cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A001700 a(n) = binomial(2*n+1, n+1): number of ways to put n+1 indistinguishable balls into n+1 distinguishable boxes = number of (n+1)-st degree monomials in n+1 variables = number of monotone maps from 1..n+1 to 1..n+1.

Original entry on oeis.org

1, 3, 10, 35, 126, 462, 1716, 6435, 24310, 92378, 352716, 1352078, 5200300, 20058300, 77558760, 300540195, 1166803110, 4537567650, 17672631900, 68923264410, 269128937220, 1052049481860, 4116715363800, 16123801841550, 63205303218876, 247959266474052
Offset: 0

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To show for example that C(2n+1, n+1) is the number of monotone maps from 1..n + 1 to 1..n + 1, notice that we can describe such a map by a nondecreasing sequence of length n + 1 with entries from 1 to n + 1. The number k of increases in this sequence is anywhere from 0 to n. We can specify these increases by throwing k balls into n+1 boxes, so the total is Sum_{k = 0..n} C((n+1) + k - 1, k) = C(2*n+1, n+1).
Also number of ordered partitions (or compositions) of n + 1 into n + 1 parts. E.g., a(2) = 10: 003, 030, 300, 012, 021, 102, 120, 210, 201, 111. - Mambetov Bektur (bektur1987(AT)mail.ru), Apr 17 2003
Also number of walks of length n on square lattice, starting at origin, staying in first and second quadrants. - David W. Wilson, May 05 2001. (E.g., for n = 2 there are 10 walks, all starting at 0, 0: 0, 1 -> 0, 0; 0, 1 -> 1, 1; 0, 1 -> 0, 2; 1, 0 -> 0, 0; 1, 0 -> 1, 1; 1, 0 -> 2, 0; 1, 0 -> 1, -1; -1, 0 -> 0, 0; -1, 0 -> -1, 1; -1, 0-> -2, 0.)
Also total number of leaves in all ordered trees with n + 1 edges.
Also number of digitally balanced numbers [A031443] from 2^(2*n+1) to 2^(2*n+2). - Naohiro Nomoto, Apr 07 2001
Also number of ordered trees with 2*n + 2 edges having root of even degree and nonroot nodes of outdegree 0 or 2. - Emeric Deutsch, Aug 02 2002
Also number of paths of length 2*d(G) connecting two neighboring nodes in optimal chordal graph of degree 4, G(2*d(G)^2 + 2*d(G) + 1, 2d(G) + 1), where d(G) = diameter of graph G. - S. Bujnowski (slawb(AT)atr.bydgoszcz.pl), Feb 11 2002
Define an array by m(1, j) = 1, m(i, 1) = i, m(i, j) = m(i, j-1) + m(i-1, j); then a(n) = m(n, n), diagonal of A165257 - Benoit Cloitre, May 07 2002
Also the numerator of the constant term in the expansion of cos^(2*n)(x) or sin^(2*n)(x) when the denominator is 2^(2*n-1). - Robert G. Wilson v
Consider the expansion of cos^n(x) as a linear combination of cosines of multiple angles. If n is odd, then the expansion is a combination of a*cos((2*k-1)*x)/2^(n-1) for all 2*k - 1 <= n. If n is even, then the expansion is a combination of a*cos(2k*x)/2^(n-1) terms plus a constant. "The constant term, [a(n)/2^(2n-1)], is due to the fact that [cos^2n(x)] is never negative, i.e., electrical engineers would say the average or 'dc value' of [cos^(2*n)(x)] is [a(n)/2^(2*n-1)]. The dc value of [cos^(2*n-1)(x)] on the other hand, is zero because it is symmetrical about the horizontal axis, i.e., it is negative and positive equally." Nahin[62] - Robert G. Wilson v, Aug 01 2002
Also number of times a fixed Dyck word of length 2*k occurs in all Dyck words of length 2*n + 2*k. Example: if the fixed Dyck word is xyxy (k = 2), then it occurs a(1) = 3 times in the 5 Dyck words of length 6 (n = 1): (xy[xy)xy], xyxxyy, xxyyxy, x(xyxy)y, xxxyyy (placed between parentheses). - Emeric Deutsch, Jan 02 2003
a(n+1) is the determinant of the n X n matrix m(i, j) = binomial(2*n-i, j). - Benoit Cloitre, Aug 26 2003
a(n-1) = (2*n)!/(2*n!*n!), formula in [Davenport] used by Gauss for the special case prime p = 4*n + 1: x = a(n-1) mod p and y = x*(2n)! mod p are solutions of p = x^2 + y^2. - Frank Ellermann. Example: For prime 29 = 4*7 + 1 use a(7-1) = 1716 = (2*7)!/(2*7!*7!), 5 = 1716 mod 29 and 2 = 5*(2*7)! mod 29, then 29 = 5*5 + 2*2.
The number of compositions of 2*n, say c_1 + c_2 + ... + c_k = 2n, satisfy that Sum_{i = 1..j} c_i < 2*j for all j = 1..k, or equivalently, the number of subsets, say S, of [2*n-1] = {1, 2, ..., 2*n-1} with at least n elements such that if 2k is in S, then there must be at least k elements in S smaller than 2k. E.g., a(2) = 3 because we can write 4 = 1 + 1 + 1 + 1 = 1 + 1 + 2 = 1 + 2 + 1. - Ricky X. F. Chen (ricky_chen(AT)mail.nankai.edu.cn), Jul 30 2006
The number of walks of length 2*n + 1 on an infinite linear lattice that begin at the origin and end at node (1). Also the number of paths on a square lattice from the origin to (n+1, n) that use steps (1,0) and (0,1). Also number of binary numbers of length 2*n + 1 with n + 1 ones and n zeros. - Stefan Hollos (stefan(AT)exstrom.com), Dec 10 2007
If Y is a 3-subset of an 2*n-set X then, for n >= 3, a(n-1) is the number of n-subsets of X having at least two elements in common with Y. - Milan Janjic, Dec 16 2007
Also the number of rankings (preferential arrangements) of n unlabeled elements onto n levels when empty levels are allowed. - Thomas Wieder, May 24 2008
Also the Catalan transform of A000225 shifted one index, i.e., dropping A000225(0). - R. J. Mathar, Nov 11 2008
With offset 1. The number of solutions in nonnegative integers to X1 + X2 + ... + Xn = n. The number of terms in the expansion of (X1 + X2 + ... + Xn)^n. The coefficient of x^n in the expansion of (1 + x + x^2 + ...)^n. The number of distinct image sets of all functions taking [n] into [n]. - Geoffrey Critzer, Feb 22 2009
The Hankel transform of the aerated sequence 1, 0, 3, 0, 10, 0, ... is 1, 3, 3, 5, 5, 7, 7, ... (A109613(n+1)). - Paul Barry, Apr 21 2009
Also the number of distinct network topologies for a network of n items with 1 to n - 1 unidirectional connections to other objects in the network. - Anthony Bachler, May 05 2010
Equals INVERT transform of the Catalan numbers starting with offset 1. E.g.: a(3) = 35 = (1, 2, 5) dot (10, 3, 1) + 14 = 21 + 14 = 35. - Gary W. Adamson, May 15 2009
The integral of 1/(1+x^2)^(n+1) is given by a(n)/2^(2*n - 1) * (x/(1 + x^2)^n*P(x) + arctan(x)), where P(x) is a monic polynomial of degree 2*n - 2 with rational coefficients. - Christiaan van de Woestijne, Jan 25 2011
a(n) is the number of Schroder paths of semilength n in which the (2,0)-steps at level 0 come in 2 colors and there are no (2,0)-steps at a higher level. Example: a(2) = 10 because, denoting U = (1,1), H = (1,0), and D = (1,-1), we have 2^2 = 4 paths of shape HH, 2 paths of shape HUD, 2 paths of shape UDH, and 1 path of each of the shapes UDUD and UUDD. - Emeric Deutsch, May 02 2011
a(n) is the number of Motzkin paths of length n in which the (1,0)-steps at level 0 come in 3 colors and those at a higher level come in 2 colors. Example: a(3)=35 because, denoting U = (1,1), H = (1,0), and D = (1,-1), we have 3^3 = 27 paths of shape HHH, 3 paths of shape HUD, 3 paths of shape UDH, and 2 paths of shape UHD. - Emeric Deutsch, May 02 2011
Also number of digitally balanced numbers having length 2*(n + 1) in binary representation: a(n) = #{m: A070939(A031443(m)) = 2*(n + 1)}. - Reinhard Zumkeller, Jun 08 2011
a(n) equals 2^(2*n + 3) times the coefficient of Pi in 2F1([1/2, n+2]; [3/2]; -1). - John M. Campbell, Jul 17 2011
For positive n, a(n) equals 4^(n+2) times the coefficient of Pi^2 in Integral_{x = 0..Pi/2} x sin^(2*n + 2)x. - John M. Campbell, Jul 19 2011 [Apparently, the contributor means Integral_{x = 0..Pi/2} x * (sin(x))^(2*n + 2).]
a(n-1) = C(2*n, n)/2 is the number of ways to assign 2*n people into 2 (unlabeled) groups of size n. - Dennis P. Walsh, Nov 09 2011
Equals row sums of triangle A205945. - Gary W. Adamson, Feb 01 2012
a(n-1) gives the number of n-regular sequences defined by Erdős and Gallai in 1960 in connection with the degree sequences of simple graphs. - Matuszka Tamás, Mar 06 2013
a(n) is the sum of falling diagonals of squares in the comment in A085812 (equivalent to the Cloitre formula of Aug 2002). - John Molokach, Sep 26 2013
For n > 0: largest terms of Zigzag matrices as defined in A088961. - Reinhard Zumkeller, Oct 25 2013
Also the number of different possible win/loss round sequences (from the perspective of the eventual winner) in a "best of 2*n + 1" two-player game. For example, a(2) = 10 means there are 10 different win/loss sequences in a "best of 5" game (like a tennis match in which the first player to win 3 sets, out of a maximum of 5, wins the match); the 10 sequences are WWW, WWLW, WWLLW, WLWW, WLWLW, WLLWW, LWWW, LWWLW, LWLWW, LLWWW. See also A072600. - Philippe Beaudoin, May 14 2014; corrected by Jon E. Schoenfield, Nov 23 2014
When adding 1 to the beginning of the sequence: Convolving a(n)/2^n with itself equals 2^(n+1). For example, when n = 4: convolving {1, 1/1, 3/2, 10/4, 35/8, 126/16} with itself is 32 = 2^5. - Bob Selcoe, Jul 16 2014
From Tom Copeland, Nov 09 2014: (Start)
The shifted array belongs to a family of arrays associated to the Catalan A000108 (t = 1), and Riordan, or Motzkin sums A005043 (t = 0), with the o.g.f. [1 - sqrt(1 - 4x/(1 + (1 - t)x))]/2 and inverse x*(1 - x)/[1 + (t - 1)*x*(1 - x)]. See A091867 for more info on this family. Here is t = -3 (mod signs in the results).
Let C(x) = [1 - sqrt(1-4x)]/2, an o.g.f. for the Catalan numbers A000108, with inverse Cinv(x) = x*(1-x) and P(x,t) = x/(1 + t*x) with inverse P(x, -t).
O.g.f: G(x) = [-1 + sqrt(1 + 4*x/(1 - 4*x))]/2 = -C[P(-x, 4)].
Inverse o.g.f: Ginv(x) = x*(1 + x)/(1 + 4*x*(1 + x)) = -P(Cinv(-x), -4) (shifted signed A001792). A088218(x) = 1 + G(x).
Equals A001813/2 omitting the leading 1 there. (End)
Placing n distinguishable balls into n indistinguishable boxes gives A000110(n) (the number of set partitions). - N. J. A. Sloane, Jun 19 2015
The sequence is the INVERTi transform of A049027: (1, 4, 17, 74, 326, ...). - Gary W. Adamson, Jun 23 2015
a(n) is the number of compositions of 2*n + 2 such that the sum of the elements at odd positions is equal to the sum of the elements at even positions. a(2) = 10 because there are 10 such compositions of 6: (3, 3), (1, 3, 2), (2, 3, 1), (1, 1, 2, 2), (1, 2, 2, 1), (2, 2, 1, 1), (2, 1, 1, 2), (1, 2, 1, 1, 1), (1, 1, 1, 2, 1), (1, 1, 1, 1, 1, 1). - Ran Pan, Oct 08 2015
a(n-1) is also the Schur function of the partition (n) of n evaluated at x_1 = x_2 = ... = x_n = 1, i.e., the number of semistandard Young tableaux of shape (n) (weakly increasing rows with n boxes with numbers from {1, 2, ..., n}). - Wolfdieter Lang, Oct 11 2015
Also the number of ordered (rooted planar) forests with a total of n+1 edges and no trivial trees. - Nachum Dershowitz, Mar 30 2016
a(n) is the number of sets (i1,...in) of length n so that n >= i1 >= i2 >= ...>= in >= 1. For instance, n=3 as there are only 10 such sets (3,3,3) (3,3,2) (3,3,1) (3,2,2) (3,2,1) (3,1,1) (2,2,2) (2,2,1) (2,1,1) (1,1,1,) 3,2,1 is each used 10 times respectively. - Anton Zakharov, Jul 04 2016
The repeated middle term in the odd rows of Pascal's triangle, or half the central binomial coefficient in the even rows of Pascal's triangle, n >= 2. - Enrique Navarrete, Feb 12 2018
a(n) is the number of walks of length 2n+1 from the origin with steps (1,1) and (1,-1) that stay on or above the x-axis. Equivalently, a(n) is the number of walks of length 2n+1 from the origin with steps (1,0) and (0,1) that stay in the first octant. - Alexander Burstein, Dec 24 2019
Total number of nodes summed over all Dyck paths of semilength n. - Alois P. Heinz, Mar 08 2020
a(n-1) is the determinant of the n X n matrix m(i, j) = binomial(n+i-1, j). - Fabio Visonà, May 21 2022
Let X_i be iid standard Gaussian random variable N(0,1), and S_n be the partial sum S_n = X_1+...+X_n. Then P(S_1>0,S_2>0,...,S_n>0) = a(n+1)/2^(2n-1) = a(n+1) / A004171(n+1). For example, P(S_1>0) = 1/2, P(S_1>0,S_2>0) = 3/8, P(S_1>0,S_2>0,S_3>0) = 5/16, etc. This probability is also equal to the volume of the region x_1 > 0, x_2 > -x_1, x_3 > -(x_1+x_2), ..., x_n > -(x_1+x_2+...+x_(n-1)) in the hypercube [-1/2, 1/2]^n. This also holds for the Cauchy distribution and other stable distributions with mean 0, skew 0 and scale 1. - Xiaohan Zhang, Nov 01 2022
a(n) is the number of parking functions of size n+1 avoiding the patterns 132, 213, and 321. - Lara Pudwell, Apr 10 2023
Number of vectors in (Z_>=0)^(n+1) such that the sum of the components is n+1. binomial(2*n-1, n) provides this property for n. - Michael Richard, Jun 12 2023
Also number of discrete negations on the finite chain L_n={0,1,...,n-1,n}, i.e., monotone decreasing unary operators such that N(0)=n and N(n)=0. - Marc Munar, Oct 10 2023
a(n) is the number of Dyck paths of semilength n+1 having one of its peaks marked. - Juan B. Gil, Jan 03 2024
a(n) is the dimension of the (n+1)-st symmetric power of an (n+1)-dimensional vector space. - Mehmet A. Ates, Feb 15 2024
a(n) is the independence number of the twisted odd graph O^(sigma)(n+2). - _Miquel A. Fiol, Aug 26 2024
a(n) is the number of non-descending sequences with length n and the last number is less or equal to n. a(n) is also the number of integer partitions (of any positive integer) with length n and largest part is less or equal to n. - Zlatko Damijanic, Dec 06 2024
a(n) is the number of triangulations of a once-punctured (n+1)-gon [from Fontaine & Plamondon's Theorem 3.6]. - Esther Banaian, May 06 2025

