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.

Showing 1-10 of 103 results. Next

A171651 Triangle T, read by rows : T(n,k) = A007318(n,k)*A005773(n+1-k).

Original entry on oeis.org

1, 2, 1, 5, 4, 1, 13, 15, 6, 1, 35, 52, 30, 8, 1, 96, 175, 130, 50, 10, 1, 267, 576, 525, 260, 75, 12, 1, 750, 1869, 2016, 1225, 455, 105, 14, 1, 2123, 6000, 7476, 5376, 2450, 728, 140, 16, 1, 6046, 19107, 27000, 22428, 12096, 4410, 1092, 180, 18, 1
Offset: 0

Views

Author

Philippe Deléham, Dec 14 2009

Keywords

Examples

			Triangle begins:
   1;
   2,   1;
   5,   4,  1;
  13,  15,  6, 1;
  35,  52, 30, 8, 1;
  ...
		

Crossrefs

Programs

  • Maple
    b:= proc(u, d, t) option remember; `if`(u=0 and d=0, 1/2,
          expand(`if`(u=0, 0, b(u-1, d, 2)*`if`(t=3, x, 1))
          +`if`(d=0, 0, b(u, d-1, `if`(t=2, 3, 1)))))
        end:
    T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n+1$2, 1)):
    seq(T(n), n=0..12);  # Alois P. Heinz, Apr 29 2015
    # second program:
    A171651:= (n, k)-> binomial(n,k)*add((-1)^(n-k-j)*binomial(n-k,j)*binomial(2*j+1,j+1),j=0..n-k): seq(print(seq(A171651(n, k), k=0..n)), n=0..9);  # Mélika Tebni, Dec 16 2023
  • Mathematica
    b[u_, d_, t_] := b[u, d, t] = If[u == 0 && d == 0, 1/2, Expand[If[u == 0, 0, b[u-1, d, 2]*If[t == 3, x, 1]] + If[d == 0, 0, b[u, d-1, If[t == 2, 3, 1]]]]];
    T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 0, n}]][b[n+1, n+1, 1] ];
    Table[T[n], {n, 0, 12}] // Flatten (* Jean-François Alcover, May 21 2016, after Alois P. Heinz *)

Formula

Sum_{k, 0<=k<=n} T(n,k)*x^k = A168491(n), A099323(n), A001405(n), A005773(n+1), A001700(n), A026378(n+1), A005573(n), A122898(n) for x = -3, -2, -1, 0, 1, 2, 3, 4 respectively.
E.g.f. of column k: exp(x)*(BesselI(0,2*x)+BesselI(1,2*x))*x^k / k!. - Mélika Tebni, Dec 16 2023

Extensions

Corrected by Philippe Deléham, Dec 18 2009

A171488 Riordan array (f(x), x*f(x)) where f(x) is the g.f. of A005773(n+1)= 1,2,5,13,35,96,267,...

Original entry on oeis.org

1, 2, 1, 5, 4, 1, 13, 14, 6, 1, 35, 46, 27, 8, 1, 96, 147, 107, 44, 10, 1, 267, 462, 396, 204, 65, 12, 1, 750, 1437, 1404, 858, 345, 90, 14, 1, 2123, 4438, 4835, 3388, 1625, 538, 119, 16, 1, 6046, 13637, 16305, 12802, 7072, 2805, 791, 152, 18, 1
Offset: 0

Views

Author

Philippe Deléham, Dec 10 2009

Keywords

Comments

Equal to A064189*B = B*A054336 = B^(-1)*A035324, B = A007318.

Examples

			Triangle T(n,k) (0<=k<=n) begins:
   1;
   2,   1;
   5,   4,   1;
  13,  14,   6,  1;
  35,  46,  27,  8,  1;
  96, 147, 107, 44, 10, 1;
  ...
		

