*n) = A001448(n), a(2*n+1) = 2*A002458(n).
n*a(n) + 2*(12*n)*a(n1)=0.
a(n) = 2^n/n! * Product_{k=0..n1} (2*k+1).
a(n) = a(n1)*(42/n) = Product_{k=1..n} (42/k) = 4*a(n1) + A002420(n) = A000142(2*n)/(A000142(n)^2) = A001813(n)/A000142(n) = sqrt(A002894(n)) = A010050(n)/A001044(n) = (n+1)*A000108(n) = A005408(n1)*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 nth moment of a positive function on the interval[0, 4], in Maple notation: a(n) = Integral_{x=0..4}(x^n*((x*(4x))^(1/2))/Pi), 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 1985, 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} binomial(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..(n1)} a(k)*a(nk+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((2n2k+1)*Pi/2).  Paul Barry, Nov 02 2004
a(n1) = (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, n1) + 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..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^(nk+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..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/(12x2x^2/(12xx^2/(12xx^2/(12xx^2/(1... (continued fraction);
G.f.: 1/(12x/(1x/(1x/(1x/(1... (continued fraction). (End)
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(14*x*B(x)^2).  Vladimir Kruchinin, Jan 16 2011
a(n) = (4)^n*sqrt(Pi)/(gamma((1/2n))*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.: hypergeometric([1/2],[1],4*x).  Wolfdieter Lang, Jan 13 2012
a(n) = 2*Sum_{k=0..n1} a(k)*A000108(nk1).  Alzhekeyev Ascar M, Mar 09 2012
G.f.: 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, 1step).  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 nth harmonic number.  Gary Detlefs, Mar 19 2013
G.f.: Q(0)*(14*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,0,1/2n,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,nk)/(2*k+1)).  Mircea Merca, Nov 12 2013
a(n) = C(2*n1,n1)*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
0 = a(n)*(16*a(n+1)  6*a(n+2)) + a(n+1)*(2*a(n+1) + a(n+2)) for all n in Z.  Michael Somos, Sep 17 2014
a(n+1) = 4*a(n)  2*A000108(n). Also a(n) = 4^n*Product_{k=1..n}(11/(2*k)).  Stanislav Sykora, Aug 09 2014
G.f.: Sum_{n>=0} x^n/(1x)^(2*n+1) * Sum_{k=0..n} C(n,k)^2 * x^k.  Paul D. Hanna, Nov 08 2014
a(n) = (4)^n*binomial(1/2,n).  JeanFrançois Alcover, Feb 10 2015
a(n) = 4^n*hypergeom([n,1/2],[1],1).  Peter Luschny, May 19 2015
a(n) = Sum_{k=0..floor(n/2)} C(n,k)*C(nk,k)*2^(n2*k).  Robert FERREOL, Aug 29 2015
a(n) ~ 4^n*(22/(8*n+2)^2+21/(8*n+2)^4671/(8*n+2)^6+45081/(8*n+2)^8)/sqrt((4*n+1) *Pi).  Peter Luschny, Oct 14 2015
A(x) = 1/x * series reversion( x*(2*x + sqrt(1 + 4*x^2)) ). Compare with the o.g.f. B(x) of A098616, which satisfies B(x) = 1/x * series reversion( x*(2*x + sqrt(1  4*x^2) ). See also A214377.  Peter Bala, Oct 19 2015
a(n) = GegenbauerC(n,n,1).  Peter Luschny, May 07 2016
a(n) = gamma(1+2*n)/gamma(1+n)^2.  Andres Cicuttin, May 30 2016
Sum_{n>=0} (1)^n/a(n) = 4*(5  sqrt(5)*log(phi))/25 = 0.6278364236143983844442267..., where phi is the golden ratio.  Ilya Gutkovskiy, Jul 04 2016
From Peter Bala, Jul 22 2016: (Start)
This sequence occurs as the closedform expression for several binomial sums:
a(n) = Sum_{k = 0..2*n} (1)^(n+k)*binomial(2*n,k)*binomial(2*n + 1,k).
a(n) = 2*Sum_{k = 0..2*n1} (1)^(n+k)*binomial(2*n  1,k)*binomial(2*n,k) for n >= 1.
a(n) = 2*Sum_{k = 0..n1} binomial(n  1,k)*binomial(n,k) for n >= 1.
a(n) = Sum_{k = 0..2*n} (1)^k*binomial(2*n,k)*binomial(x + k,n)*binomial(y + k,n) = Sum_{k = 0..2*n} (1)^k*binomial(2*n,k)*binomial(x  k,n)*binomial(y  k,n) for arbitrary x and y.
For m = 3,4,5,... both Sum_{k = 0..m*n} (1)^k*binomial(m*n,k)*binomial(x + k,n)*binomial(y + k,n) and Sum_{k = 0..m*n} (1)^k*binomial(m*n,k)*binomial(x  k,n)*binomial(y  k,n) appear to equal Kronecker's delta(n,0).
a(n) = (1)^n*Sum_{k = 0..2*n} (1)^k*binomial(2*n,k)*binomial(x + k,n)*binomial(y  k,n) for arbitrary x and y.
For m = 3,4,5,... Sum_{k = 0..m*n} (1)^k*binomial(m*n,k)*binomial(x + k,n)*binomial(y  k,n) appears to equal Kronecker's delta(n,0).
a(n) = Sum_{k = 0..2n} (1)^k*binomial(2*n,k)*binomial(3*n  k,n)^2 = Sum_{k = 0..2*n} (1)^k*binomial(2*n,k)* binomial(n + k,n)^2. (Gould, Vol. 7, 5.23).
a(n) = Sum_{k = 0..n} (1)^(n+k)*binomial(2*n,n + k)*binomial(n + k,n)^2. (End)
From Ralf Steiner, Apr 07 2017: (Start)
Sum_{k>=0} a(k)/(p/q)^k = sqrt(p/(p4q)) for q in N, p in Z/{4q< (some p) <2}.
...
Sum_{k>=0} a(k)/(4)^k = 1/sqrt(2).
Sum_{k>=0} a(k)/(17/4)^k = sqrt(17).
Sum_{k>=0} a(k)/(18/4)^k = 3.
Sum_{k>=0} a(k)/5^k = sqrt(5).
Sum_{k>=0} a(k)/6^k = sqrt(3).
Sum_{k>=0} a(k)/8^k = sqrt(2).
...
Sum_{k>=0} a(k)/(p/q)^k = sqrt(p/(p4q)) for p>4q.(End)
BoasBuck recurrence: a(n) = (2/n)*Sum_{k=0..n1} 4^(nk1)*a(k), n >= 1, a(0) = 1. Proof from a(n) = A046521(n, 0). See a comment there.  Wolfdieter Lang, Aug 10 2017
a(n) = Sum_{k = 0..n} (1)^k * binomial(2*n+1, k) for n in N.  Rene Adad, Sep 30 2017
a(n) = A034870(n,n).  Franck Maminirina Ramaharo, Nov 26 2018