Examples

			There are a(2)=10 ways to put 3 indistinguishable balls into 3 distinguishable boxes, namely, (OOO)()(), ()(OOO)(), ()()(OOO), (OO)(O)(), (OO)()(O), (O)(OO)(), ()(OO)(O), (O)()(OO), ()(O)(OO), and (O)(O)(O). - _Dennis P. Walsh_, Apr 11 2012
a(2) = 10: Semistandard Young tableaux for partition (3) of 3 (the indeterminates x_i, i = 1, 2, 3 are omitted and only their indices are given): 111, 112, 113, 122, 123, 133, 222, 223, 233, 333. - _Wolfdieter Lang_, Oct 11 2015
		

References

  • H. Davenport, The Higher Arithmetic. Cambridge Univ. Press, 7th ed., 1999, ch. V.3 (p. 122).
  • A. Frosini, R. Pinzani, and S. Rinaldi, About half the middle binomial coefficient, Pure Math. Appl., 11 (2000), 497-508.
  • Charles Jordan, Calculus of Finite Differences, Chelsea 1965, p. 449.
  • J. C. P. Miller, editor, Table of Binomial Coefficients. Royal Society Mathematical Tables, Vol. 3, Cambridge Univ. Press, 1954.
  • Paul J. Nahin, "An Imaginary Tale, The Story of [Sqrt(-1)]," Princeton University Press, Princeton, NJ 1998, p. 62.
  • L. W. Shapiro and C. J. Wang, Generating identities via 2 X 2 matrices, Congressus Numerantium, 205 (2010), 33-46.
  • 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).