Crossrefs

Programs

  • Maxima
    T(n,k)=((k+1)*sum(binomial(2*j+k,j)*(-1)^j*3^(n-j-k)*binomial(n+1,j+k+1),j,0,n-k))/(n+1); /* Vladimir Kruchinin Sep 30 2020 */

Formula

Sum_{k, 0<=k<=n} T(n,k)*x^k = A005043(n), A001006(n), A005773(n+1), A059738(n) for x = -2, -1, 0, 1 respectively.
T(n,k) = T(n-1,k-1) + 2*T(n-1,k) + sum_{i, i>=0} T(n-1,k+1+i)*(-1)^i. - Philippe Deléham, Feb 23 2012
T(n,k) = (k+1)*Sum_{j=0..n-k} C(2*j+k,j)*(-1)^j*3^(n-j-k)*C(n+1,j+k+1)/(n+1). - Vladimir Kruchinin Sep 30 2020

A383527 Partial sums of A005773.

Original entry on oeis.org

1, 2, 4, 9, 22, 57, 153, 420, 1170, 3293, 9339, 26642, 76363, 219728, 634312, 1836229, 5328346, 15494125, 45137995, 131712826, 384900937, 1126265986, 3299509114, 9676690939, 28407473191, 83470059532, 245465090758, 722406781935, 2127562036990, 6270020029353
Offset: 0

Views

Author

Mélika Tebni, Apr 29 2025

Keywords

Comments

For p prime of the form 4*k+3 (A002145), a(p) == 0 (mod p).
For p Pythagorean prime (A002144), a(p) - 2 == 0 (mod p).
a(n) (mod 2) = A010059(n).
a(A000069(n+1)) is even.
a(A001969(n+1)) is odd.

Crossrefs

Programs

  • Maple
    gf := (1 + sqrt((1 + x) / (1 - 3*x))) / (2*(1 - x)):
    a := n-> coeff(series(gf, x, n+1), x, n):
    seq(a(n), n = 0 .. 29);
    # Recurrence:
    a:= proc(n) option remember; `if`(n<=2, 2^n, 3*a(n-1) - (6/n-1)*a(n-2) + (6/n-3)*a(n-3)) end:
    seq(a(n), n = 0 .. 29);
  • Mathematica
    Module[{a, n}, RecurrenceTable[{a[n] == 3*a[n-1] - (6-n)*a[n-2]/n + 3*(2-n)*a[n-3]/n, a[0] == 1, a[1] == 2, a[2] == 4}, a, {n, 0, 30}]] (* Paolo Xausa, May 05 2025 *)
  • Python
    from math import comb as C
    def a(n):
      return sum(C(n, k)*abs(sum((-1)**j*C(k, j) for j in range(k//2 + 1))) for k in range(n + 1))
    print([a(n) for n in range(30)])

Formula

First differences of A211278.
a(n) = Sum_{k=0..n} A167630(n, k).
Binomial transform of A210736 (see Python program).
G.f.: (1 + sqrt((1 + x) / (1 - 3*x))) / (2*(1 - x)).
E.g.f.: (Integral_{x=-oo..oo} BesselI(0,2*x) dx + (1 + BesselI(0,2*x)) / 2)*exp(x).
Recurrence: n*a(n) = 3*n*a(n-1) - (6-n)*a(n-2) + 3*(2-n)*a(n-3). If n <= 2, a(n) = 2^n.
a(n) ~ 3^(n + 1/2) / (2*sqrt(Pi*n)). - Vaclav Kotesovec, May 02 2025
From Mélika Tebni, May 09 2025: (Start)
a(n) = A257520(n) + A097893(n-1) for n > 0.
a(n) = Sum_{j=0..n}(Sum_{k=0..j} A122896(j, k)).
a(n+2) - 3*a(n+1) + 2*a(n) = A005774(n).
a(n+2) - 4*a(n+1) + 4*a(n) - a(n-1) = A005775(n) for n >= 3. (End)

A144025 Eigentriangle by rows, A001006(n-k)*A005773(k); 0<=k<=n.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 4, 2, 2, 5, 9, 4, 4, 5, 13, 21, 9, 8, 10, 13, 35, 51, 21, 18, 20, 26, 35, 96, 127, 51, 42, 45, 52, 70, 96, 267, 323, 127, 102, 105, 117, 140, 192, 267, 750, 835, 323, 254, 255, 273, 315, 384, 534, 750, 2123, 2188, 835, 646, 635, 663, 735, 864, 1068, 1500, 2123, 6046
Offset: 0

Views

Author

Gary W. Adamson, Sep 07 2008

Keywords

Comments

Left border = Motzkin numbers, A001006.
Right border = A005773.
Row sums = A005773 shifted: (1, 2, 5, 13, 35, 96, 267,...).
Sum of n-th row terms = rightmost term of next row.