EXAMPLE

G.f.: 1 + 2*x + 6*x^2 + 20*x^3 + 70*x^4 + 252*x^5 + 924*x^6 + ...
For n=2, a(2) = 4!/(2!)^2 = 24/4 = 6, and this is the middle coefficient of the binomial expansion (a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4.  Michael B. Porter, Jul 06 2016


MAPLE

A000984 := n> binomial(2*n, n); seq(A000984(n), n=0..30);
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:=(1sqrt(1z))*4^n/sqrt(1z): 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


MATHEMATICA

Table[Binomial[2n, n], {n, 0, 24}] (* Alonso del Arte, Nov 10 2005 *)
CoefficientList[Series[1/Sqrt[14x], {x, 0, 25}], x] (* Harvey P. Dale, Mar 14 2011 *)


PROG

(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
(Haskell)
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 */
(Python)
from __future__ import division
A000984_list, b = [1], 1
for n in range(10**3):
b = b*(4*n+2)//(n+1)
A000984_list.append(b) # Chai Wah Wu, Mar 04 2016
(GAP) List([1..1000], n > Binomial(2*n, n)); # Muniru A Asiru, Jan 30 2018


CROSSREFS

Cf. A000108, A002420, A002457, A030662, A002144, A135091, A152229, A158815, A081696, A205946, A182400. Differs from A071976 at 10th 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.
Cf. A258290 (arithmetic derivative). Cf. A098616, A214377.
See A261009 for a conjecture about this sequence.
Cf. A046521 (first column).
The Apérylike numbers [or Apérylike sequences, Aperylike numbers, Aperylike sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Aperylike" is not welldefined.)
Sequence in context: A302645 A071976 A302646 * A087433 A119373 A151284
Adjacent sequences: A000981 A000982 A000983 * A000985 A000986 A000987


KEYWORD

nonn,easy,core,nice,walk,changed


AUTHOR

N. J. A. Sloane


STATUS

approved