Crossrefs

Equals A000984(n+1)/2.
a(n) = (2*n+1)*Catalan(n) [A000108] = A035324(n+1, 1) (first column of triangle).
Row sums of triangles A028364, A050166, A039598.
Bisections: a(2*k) = A002458(k), a(2*k+1) = A001448(k+1)/2, k >= 0.
Other versions of the same sequence: A088218, A110556, A138364.
Diagonals 1 and 2 of triangle A100257.
Second row of array A102539.
Column of array A073165.
Row sums of A103371. - Susanne Wienand, Oct 22 2011
Cf. A002054: C(2*n+1, n-1). - Bruno Berselli, Jan 20 2014

Programs

  • GAP
    List([0..30],n->Binomial(2*n+1,n+1)); # Muniru A Asiru, Feb 26 2019
  • Haskell
    a001700 n = a007318 (2*n+1) (n+1)  -- Reinhard Zumkeller, Oct 25 2013
    
  • Magma
    [Binomial(2*n, n)/2: n in [1..40]]; // Vincenzo Librandi, Nov 10 2014
    
  • Maple
    A001700 := n -> binomial(2*n+1,n+1); seq(A001700(n), n=0..20);
    A001700List := proc(m) local A, P, n; A := [1]; P := [1];
    for n from 1 to m - 2 do P := ListTools:-PartialSums([op(P), 2*P[-1]]);
    A := [op(A), P[-1]] od; A end: A001700List(27); # Peter Luschny, Mar 24 2022
  • Mathematica
    Table[ Binomial[2n + 1, n + 1], {n, 0, 23}]
    CoefficientList[ Series[2/((Sqrt[1 - 4 x] + 1)*Sqrt[1 - 4 x]), {x, 0, 22}], x] (* Robert G. Wilson v, Aug 08 2011 *)
  • Maxima
    B(n,a,x):=coeff(taylor(exp(x*t)*(t/(exp(t)-1))^a,t,0,20),t,n)*n!;
    makelist((-1)^(n)*B(n,n+1,-n-1)/n!,n,0,10); /* Vladimir Kruchinin, Apr 06 2016 */
    
  • PARI
    a(n)=binomial(2*n+1,n+1)
    
  • PARI
    z='z+O('z^50); Vec((1/sqrt(1-4*z)-1)/(2*z)) \\ Altug Alkan, Oct 11 2015
    
  • Python
    from _future_ import division
    A001700_list, b = [], 1
    for n in range(10**3):
        A001700_list.append(b)
        b = b*(4*n+6)//(n+2) # Chai Wah Wu, Jan 26 2016
    
  • Sage
    [rising_factorial(n+1,n+1)/factorial(n+1) for n in (0..22)] # Peter Luschny, Nov 07 2011
    

Formula

a(n-1) = binomial(2*n, n)/2 = A000984(n)/2 = (2*n)!/(2*n!*n!).
D-finite with recurrence: a(0) = 1, a(n) = 2*(2*n+1)*a(n-1)/(n+1) for n > 0.
G.f.: (1/sqrt(1 - 4*x) - 1)/(2*x).
L.g.f.: log((1 - sqrt(1 - 4*x))/(2*x)) = Sum_{n >= 0} a(n)*x^(n+1)/(n+1). - Vladimir Kruchinin, Aug 10 2010
G.f.: 2F1([1, 3/2]; [2]; 4*x). - Paul Barry, Jan 23 2009
G.f.: 1/(1 - 2*x - x/(1 - x/(1 - x/(1 - x/(1 - ... (continued fraction). - Paul Barry, May 06 2009
G.f.: c(x)^2/(1 - x*c(x)^2), c(x) the g.f. of A000108. - Paul Barry, Sep 07 2009
O.g.f.: c(x)/sqrt(1 - 4*x) = (2 - c(x))/(1 - 4*x), with c(x) the o.g.f. of A000108. Added second formula. - Wolfdieter Lang, Sep 02 2012
Convolution of A000108 (Catalan) and A000984 (central binomial): Sum_{k=0..n} C(k)*binomial(2*(n-k), n-k), C(k) Catalan. - Wolfdieter Lang, Dec 11 1999
a(n) = Sum_{k=0..n} C(n+k, k). - Benoit Cloitre, Aug 20 2002
a(n) = Sum_{k=0..n} C(n, k)*C(n+1, k+1). - Benoit Cloitre, Oct 19 2002
a(n) = Sum_{k = 0..n+1} binomial(2*n+2, k)*cos((n - k + 1)*Pi). - Paul Barry, Nov 02 2004
a(n) = 4^n*binomial(n+1/2, n)/(n+1). - Paul Barry, May 10 2005
E.g.f.: Sum_{n >= 0} a(n)*x^(2*n + 1)/(2*n + 1)! = BesselI(1, 2*x). - Michael Somos, Jun 22 2005
E.g.f. in Maple notation: exp(2*x)*(BesselI(0, 2*x) + BesselI(1, 2*x)). Integral representation as n-th moment of a positive function on [0, 4]: a(n) = Integral_{x = 0..4} x^n * (x/(4 - x))^(1/2)/(2*Pi) dx, n >= 0. This representation is unique. - Karol A. Penson, Oct 11 2001
Narayana transform of [1, 2, 3, ...]. Let M = the Narayana triangle of A001263 as an infinite lower triangular matrix and V = the Vector [1, 2, 3, ...]. Then A001700 = M * V. - Gary W. Adamson, Apr 25 2006
a(n) = A122366(n,n). - Reinhard Zumkeller, Aug 30 2006
a(n) = C(2*n, n) + C(2*n, n-1) = A000984(n) + A001791(n). - Zerinvary Lajos, Jan 23 2007
a(n-1) = (n+1)*(n+2)*...*(2*n-1)/(n-1)! (product of n-1 consecutive integers, divided by (n-1)!). - Jonathan Vos Post, Apr 09 2007; [Corrected and shortened by Giovanni Ciriani, Mar 26 2019]
a(n-1) = (2*n - 1)!/(n!*(n - 1)!). - William A. Tedeschi, Feb 27 2008
a(n) = (2*n + 1)*A000108(n). - Paul Barry, Aug 21 2007
Binomial transform of A005773 starting (1, 2, 5, 13, 35, 96, ...) and double binomial transform of A001405. - Gary W. Adamson, Sep 01 2007
Row sums of triangle A132813. - Gary W. Adamson, Sep 01 2007
Row sums of triangle A134285. - Gary W. Adamson, Nov 19 2007
a(n) = 2*A000984(n) - A000108(n), that is, a(n) = 2*C(2*n, n) - n-th Catalan number. - Joseph Abate, Jun 11 2010
Conjectured: 4^n GaussHypergeometric(1/2,-n; 2; 1) -- Solution for the path which stays in the first and second quadrant. - Benjamin Phillabaum, Feb 20 2011
a(n)= Sum_{k=0..n} A038231(n,k) * (-1)^k * A000108(k). - Philippe Deléham, Nov 27 2009
Let A be the Toeplitz matrix of order n defined by: A[i,i-1] = -1, A[i,j] = Catalan(j-i), (i <= j), and A[i,j] = 0, otherwise. Then, for n >= 1, a(n) = (-1)^n * charpoly(A,-2). - Milan Janjic, Jul 08 2010
a(n) is the upper left term of M^(n+1), where M is the infinite matrix in which a column of (1,2,3,...) is prepended to an infinite lower triangular matrix of all 1's and the rest zeros, as follows:
1, 1, 0, 0, 0, ...
2, 1, 1, 0, 0, ...
3, 1, 1, 1, 0, ...
4, 1, 1, 1, 1, ...
...
Alternatively, a(n) is the upper left term of M^n where M is the infinite matrix:
3, 1, 0, 0, 0, ...
1, 1, 1, 0, 0, ...
1, 1, 1, 1, 0, ...
1, 1, 1, 1, 1, ...
...
- Gary W. Adamson, Jul 14 2011
a(n) = (n + 1)*hypergeom([-n, -n], [2], 1). - Peter Luschny, Oct 24 2011
a(n) = Pochhammer(n+1, n+1)/(n+1)!. - Peter Luschny, Nov 07 2011
E.g.f.: 1 + 6*x/(U(0) - 6*x); U(k) = k^2 + (4*x + 3)*k + 6*x + 2 - 2*x*(k + 1)*(k + 2)*(2*k + 5)/U(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 18 2011
a(n) = 2*A000984(n) - A000108(n). [Abate & Whitt]
a(n) = 2^(2*n+1)*binomial(n+1/2, -1/2). - Peter Luschny, May 06 2014
For n > 1: a(n-1) = A166454(2*n, n), central terms in A166454. - Reinhard Zumkeller, Mar 04 2015
a(n) = 2*4^n*Gamma(3/2 + n)/(sqrt(Pi)*Gamma(2+n)). - Peter Luschny, Dec 14 2015
a(n) ~ 2*4^n*(1 - (5/8)/n + (73/128)/n^2 - (575/1024)/n^3 + (18459/32768)/n^4)/sqrt(n*Pi). - Peter Luschny, Dec 16 2015
a(n) = (-1)^(n)*B(n, n+1, -n-1)/n!, where B(n,a,x) is a generalized Bernoulli polynomial. - Vladimir Kruchinin, Apr 06 2016
a(n) = Gamma(2 + 2*n)/(n!*Gamma(2 + n)). Andres Cicuttin, Apr 06 2016
a(n) = (n + (n + 1))!/(Gamma(n)*Gamma(1 + n)*A002378(n)), for n > 0. Andres Cicuttin, Apr 07 2016
From Ilya Gutkovskiy, Jul 04 2016: (Start)
Sum_{n >= 0} 1/a(n) = 2*(9 + 2*sqrt(3)*Pi)/27 = A248179.
Sum_{n >= 0} (-1)^n/a(n) = 2*(5 + 4*sqrt(5)*arcsinh(1/2))/25 = 2*(5*A145433 - 1).
Sum_{n >= 0} (-1)^n*a(n)/n! = BesselI(2,2)*exp(-2) = A229020*A092553. (End)
Conjecture: a(n) = Sum_{k=2^n..2^(n+1)-1} A178244(k). - Mikhail Kurkov, Feb 20 2021
a(n-1) = 1 + (1/n)*Sum_{t=1..n/2} (2*cos((2*t-1)*Pi/(2*n)))^(2*n). - Greg Dresden, Oct 11 2022
a(n) = Product_{1 <= i <= j <= n} (i + j + 1)/(i + j - 1). Cf. A006013. - Peter Bala, Feb 21 2023
Sum_{n >= 0} a(n)*x^(n+1)/(n+1) = x + 3*x^2/2 + 10*x^3/3 + 35*x^4/4 + ... = the series reversion of exp(-x)*(1 - exp(-x)). - Peter Bala, Sep 06 2023