Examples

			First few rows of the triangle =
    1;
    1,   1;
    2,   1,   2;
    4,   2,   2,   5;
    9,   4,   4,   5,  13;
   21,   9,   8,  10,  13,  35;
   51,  21,  18,  20,  26,  35,  96;
  127,  51,  42,  45,  52,  70,  96, 267;
  323, 127, 102, 105, 117, 140, 192, 267, 750;
  835, 323, 254, 255, 273, 315, 384, 534, 750, 2123;
  ...
Row 3 = (4, 2, 2, 5) = termwise product of (4, 2, 1, 1) and the first 4 terms of A005773: (1, 1, 2, 5) = (4*1, 2*1, 1*2, 1*5). (4, 2, 1, 1) = the first 4 terms of A001066, reversed.
		

Crossrefs

Formula

Eigentriangle by rows: T(n,k) = A001006(n-k)*A005773(k); 0<=k<=n.

Extensions

a(53) corrected by Georg Fischer, Apr 29 2025

A171670 Triangle T read by rows : T(n,k)= A007318(n,k)*A005773(n-k).

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 5, 6, 3, 1, 13, 20, 12, 4, 1, 35, 65, 50, 20, 5, 1, 96, 210, 195, 100, 30, 6, 1, 267, 672, 735, 455, 175, 42, 7, 1, 750, 2136, 2688, 1960, 910, 280, 56, 8, 1, 2123, 6750, 9612, 8064, 4410, 1638, 420, 72, 9, 1
Offset: 0

Views

Author

Philippe Deléham, Dec 14 2009

Keywords

Examples

			Triangle begins : 1 ; 1,1 ; 2,2,1 ; 5,6,3,1 ; 13,20,12,4,1 ; 35,65,50,20,5,1 ; ...
		

Crossrefs

Formula

Sum_{k, 0<=k<=n} T(n,k)= A024718(n).

A204851 Triangle by rows relating to A005773.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 5, 5, 2, 1, 13, 13, 6, 2, 1, 35, 35, 16, 7, 2, 1, 96, 96, 45, 19, 8, 2, 1, 267, 267, 126, 56, 22, 9, 2, 1, 750, 750, 357, 160, 68, 25, 10, 2, 1, 2123, 2123, 1016, 463, 198, 81, 28, 11, 2, 1, 6046, 6046, 2907, 1337, 586, 240, 95, 31, 12, 2, 1
Offset: 0

Views

Author

Gary W. Adamson, Jan 19 2012

Keywords

Comments

Left border = A005773. Row sums = A005773 with offset 1

Examples

			First few rows of the triangle =
1;
1, 1;
2, 2, 1;
5, 5, 2, 1;
13, 13, 6, 2, 1;
35, 35, 16, 7, 2, 1;
96, 96, 45, 19, 8, 2, 1;
267, 267, 126, 56, 22, 9, 2, 1;
750, 750, 357, 160, 68, 25, 10, 2, 1;
2123, 2123, 1016, 463, 198, 81, 28, 11, 2, 1;
6046, 6046, 2907, 1337, 586, 240, 95, 31, 12, 2, 1;
...
Row 3 = (5, 5, 2, 1) since the top row of M^3 = [5, 5, 2, 1, 0, 0, 0,...]
		

Crossrefs

Formula

n-th row of the triangle is the top row of M^n, deleting the zeros; where M = the infinite square production matrix:
1, 1, 0, 0, 0, 0, 0,...
1, 1, 1, 0, 0, 0, 0,...
1, 1, 0, 1, 0, 0, 0,...
1, 1, 1, 0, 1, 0, 0,...
1, 1, 1, 1, 0, 1, 0,...
1, 1, 1, 1, 1, 0, 1,...
1, 1, 1, 1, 1, 1, 0,...
...

A001006 Motzkin numbers: number of ways of drawing any number of nonintersecting chords joining n (labeled) points on a circle.