Extensions

Name corrected by Paul S. Coombes, Jan 11 2012
Name corrected by Robert Tanniru, Feb 01 2014

A088218 Total number of leaves in all rooted ordered trees with n edges.

Original entry on oeis.org

1, 1, 3, 10, 35, 126, 462, 1716, 6435, 24310, 92378, 352716, 1352078, 5200300, 20058300, 77558760, 300540195, 1166803110, 4537567650, 17672631900, 68923264410, 269128937220, 1052049481860, 4116715363800, 16123801841550, 63205303218876, 247959266474052
Offset: 0

Views

Author

Michael Somos, Sep 24 2003

Keywords

Comments

Essentially the same as A001700, which has more information.
Note that the unique rooted tree with no edges has no leaves, so a(0)=1 is by convention. - Michael Somos, Jul 30 2011
Number of ordered partitions of n into n parts, allowing zeros (cf. A097070) is binomial(2*n-1,n) = a(n) = essentially A001700. - Vladeta Jovovic, Sep 15 2004
Hankel transform is A000027; example: Det([1,1,3,10;1,3,10,35;3,10,35,126; 10,35,126,462]) = 4. - Philippe Deléham, Apr 13 2007
a(n) is the number of functions f:[n]->[n] such that for all x,y in [n] if xA045992(n). - Geoffrey Critzer, Apr 02 2009
Hankel transform of the aeration of this sequence is A000027 doubled: 1,1,2,2,3,3,... - Paul Barry, Sep 26 2009
The Fi1 and Fi2 triangle sums of A039599 are given by the terms of this sequence. For the definitions of these triangle sums see A180662. - Johannes W. Meijer, Apr 20 2011
Alternating row sums of Riordan triangle A094527. See the Philippe Deléham formula. - Wolfdieter Lang, Nov 22 2012
(-2)*a(n) is the Z-sequence for the Riordan triangle A110162. For the notion of Z- and A-sequences for Riordan arrays see the W. Lang link under A006232 with details and references. - Wolfdieter Lang, Nov 22 2012
From Gus Wiseman, Jun 27 2021: (Start)
Also the number of integer compositions of 2n with alternating (or reverse-alternating) sum 0 (ranked by A344619). This is equivalent to Ran Pan's comment at A001700. For example, the a(0) = 1 through a(3) = 10 compositions are:
() (11) (22) (33)
(121) (132)
(1111) (231)
(1122)
(1221)
(2112)
(2211)
(11121)
(12111)
(111111)
For n > 0, a(n) is also the number of integer compositions of 2n with alternating sum 2.
(End)
Number of terms in the expansion of (x_1+x_2+...+x_n)^n. - César Eliud Lozada, Jan 08 2022

Examples

			G.f. = 1 + x + 3*x^2 + 10*x^3 + 35*x^4 + 126*x^5 + 462*x^6 + 1716*x^7 + ...
The five rooted ordered trees with 3 edges have 10 leaves.
..x........................
..o..x.x..x......x.........
..o...o...o.x..x.o..x.x.x..
..r...r....r....r.....r....
		

References

  • L. W. Shapiro and C. J. Wang, Generating identities via 2 X 2 matrices, Congressus Numerantium, 205 (2010), 33-46.

Crossrefs

Same as A001700 modulo initial term and offset.
First differences are A024718.
Main diagonal of A071919 and of A305161.
A signed version is A110556.
A000041 counts partitions of 2n with alternating sum 0, ranked by A000290.
A003242 counts anti-run compositions.
A025047 counts wiggly compositions (ascend: A025048, descend: A025049).
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A106356 counts compositions by number of maximal anti-runs.
A124754 gives the alternating sum of standard compositions.
A345197 counts compositions by sum, length, and alternating sum.
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0: counted by A088218 (this sequence), ranked by A344619/A344619.
- k = 1: counted by A000984, ranked by A345909/A345911.
- k = -1: counted by A001791, ranked by A345910/A345912.
- k = 2: counted by A088218 (this sequence), ranked by A345925/A345922.
- k = -2: counted by A002054, ranked by A345924/A345923.
- k >= 0: counted by A116406, ranked by A345913/A345914.
- k <= 0: counted by A058622(n-1), ranked by A345915/A345916.
- k > 0: counted by A027306, ranked by A345917/A345918.
- k < 0: counted by A294175, ranked by A345919/A345920.
- k != 0: counted by A058622, ranked by A345921/A345921.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd: counted by A000302, ranked by A053738/A053738.

Programs

  • Magma
    [Binomial(2*n-1, n): n in [0..30]]; // Vincenzo Librandi, Aug 07 2014
  • Maple
    seq(binomial(2*n-1, n),n=0..24); # Peter Luschny, Sep 22 2014
  • Mathematica
    a[ n_] := SeriesCoefficient[(1 - x)^-n, {x, 0, n}];
    c = (1 - (1 - 4 x)^(1/2))/(2 x);CoefficientList[Series[1/(1-(c-1)),{x,0,20}],x] (* Geoffrey Critzer, Dec 02 2010 *)
    Table[Binomial[2 n - 1, n], {n, 0, 20}] (* Vincenzo Librandi, Aug 07 2014 *)
    a[ n_] := If[ n < 0, 0, With[ {m = 2 n}, m! SeriesCoefficient[ (1 + BesselI[0, 2 x]) / 2, {x, 0, m}]]]; (* Michael Somos, Nov 22 2014 *)
  • PARI
    {a(n) = sum( i=0, n, binomial(n+i-2,i))};
    
  • PARI
    {a(n) = if( n<0, 0, polcoeff( (1 + 1 / sqrt(1 - 4*x + x * O(x^n))) / 2, n))};
    
  • PARI
    {a(n) = if( n<0, 0, polcoeff( 1 / (1 - x + x * O(x^n))^n, n))};
    
  • PARI
    {a(n) = if( n<0, 0, binomial( 2*n - 1, n))};
    
  • PARI
    {a(n) = if( n<1, n==0, polcoeff( subst((1 - x) / (1 - 2*x), x, serreverse( x - x^2 + x * O(x^n))), n))};
    
  • Sage
    def A088218(n):
        return rising_factorial(n,n)/falling_factorial(n,n)
    [A088218(n) for n in (0..24)]  # Peter Luschny, Nov 21 2012
    

Formula

G.f.: (1 + 1 / sqrt(1 - 4*x)) / 2.
a(n) = binomial(2*n - 1, n).
a(n) = (n+1)*A000108(n)/2, n>=1. - B. Dubalski (dubalski(AT)atr.bydgoszcz.pl), Feb 05 2002 (in A060150)
a(n) = (0^n + C(2n, n))/2. - Paul Barry, May 21 2004
a(n) is the coefficient of x^n in 1 / (1 - x)^n and also the sum of the first n coefficients of 1 / (1 - x)^n. Given B(x) with the property that the coefficient of x^n in B(x)^n equals the sum of the first n coefficients of B(x)^n, then B(x) = B(0) / (1 - x).
G.f.: 1 / (2 - C(x)) = (1 - x*C(x))/sqrt(1-4*x) where C(x) is g.f. for Catalan numbers A000108. Second equation added by Wolfdieter Lang, Nov 22 2012.
From Paul Barry, Nov 02 2004: (Start)
a(n) = Sum_{k=0..n} binomial(2*n, k)*cos((n-k)*Pi);
a(n) = Sum_{k=0..n} binomial(n, (n-k)/2)*(1+(-1)^(n-k))*cos(k*Pi/2)/2 (with interpolated zeros);
a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)*cos((n-2*k)*Pi/2) (with interpolated zeros); (End)
a(n) = A110556(n)*(-1)^n, central terms in triangle A110555. - Reinhard Zumkeller, Jul 27 2005
a(n) = Sum_{0<=k<=n} A094527(n,k)*(-1)^k. - Philippe Deléham, Mar 14 2007
From Paul Barry, Mar 29 2010: (Start)
G.f.: 1/(1-x/(1-2x/(1-(1/2)x/(1-(3/2)x/(1-(2/3)x/(1-(4/3)x/(1-(3/4)x/(1-(5/4)x/(1-... (continued fraction);
E.g.f.: (of aerated sequence) (1 + Bessel_I(0, 2*x))/2. (End)
a(n + 1) = A001700(n). a(n) = A024718(n) - A024718(n - 1).
E.g.f.: E(x) = 1+x/(G(0)-2*x) ; G(k) = (k+1)^2+2*x*(2*k+1)-2*x*(2*k+3)*((k+1)^2)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Dec 21 2011
a(n) = Sum_{k=0..n}(-1)^k*binomial(2*n,n+k). - Mircea Merca, Jan 28 2012
a(n) = rf(n,n)/ff(n,n), where rf is the rising factorial and ff the falling factorial. - Peter Luschny, Nov 21 2012
D-finite with recurrence: n*a(n) +2*(-2*n+1)*a(n-1) = 0. - R. J. Mathar, Dec 04 2012
a(n) = hypergeom([1-n,-n],[1],1). - Peter Luschny, Sep 22 2014
G.f.: 1 + x/W(0), where W(k) = 4*k+1 - (4*k+3)*x/(1 - (4*k+1)*x/(4*k+3 - (4*k+5)*x/(1 - (4*k+3)*x/W(k+1) ))) ; (continued fraction). - Sergei N. Gladkovskii, Nov 13 2014
a(n) = A000984(n) + A001791(n). - Gus Wiseman, Jun 28 2021
E.g.f.: (1 + exp(2*x) * BesselI(0,2*x)) / 2. - Ilya Gutkovskiy, Nov 03 2021
From Amiram Eldar, Mar 12 2023: (Start)
Sum_{n>=0} 1/a(n) = 5/3 + 4*Pi/(9*sqrt(3)).
Sum_{n>=0} (-1)^n/a(n) = 3/5 - 8*log(phi)/(5*sqrt(5)), where phi is the golden ratio (A001622). (End)
a(n) ~ 2^(2*n-1)/sqrt(n*Pi). - Stefano Spezia, Apr 17 2024

A002894 a(n) = binomial(2n, n)^2.

Original entry on oeis.org

1, 4, 36, 400, 4900, 63504, 853776, 11778624, 165636900, 2363904400, 34134779536, 497634306624, 7312459672336, 108172480360000, 1609341595560000, 24061445010950400, 361297635242552100, 5445717990022688400, 82358080713306090000, 1249287673091590440000
Offset: 0

Views

Author

Keywords

Comments

a(n) is the number of monotonic paths (only moving N and E) in the lattice [0..2n] X [0..2n] that contain the points (0,0), (n,n) and (2n,2n). - Joe Keane (jgk(AT)jgk.org), Jun 06 2002
This is the Taylor expansion of a special point on a curve described by Beauville. - Matthijs Coster, Apr 28 2004
Expansion of K(k) / (Pi/2) in powers of m/16 = (k/4)^2, where K(k) is the complete elliptic integral of the first kind evaluated at k. - Michael Somos, Mar 04 2003
Square lattice walks that start and end at origin after 2n steps. - Gareth McCaughan and Michael Somos, Jun 12 2004
If A is a random matrix in USp(4) (4 X 4 complex matrices that are unitary and symplectic) then a(n)=E[(tr(A^k))^{2n}] for any k > 4. - Andrew V. Sutherland, Apr 01 2008
From R. H. Hardin, Feb 03 2016 and R. J. Mathar, Feb 18 2016: (Start)
Also, number of 2 X (2n) arrays of permutations of 2n copies of 0 or 1 with row sums equal.
For example, some solutions for n=3:
0 1 0 1 0 1 0 1 0 1 1 0 0 0 1 0 1 1 1 1 1 0 0 0
1 0 0 0 1 1 1 1 0 1 0 0 0 0 0 1 1 1 0 0 1 1 0 1
There is a simple combinatorial argument to show that this is a(n): We have 2n copies of 0's and 1's and need equal row sums. Therefore there must be n 1's in each of the two rows. Otherwise there are no constraints, so there are C(2n,n) ways of placing the 1's in the first row and independently C(2n,n) ways of placing the 1's in the second. The product is clearly C(2n,n)^2. (End)
Also the even part of the bisection of A241530. One half of the odd part is given in A000894. - Wolfdieter Lang, Sep 06 2016
From Peter Bala, Jan 26 2018: (Start)
Let S = {[1,0,0], [0,1,0], [1,0,1], [0,1,1]} be a set of four column vectors. Then a(n) equals the number of 3 X k arrays whose columns belong to the set S and whose row sums are all equal to n (apply Eger, Theorem 3). An example is given below. Equivalently, a(n) equals the number of lattice paths from (0,0,0) to (n,n,n) using steps (1,0,0), (0,1,0), (1,0,1) and (0,1,1).
The o.g.f. for the sequence equals the diagonal of the rational function 1/(1 - (x + y + x*z + y*z)).
Row sums of A069466. (End)
Also, the constant term in the expansion of (x + 1/x + y + 1/y)^(2n). - Christopher J. Smyth, Sep 26 2018
Number of ways to place 2n^2 nonattacking pawns on a 2n x 2n board. - Tricia Muldoon Brown, Dec 12 2018
For n>0, a(n) is the number of Littlewood polynomials of degree 4n-1 that have a closed Lill path. A polynomial p(x) has a closed Lill path if and only if p(x) is divisible by x^(2)+1. - Raul Prisacariu, Aug 28 2024

Examples

			G.f. = 1 + 4*x + 36*x^2 + 400*x^3 + 4900*x^4 + 63504*x^5 + 853776*x^6 + ... - _Michael Somos_, Aug 06 2014
From _Peter Bala_, Jan 26 2018: (Start)
a(2) = 36: The thirty six 3 x k arrays with columns belonging to the set of column vectors S = {[1,0,0], [0,1,0], [1,0,1], [0,1,1]} and having all row sums equal to 2 are the 6 distinct arrays obtained by permuting the columns of
  /1 1 0 0\
  |0 0 1 1|,
  \0 0 1 1/
the 6 distinct arrays obtained by permuting the columns of
  /0 0 1 1\
  |1 1 0 0|
  \0 0 1 1/
and the 24 arrays obtained by permuting the columns of
  /1 0 1 0\
  |0 1 0 1|. (End)
  \0 0 1 1/
		

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. 591,828.
  • J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 8.
  • Matthijs Coster, Over 6 families van krommen [On 6 families of curves], Master's Thesis (unpublished), Aug 26 1983.
  • 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): 339-358.
  • 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).