Original entry on oeis.org

1, 1, 2, 4, 9, 21, 51, 127, 323, 835, 2188, 5798, 15511, 41835, 113634, 310572, 853467, 2356779, 6536382, 18199284, 50852019, 142547559, 400763223, 1129760415, 3192727797, 9043402501, 25669818476, 73007772802, 208023278209, 593742784829, 1697385471211
Offset: 0

Views

Author

Keywords

Comments

Number of 4321-, (3412,2413)-, (3412,3142)- and 3412-avoiding involutions in S_n.
Number of sequences of length n-1 consisting of positive integers such that the first and last elements are 1 or 2 and the absolute difference between any 2 consecutive elements is 0 or 1. - Jon Perry, Sep 04 2003
From David Callan, Jul 15 2004: (Start)
Also number of Motzkin n-paths: paths from (0,0) to (n,0) in an n X n grid using only steps U = (1,1), F = (1,0) and D = (1,-1).
Number of Dyck n-paths with no UUU. (Given such a Dyck n-path, change each UUD to U, then change each remaining UD to F. This is a bijection to Motzkin n-paths. Example with n=5: U U D U D U U D D D -> U F U D D.)
Number of Dyck (n+1)-paths with no UDU. (Given such a Dyck (n+1)-path, mark each U that is followed by a D and each D that is not followed by a U. Then change each unmarked U whose matching D is marked to an F. Lastly, delete all the marked steps. This is a bijection to Motzkin n-paths. Example with n=6 and marked steps in small type: U U u d D U U u d d d D u d -> U U u d D F F u d d d D u d -> U U D F F D.) (End)
a(n) is the number of strings of length 2n+2 from the following recursively defined set: L contains the empty string and, for any strings a and b in L, we also find (ab) in L. The first few elements of L are e, (), (()), ((())), (()()), (((()))), ((()())), ((())()), (()(())) and so on. This proves that a(n) is less than or equal to C(n), the n-th Catalan number. See Orrick link (2024). - Saul Schleimer (saulsch(AT)math.rutgers.edu), Feb 23 2006 (Additional linked comment added by William P. Orrick, Jun 13 2024.)
a(n) = number of Dyck n-paths all of whose valleys have even x-coordinate (when path starts at origin). For example, T(4,2)=3 counts UDUDUUDD, UDUUDDUD, UUDDUDUD. Given such a path, split it into n subpaths of length 2 and transform UU->U, DD->D, UD->F (there will be no DUs for that would entail a valley with odd x-coordinate). This is a bijection to Motzkin n-paths. - David Callan, Jun 07 2006
Also the number of standard Young tableaux of height <= 3. - Mike Zabrocki, Mar 24 2007
a(n) is the number of RNA shapes of size 2n+2. RNA Shapes are essentially Dyck words without "directly nested" motifs of the form A[[B]]C, for A, B and C Dyck words. The first RNA Shapes are []; [][]; [][][], [[][]]; [][][][], [][[][]], [[][][]], [[][]][]; ... - Yann Ponty (ponty(AT)lri.fr), May 30 2007
The sequence is self-generated from top row A going to the left starting (1,1) and bottom row = B, the same sequence but starting (0,1) and going to the right. Take dot product of A and B and add the result to n-th term of A to get the (n+1)-th term of A. Example: a(5) = 21 as follows: Take dot product of A = (9, 4, 2, 1, 1) and (0, 1, 1, 2, 4) = (0, + 4 + 2 + 2 + 4) = 12; which is added to 9 = 21. - Gary W. Adamson, Oct 27 2008
Equals A005773 / A005773 shifted (i.e., (1,2,5,13,35,96,...) / (1,1,2,5,13,35,96,...)). - Gary W. Adamson, Dec 21 2008
Starting with offset 1 = iterates of M * [1,1,0,0,0,...], where M = a tridiagonal matrix with [0,1,1,1,...] in the main diagonal and [1,1,1,...] in the super and subdiagonals. - Gary W. Adamson, Jan 07 2009
a(n) is the number of involutions of {1,2,...,n} having genus 0. The genus g(p) of a permutation p of {1,2,...,n} is defined by g(p)=(1/2)[n+1-z(p)-z(cp')], where p' is the inverse permutation of p, c = 234...n1 = (1,2,...,n), and z(q) is the number of cycles of the permutation q. Example: a(4)=9; indeed, p=3412=(13)(24) is the only involution of {1,2,3,4} with genus > 0. This follows easily from the fact that a permutation p of {1,2,...,n} has genus 0 if and only if the cycle decomposition of p gives a noncrossing partition of {1,2,...,n} and each cycle of p is increasing (see Lemma 2.1 of the Dulucq-Simion reference). [Also, redundantly, for p=3412=(13)(24) we have cp'=2341*3412=4123=(1432) and so g(p)=(1/2)(4+1-2-1)=1.] - Emeric Deutsch, May 29 2010
Let w(i,j,n) denote walks in N^2 which satisfy the multivariate recurrence w(i,j,n) = w(i, j + 1, n - 1) + w(i - 1, j, n - 1) + w(i + 1, j - 1, n - 1) with boundary conditions w(0,0,0) = 1 and w(i,j,n) = 0 if i or j or n is < 0. Then a(n) = Sum_{i = 0..n, j = 0..n} w(i,j,n) is the number of such walks of length n. - Peter Luschny, May 21 2011
a(n)/a(n-1) tends to 3.0 as N->infinity: (1+2*cos(2*Pi/N)) relating to longest odd N regular polygon diagonals, by way of example, N=7: Using the tridiagonal generator [cf. comment of Jan 07 2009], for polygon N=7, we extract an (N-1)/2 = 3 X 3 matrix, [0,1,0; 1,1,1; 0,1,1] with an e-val of 2.24697...; the longest Heptagon diagonal with edge = 1. As N tends to infinity, the diagonal lengths tend to 3.0, the convergent of the sequence. - Gary W. Adamson, Jun 08 2011
Number of (n+1)-length permutations avoiding the pattern 132 and the dotted pattern 23\dot{1}. - Jean-Luc Baril, Mar 07 2012
Number of n-length words w over alphabet {a,b,c} such that for every prefix z of w we have #(z,a) >= #(z,b) >= #(z,c), where #(z,x) counts the letters x in word z. The a(4) = 9 words are: aaaa, aaab, aaba, abaa, aabb, abab, aabc, abac, abca. - Alois P. Heinz, May 26 2012
Number of length-n restricted growth strings (RGS) [r(1), r(2), ..., r(n)] such that r(1)=1, r(k)<=k, and r(k)!=r(k-1); for example, the 9 RGS for n=4 are 1010, 1012, 1201, 1210, 1212, 1230, 1231, 1232, 1234. - Joerg Arndt, Apr 16 2013
Number of length-n restricted growth strings (RGS) [r(1), r(2), ..., r(n)] such that r(1)=0, r(k)<=k and r(k)-r(k-1) != 1; for example, the 9 RGS for n=4 are 0000, 0002, 0003, 0004, 0022, 0024, 0033, 0222, 0224. - Joerg Arndt, Apr 17 2013
Number of (4231,5276143)-avoiding involutions in S_n. - Alexander Burstein, Mar 05 2014
a(n) is the number of increasing unary-binary trees with n nodes that have an associated permutation that avoids 132. For more information about unary-binary trees with associated permutations, see A245888. - Manda Riehl, Aug 07 2014
a(n) is the number of involutions on [n] avoiding the single pattern p, where p is any one of the 8 (classical) patterns 1234, 1243, 1432, 2134, 2143, 3214, 3412, 4321. Also, number of (3412,2413)-, (3412,3142)-, (3412,2413,3142)-avoiding involutions on [n] because each of these 3 sets actually coincides with the 3412-avoiding involutions on [n]. This is a complete list of the 8 singles, 2 pairs, and 1 triple of 4-letter classical patterns whose involution avoiders are counted by the Motzkin numbers. (See Barnabei et al. 2011 reference.) - David Callan, Aug 27 2014
From Tony Foster III, Jul 28 2016: (Start)
A series created using 2*a(n) + a(n+1) has Hankel transform of F(2n), offset 3, F being the Fibonacci bisection, A001906 (empirical observation).
A series created using 2*a(n) + 3*a(n+1) + a(n+2) gives the Hankel transform of Sum_{k=0..n} k*Fibonacci(2*k), offset 3, A197649 (empirical observation). (End)