Crossrefs

Row sums of A069466.
Row 2 of A268367 (even terms).
Equals 4*A060150.
Cf. A000984, A000515, A010370, A054474 (INVERTi transform), A172390, A000897, A002897, A006480, A008977, A186420, A188662, A000894, A241530, A002898 (walks hex lattice).

Programs

  • Magma
    [Binomial(2*n, n)^2: n in [0..20]]; // Vincenzo Librandi, Aug 07 2014
  • Maple
    A002894 := n-> binomial(2*n,n)^2.
  • Mathematica
    CoefficientList[Series[Hypergeometric2F1[1/2, 1/2, 1, 16x], {x, 0, 20}], x]
    Table[Binomial[2n,n]^2,{n,0,20}] (* Harvey P. Dale, Jul 06 2011 *)
    a[ n_] := SeriesCoefficient[ EllipticK[16 x] / (Pi/2), {x, 0, n}]; (* Michael Somos, Aug 06 2014 *)
    a[n_] := 16^n HypergeometricPFQ[{1/2, -2 n, 2 n + 1}, {1, 1}, 1];
    Table[a[n], {n, 0, 19}] (* Peter Luschny, Mar 14 2018 *)
  • PARI
    {a(n) = binomial(2*n, n)^2};
    
  • PARI
    {a(n) = if( n<0, 0, polcoeff( polcoeff( polcoeff( 1 / (1 - x * (y + z + 1/y + 1/z)) + x * O(x^(2*n)), 2*n), 0), 0))}; /* Michael Somos, Jun 12 2004 */
    
  • Sage
    [binomial(2*n, n)**2 for n in range(17)]  # Zerinvary Lajos, Apr 21 2009
    

Formula

D-finite with recurrence: (n+1)^2*a(n+1) = 4*(2*n + 1)^2*a(n). - Matthijs Coster, Apr 28 2004
a(n) ~ Pi^(-1)*n^(-1)*2^(4*n). - Joe Keane (jgk(AT)jgk.org), Jun 06 2002
G.f.: F(1/2, 1/2; 1; 16*x) = 1 / AGM(1, (1 - 16*x)^(1/2)) = K(4*sqrt(x)) / (Pi/2), where AGM(x, y) is the arithmetic-geometric mean of Gauss and Legendre. - Michael Somos, Mar 04 2003
G.f.: 2*EllipticK(4*sqrt(x))/Pi, using Maple's convention for elliptic integrals.
E.g.f.: Sum_{n>=0} a(n)*x^(2*n)/(2*n)! = BesselI(0, 2x)^2.
a(n) = A000984(n)^2. - Jonathan Vos Post, Jun 17 2007
E.g.f.: (BesselI(0, 2*x))^2 = 1+2*x^2/(U(0)-2*x^2); U(k) = 2*x^2*(2*k+1)+(k+1)^3-2*x^2*(2*k+3)*(k+1)^3/U(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 23 2011
In generally, for (BesselI(b, 2x))^2=((x^(2*b))/(GAMMA(b+1))^2)*(1+(2*x^2)*(2*b+1)/(Q(0)-(2*x^2)*(2*b+1)); Q(k)=(2*x^2)*(2*k+2*b+1)+(k+1)*(k+b+1)*(k+2*b+1)-(2*x^2)*(k+1)*(k+b+1)*(k+2*b+1)*(2*k+2*b+3)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Nov 23 2011
G.f.: G(0)/2, where G(k)= 1 + 1/(1 - 4*(2*k+1)^2*x*(1+4*x)^2/(4*(2*k+1)^2*x*(1+4*x)^2 + (k+1)^2*(1+4*x)^2/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 01 2013
0 = +a(n)*(+393216*a(n+2) -119040*a(n+3) +6860*a(n+4)) +a(n+1)*(-16128*a(n+2) +6928*a(n+3) -465*a(n+4)) +a(n+2)*(+36*a(n+2) -63*a(n+3) +6*a(n+4)) for all n in Z. - Michael Somos, Aug 06 2014
Integral representation as the n-th moment of a positive function W(x) on (0,16), in Maple notation, W(x) = EllipticK(sqrt(1-x/16))/(2*Pi^2*sqrt(x)); a(n) = Integral_{x=0..16} x^n*W(x) dx, n>=0. The function W(x) is singular at x=0 and W(16) = 1/(16*Pi). This representation is unique since W(x) is the solution of the Hausdorff moment problem. - Stanley Smith and Karol A. Penson, Jun 19 2015
a(n) ~ 16^n*(2-2/(8*n+2)^2+21/(8*n+2)^4-671/(8*n+2)^6+45081/(8*n+2)^8)^2/((4*n+1)* Pi). - Peter Luschny, Oct 14 2015
a(n) = binomial(2*n,n)*binomial(2*n,n) = ( [x^n](1 + x)^(2*n) ) *( [x^n](1 + x)^(2*n) ) = [x^n](F(x)^(4*n)), where F(x) = 1 + x + x^2 + 4*x^3 + 20*x^4 + 120*x^5 + 798*x^6 + 5697*x^7 + ... appears to have integer coefficients. For similar results see A000897, A002897, A006480, A008977, A186420 and A188662. - Peter Bala, Jul 14 2016
a(n) = Sum_{k = 0..n} binomial(2*n + k,k)*binomial(n,k)^2. Cf. A005258(n) = Sum_{k = 0..n} binomial(n + k,k)*binomial(n,k)^2. - Peter Bala, Jul 27 2016
a(n) = A241530(2*n), n >= 0. - Wolfdieter Lang, Sep 06 2016
E.g.f.: 2F2(1/2,1/2; 1,1; 16*x). - Ilya Gutkovskiy, Jan 23 2018
a(n) = 16^n*hypergeom([1/2, -2*n, 2*n + 1], [1, 1], 1). - Peter Luschny, Mar 14 2018
The right-hand side of the binomial coefficient identity Sum_{k = 0..n} C(n,k)*C(n+k,k)*C(2*n+2*k,n+k)*(-4)^(n-k) = a(n). - Peter Bala, Mar 16 2018
a(n) = [x^n] (1 - x)^(2*n) * P(2*n,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. Compare with A245086(n) = [x^n] (1 - x)^(2*n) * P(n,(1 + x)/(1 - x)). - Peter Bala, Mar 23 2022
a(n) = Sum_{k=0..n} multinomial(2n [k k (n-k) (n-k)]), which is another way to count random walks on Z^2, with steps of (0,+-1) or (+-1,0), that return to the point of origin after 2n steps (not necessarily for the first time), as is C(2n,n)^2. - Shel Kaphan, Jan 12 2023
0 = a(n)*(+393216*a(n+2) -119040*a(n+3) +6860*a(n+4)) +a(n+1)*(-16128*a(n+2) +6928*a(n+3) -465*a(n+4)) +a(n+2)*(+36*a(n+2) -63*a(n+3) +6*a(n+4)) for n>=0. - Michael Somos, May 30 2023
From Peter Bala, Sep 12 2023: (Start)
Right-hand side of the binomial coefficient identities
1) Sum_{k = 0..n} (-1)^(n+k) * C(n,k)*C(n+k,n)*C(2*n+k,n) = a(n).
2) 2*Sum_{k = 0..n} (-1)^(n+k) * C(n,k)*C(n+k-1,n)*C(2*n+k-1,n) = a(n) for n >= 1.
3) (4/3)*Sum_{k = 0..n} (-1)^(n+k) * C(n,k)*C(n+k,n)*C(2*n+k-1,n) = a(n) for n >= 1. (End)

Extensions

Edited by N. J. A. Sloane, Feb 18 2016

A005566 Number of walks of length n on square lattice, starting at origin, staying in first quadrant.

Original entry on oeis.org

1, 2, 6, 18, 60, 200, 700, 2450, 8820, 31752, 116424, 426888, 1585584, 5889312, 22084920, 82818450, 312869700, 1181952200, 4491418360, 17067389768, 65166397296, 248817153312, 953799087696, 3656229836168, 14062422446800, 54086240180000, 208618354980000
Offset: 0

Views

Author

Keywords

Comments

a(n) is the number of involutions of length 2n which are invariant under the reverse-complement map and have no decreasing subsequences of length 5. - Eric S. Egge, Oct 21 2008

Examples

			G.f. = 1 + 2*x + 6*x^2 + 18*x^3 + 60*x^4 + 200*x^5 + 700*x^6 + 2450*x^7 + ... - _Michael Somos_, Oct 17 2019
		

References

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

Crossrefs

a(2*n) = A000894(n), a(2*n+1) = 2*A060150(n+1).

Programs

  • Magma
    [Binomial(n, Floor(n/2))*Binomial(n+1, Floor((n+1)/2)): n in [0..30]]; // Vincenzo Librandi, Feb 18 2015
  • Mathematica
    f[n_] := Binomial[n, Floor[n/2]] Binomial[n + 1, Floor[(n + 1)/2]]; Array[f, 25, 0] (* Robert G. Wilson v *)

Formula

a(n) = binomial(n, floor(n/2))*binomial(n+1, floor((n+1)/2)).
E.g.f.: (BesselI(0, 2*x)+BesselI(1, 2*x))^2. - Vladeta Jovovic, Apr 28 2003
EXPCONV of A001405 with itself, i.e., a(n) = sum_{k=0}^n binomial(n,k)*A001405(k)*A001405(n-k). - Max Alekseyev, May 18 2006
G.f.: (16*x^2-1)*hypergeom([3/2, 3/2],[2],16*x^2) + (1/(2x)+2)*hypergeom([1/2, 1/2],[1],16*x^2) - 1/(2x). - Mark van Hoeij, Oct 13 2009
G.f.: (hypergeom([1/2,1/2],[1],16*x^2) - 1)/(2*x) + hypergeom([1/2,3/2],[2],16*x^2). - Mark van Hoeij, Aug 14 2014
a(n) = A241530(n)*2*floor(n/2)/(floor(n/2)+1). - Peter Luschny, Apr 25 2014
D-finite with recurrence (n+2)*(n+1)*a(n) +4*(-2*n-1)*a(n-1) -16*n*(n-1)*a(n-2)=0. - R. J. Mathar, Mar 07 2015
0 = a(n)*(+16*a(n+2) -6*a(n+3)) +a(n+1)*(-2*a(n+2) +a(n+3)) if n >= 0. - Michael Somos, Oct 17 2019
a(n) = binomial(floor(n + 1/2), floor(n/2)) * binomial(ceiling(n + 1/2), ceiling(n/2)). - Peter Luschny, Dec 14 2024

Extensions

Additional comments from David W. Wilson, May 05 2001
a(25)-a(26) from Vincenzo Librandi, Feb 18 2015

A018224 a(n) = binomial(n, floor(n/2))^2 = A001405(n)^2.

Original entry on oeis.org

1, 1, 4, 9, 36, 100, 400, 1225, 4900, 15876, 63504, 213444, 853776, 2944656, 11778624, 41409225, 165636900, 590976100, 2363904400, 8533694884, 34134779536, 124408576656, 497634306624, 1828114918084, 7312459672336, 27043120090000, 108172480360000, 402335398890000
Offset: 0

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Author

Keywords

Comments

a(n) is also the number of rooted two-vertex (or, dually, two-face) regular planar maps of valency n+1. - Valery A. Liskovets, Oct 19 2005
If A is a random matrix in USp(4) (4 X 4 complex matrices that are unitary and symplectic), then a(n)=(-1)^n*E[(tr(A^4))^n]. - Andrew V. Sutherland, Apr 01 2008
Number of square lattice walks with unit steps in all four directions (NSWE), starting at the origin, ending on the y-axis, and never going below the x-axis. Row sums of A378061. - Peter Luschny, Dec 08 2024

Examples

			The 9 lattice walks defined in the comments: 'NNN', 'NNS', 'NSN', 'NWE', 'NEW', 'WNE', 'WEN', 'ENW', 'EWN'.
		

Crossrefs

Programs

  • Maple
    s := x -> (1+x)*EllipticK(x)/(x*Pi/2)-1/x:
    seq(4^n*coeff(series(s(x),x,n+2),x,n),n=0..23); # Peter Luschny, Oct 14 2015
  • Mathematica
    (* Note that Mathematica uses a different definition of the EllipticK function. *)
    CoefficientList[Series[(-Pi + (2 + 8 x) EllipticK[16 x^2])/(4 Pi x), {x,0,23}], x] (* Peter Luschny, Oct 14 2015 *)
    Table[Binomial[n,Floor[n/2]]^2,{n,0,30}] (* Harvey P. Dale, Dec 02 2022 *)
  • PARI
    vector(50, n, n--; binomial(n, n\2)^2) \\ Altug Alkan, Oct 14 2015

Formula

E.g.f.: BesselI(0, 2*x)*(BesselI(0, 2*x)+BesselI(1, 2*x)). - Vladeta Jovovic, Jun 12 2005
G.f. (1+1/(4*x))*hypergeom([1/2, 1/2],[1],16*x^2)-1/(4*x). - Mark van Hoeij, Oct 13 2009
a(n) = (n!/(floor(n/2)!*floor((n+1)/2)!))^2. - Peter Luschny, Apr 29 2014
a(n) = A056040(n) * A056040(n+1) / (n+1). - Peter Luschny, Apr 29 2014
a(n) = 4^n*[x^n]((1+x)*EllipticK(x)/(x*Pi/2)-1/x). - Peter Luschny, Oct 14 2015
a(n) ~ 4^n*((2*n+3)/(2*n+1))^((-1)^n/2)/((n+1)*Pi/2). - Peter Luschny, Oct 14 2015
a(n) = Sum_{k=0..n} (-1)^k*binomial(n,k)*C(k)*binomial(2*n-2*k,n-k) where C(k) are Catalan numbers (A000108), see Prodinger. - Michel Marcus, Nov 19 2019
From Peter Bala, Jul 03 2023: (Start)
Right hand side of the binomial sum identity (1/2)*Sum_{k = 0..n+1} (-1)^k*4^(n+1-k)*binomial(n+1,k)*binomial(n+k,k)*binomial(2*k,k) = a(n).
a(n) = (1/2)*4^(n+1) * hypergeom([n+1, -n-1, 1/2], [1, 1], 1).
P-recursive:
(2*n - 1)*(n + 1)^2*a(n) = 4*(2*n^2 - 1)*a(n-1) + 16*(2*n + 1)*(n - 1)^2*a(n-2) with a(0) = a(1) = 1. (End)

A135389 Number of walks of length 2*n+2 from origin to (1,1) in a square lattice.

Original entry on oeis.org

2, 24, 300, 3920, 52920, 731808, 10306296, 147232800, 2127513960, 31031617760, 456164781072, 6749962774464, 100445874620000, 1502052155856000, 22557604697766000, 340044833169460800, 5143178101688094600
Offset: 0

Views

Author

Stefan Hollos (stefan(AT)exstrom.com), Dec 11 2007

Keywords

Comments

a(n) is the number of walks of length 2n+2 in an infinite square lattice that begin at the origin and end at (1,1) using steps (1,0), (-1,0), (0,1), (0,-1).

Examples

			G.f. = 2 + 24*x + 300*x^2 + 3920*x^3 + 731808*x^4 + 10306296*x^5 + ... - _Michael Somos_, Oct 17 2019
		

Crossrefs

Programs

  • Maple
    series( 2*hypergeom([3/2, 3/2],[3],16*x), x=0, 20);  # Mark van Hoeij, Apr 06 2013
  • Mathematica
    Table[Binomial[2n + 2, n] Binomial[2n + 2, n + 1], {n, 0, 19}] (* Alonso del Arte, Apr 06 2013 *)

Formula

a(n) = binomial(2n+2,n) * binomial(2n+2,n+1) = A001791(n+1)*A000984(n+1).
G.f.: 2*2F1(3/2,3/2; 3; 16*x). - Mark van Hoeij, Apr 06 2013
D-finite with recurrence n*(n+2)*a(n) -4*(2*n+1)^2*a(n-1)=0. - R. J. Mathar, Jul 14 2013
E.g.f.: Sum_{n>0} a(n-1) * x^(2*n)/(2*n)! = BesselI(1, 2*x)^2. - Michael Somos, Oct 17 2019

A209349 Number A(n,k) of initially rising meander words, where each letter of the cyclic k-ary alphabet occurs n times; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 5, 1, 0, 1, 1, 1, 9, 29, 1, 0, 1, 1, 1, 11, 100, 182, 1, 0, 1, 1, 1, 16, 182, 1225, 1198, 1, 0, 1, 1, 1, 19, 484, 3542, 15876, 8142, 1, 0, 1, 1, 1, 25, 902, 17956, 76258, 213444, 56620, 1, 0, 1
Offset: 0

Views

Author

Alois P. Heinz, Mar 06 2012

Keywords

Comments

In a meander word letters of neighboring positions have to be neighbors in the alphabet, where in a cyclic alphabet the first and the last letters are considered neighbors too. The words are not considered cyclic here.
A word is initially rising if it is empty or if it begins with the first letter of the alphabet that can only be followed by the second letter in this word position.
A(n,k) is also the number of (n*k-1)-step walks on k-dimensional cubic lattice from (1,0,...,0) to (n,n,...,n) with positive unit steps in all dimensions such that the indices of dimensions used in consecutive steps differ by 1 or are in the set {1,k}.

Examples

			A(0,0) = A(0,k) = A(n,0) = 1: the empty word.
A(1,1) = 1 = |{a}|.
A(2,1) = 0 = |{ }|.
A(2,2) = 1 = |{abab}|.
A(2,3) = 5 = |{abacbc, abcabc, abcacb, abcbac, abcbca}|.
A(1,4) = 1 = |{abcd}|.
A(2,4) = 9 = |{ababcdcd, abadcbcd, abadcdcb, abcbadcd, abcbcdad, abcdabcd, abcdadcb, abcdcbad, abcdcdab}|.
Square array A(n,k) begins:
1,  1,  1,    1,      1,       1,        1, ...
1,  1,  1,    1,      1,       1,        1, ...
1,  0,  1,    5,      9,      11,       16, ...
1,  0,  1,   29,    100,     182,      484, ...
1,  0,  1,  182,   1225,    3542,    17956, ...
1,  0,  1, 1198,  15876,   76258,   749956, ...
1,  0,  1, 8142, 213444, 1753522, 33779344, ...
		

Crossrefs

Rows n=0+1, 2-3 give: A000012, A209350, A240954.
Columns k=0+2, 3-7 give: A000012, A190917 = A110706/6, A060150 = A088218^2, A209351, A209352, A209353.

Programs

  • Maple
    b:= proc() option remember; local n; n:= nargs;
         `if`({args}={0}, 1,
           `if`(args[2]>0, b(args[2]-1, args[i]$i=3..n, args[1]), 0)+
           `if`(n>2 and args[n]>0, b(args[n]-1, args[i]$i=1..n-1), 0))
        end:
    A:= (n, k)-> `if`(n<2, 1, `if`(k<2, 1-k, b((n-1)$2, n$(k-2)))):
    seq(seq(A(n, d-n), n=0..d), d=0..10);
  • Mathematica
    b[args_List] := b[args] = Module[{n = Length[args]}, If[Union[args] == {0}, 1, If[args[[2]] > 0, b[Join[{args[[2]] - 1}, args[[3 ;; n]], { args[[1]]}]], 0] + If[n > 2 && args[[n]] > 0, b[Join[{args[[n]] - 1}, args[[1 ;; n - 1]]]], 0]]]; A[n_, k_] := If[n < 2, 1, If[k < 2, 1 - k, b[Join[{n - 1, n - 1}, Array[n&, k - 2]]]]]; Table[Table[A[n, d - n], {n, 0, d}], {d, 0, 10}] // Flatten (* Jean-François Alcover, Jan 21 2015, after Alois P. Heinz *)

A078817 Table by antidiagonals giving variants on Catalan sequence: T(n,k)=C(2n,n)*C(2k,k)*(2k+1)/(n+k+1).

Original entry on oeis.org

1, 3, 1, 10, 4, 2, 35, 15, 9, 5, 126, 56, 36, 24, 14, 462, 210, 140, 100, 70, 42, 1716, 792, 540, 400, 300, 216, 132, 6435, 3003, 2079, 1575, 1225, 945, 693, 429, 24310, 11440, 8008, 6160, 4900, 3920, 3080, 2288, 1430, 92378, 43758, 30888, 24024, 19404
Offset: 0

Views

Author

Henry Bottomley, Dec 07 2002

Keywords

Examples

			Rows start:
     1,     3,    10,    35,   126,   462,  1716,
     1,     4,    15,    56,   210,   792,  3003,
     2,     9,    36,   140,   540,  2079,  8008,
     5,    24,   100,   400,  1575,  6160, 24024,
    14,    70,   300,  1225,  4900, 19404, 76440,
    42,   216,   945,  3920, 15876, 63504,252252,
   132,   693,  3080, 12936, 52920,213444,853776,
etc.
		

Crossrefs

Columns include A000108 (catalan), A038629, A078818 and A078819. Rows include A001700, A001791, A007946 and A078820. Diagonals include A002894 and A060150.
Essentially a reflected version of A033820.

Programs

  • Maple
    A078817 := proc(n,k)
        binomial(2*n,n)*binomial(2*k,k)*(2*k+1)/(n+k+1) ;
    end proc: # R. J. Mathar, Dec 06 2018

Formula

T(n, k) = A000984(n)*A002457(k)/(n+k+1) = T(k, n)*(2k+1)/(2n+1).

A337900 The number of walks of length 2n on the square lattice that start from the origin (0,0) and end at the vertex (2,0).

Original entry on oeis.org

1, 16, 225, 3136, 44100, 627264, 9018009, 130873600, 1914762564, 28210561600, 418151049316, 6230734868736, 93271169290000, 1401915345465600, 21147754404155625, 320042195924198400, 4857445984927644900, 73916947787011560000, 1127482124965160372100
Offset: 1

Views

Author

R. J. Mathar, Sep 29 2020

Keywords

Examples

			a(2) = 16 counts the walks RRRL, RRLR, RLRR, LRRR, RRUD, RRDU, RDRU, RURD, RUDR, RDUR, URRD, DRRU, URDR, DRUR, UDRR, DURR of length 4.
		

Crossrefs

Cf. A002894 (at (0,0)), A060150 (at (1,0)), A135389 (at (1,1)), A337901 (at (3,0)), A337902 (at (2,1)).
Cf. A001791.

Programs

  • Maple
    egf := BesselI(0, 2*x)*BesselI(2, 2*x): ser := series(egf, x, 40):
    seq((2*n)!*coeff(ser, x, 2*n), n = 1..19);  # Peter Luschny, Dec 05 2024

Formula

a(n) = [A001791(n)]^2.
G.f.: x*4F3(3/2, 3/2, 2, 2; 1, 3, 3; 16*x).
D-finite with recurrence (n-1)^2*(n+1)^2*a(n) - 4*n^2*(2*n-1)^2*a(n-1) = 0.
a(n) = (2n)!*[x^(2n)] BesselI(0, 2x)*BesselI(2, 2x). - Peter Luschny, Dec 05 2024

A361713 a(n) = Sum_{k = 0..n-1} binomial(n,k)^2 * binomial(n+k-1,k)^2.

Original entry on oeis.org

0, 1, 17, 406, 10257, 268126, 7213166, 198978074, 5609330705, 161095277710, 4700175389142, 138986764820410, 4157185583199534, 125568602682092818, 3825026187780837266, 117376010145070696906, 3625095243230562818065, 112596592142021739522670, 3514965607470183733302470
Offset: 0

Views

Author

Peter Bala, Mar 21 2023

Keywords

Comments

Conjecture 1: the supercongruence a(p) == a(1) (mod p^5) holds for all primes p >= 7 (checked up to p = 199).
Conjecture 2: for r >= 2, the supercongruence a(p^r) == a(p^(r-1)) (mod p^(4*r+1)) holds for all primes p >= 7.
Compare with the Apéry numbers A005259(n) = Sum_{k = 0..n} binomial(n,k)^2 * binomial(n+k,k)^2, which satisfy the weaker supercongruences A005259(p^r) == A005259(p^(r-1)) (mod p^(3*r)) for all positive integers r and all primes p >= 5.

Crossrefs

Programs

  • Maple
    seq(add(binomial(n,k)^2*binomial(n+k-1,k)^2, k = 0..n-1), n = 0..25);
    # Alternative:
    A361713 := n -> hypergeom([-n, -n, n, n], [1, 1, 1], 1) - binomial(2*n - 1, n)^2:
    seq(simplify(A361713(n)), n = 0..18); # Peter Luschny, Mar 27 2023
  • Mathematica
    A361713[n_] := HypergeometricPFQ[{-n, -n, n, n}, {1, 1, 1}, 1] - Binomial[2*n-1, n]^2; Array[A361713, 20, 0] (* Paolo Xausa, Jul 11 2024 *)

Formula

a(n) = (1/3)*(A005259(n) + A005259(n-1)) - (1/4)*binomial(2*n,n)^2 = A177316(n) - A060150(n).
a(n) ~ C*(12*sqrt(2) + 17)^n/n^(3/2), where C = 1/(2^(5/4)*Pi^(3/2)).
a(n) = hypergeom([-n, -n, n, n], [1, 1, 1], 1) - binomial(2*n-1, n)^2. This is another way to write the first formula. - Peter Luschny, Mar 27 2023
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