Conjecture: (2/n)*Sum_{k=1..n} (2k+1)*a(k)^2 is an integer for each positive integer n. - Zhi-Wei Sun, Nov 16 2017
The Rubey and Stump reference proves a refinement of a conjecture of René Marczinzik, which they state as: "The number of 2-Gorenstein algebras which are Nakayama algebras with n simple modules and have an oriented line as associated quiver equals the number of Motzkin paths of length n." - Eric M. Schmidt, Dec 16 2017
Number of U_{k}-equivalence classes of Łukasiewicz paths. Łukasiewicz paths are P-equivalent iff the positions of pattern P are identical in these paths. - Sergey Kirgizov, Apr 08 2018
If tau_1 and tau_2 are two distinct permutation patterns chosen from the set {132,231,312}, then a(n) is the number of valid hook configurations of permutations of [n+1] that avoid the patterns tau_1 and tau_2. - Colin Defant, Apr 28 2019
Number of permutations of length n that are sorted to the identity by a consecutive-321-avoiding stack followed by a classical-21-avoiding stack. - Colin Defant, Aug 29 2020
From Helmut Prodinger, Dec 13 2020: (Start)
a(n) is the number of paths in the first quadrant starting at (0,0) and consisting of n steps from the infinite set {(1,1), (1,-1), (1,-2), (1,-3), ...}.
For example, denoting U=(1,1), D=(1,-1), D_ j=(1,-j) for j >= 2, a(4) counts UUUU, UUUD, UUUD_2, UUUD_3, UUDU, UUDD, UUD_2U, UDUU, UDUD.
This step set is inspired by {(1,1), (1,-1), (1,-3), (1,-5), ...}, suggested by Emeric Deutsch around 2000.
See Prodinger link that contains a bijection to Motzkin paths. (End)
Named by Donaghey (1977) after the Israeli-American mathematician Theodore Motzkin (1908-1970). In Sloane's "A Handbook of Integer Sequences" (1973) they were called "generalized ballot numbers". - Amiram Eldar, Apr 15 2021
Number of Motzkin n-paths a(n) is split into A107587(n), number of even Motzkin n-paths, and A343386(n), number of odd Motzkin n-paths. The value A107587(n) - A343386(n) can be called the "shadow" of a(n) (see A343773). - Gennady Eremin, May 17 2021
Conjecture: If p is a prime of the form 6m+1 (A002476), then a(p-2) is divisible by p. Currently, no counterexample exists for p < 10^7. Personal communication from Robert Gerbicz: mod such p this is equivalent to A066796 with comment: "Every A066796(n) from A066796((p-1)/2) to A066796(p-1) is divisible by prime p of form 6m+1". - Serge Batalov, Feb 08 2022
From Rob Burns, Nov 11 2024: (Start)
The conjecture is proved in the 2017 paper by Rob Burns in the Links below. The result is contained in Tables 4 and 5 of the paper, which show that a(p-2) == 0 (mod p) when p == 1 (mod 6) and a(p-2) == -1 (mod p) when p == -1 (mod 6).
In fact, the 2017 paper by Burns establishes more general congruences for a(p^k - 2) where k >= 1.
If p == 1 (mod 6) then a(p^k - 2) == 0 (mod p) for k >= 1.
If p == -1 (mod 6) then a(p^k - 2) == -1 (mod p) when k is odd and a(p^k - 2) == 0 (mod p) when k is even.
These are consequences of the transitions provided in Tables 4, 5 and 6 of the paper.
The 2024 paper by Nadav Kohen also proves the conjecture. Proposition 6 of the paper states that a prime p divides a(p-2) if and only if p = (1 mod 3). (End)
From Peter Bala, Feb 10 2022: (Start)
Conjectures:
(1) For prime p == 1 (mod 6) and n, r >= 1, a(n*p^r - 2) == -A005717(n-1) (mod p), where we take A005717(0) = 0 to match Batalov's conjecture above.
(2) For prime p == 5 (mod 6) and n >= 1, a(n*p - 2) == -A005773(n) (mod p).
(3) For prime p >= 3 and k >= 1, a(n + p^k) == a(n) (mod p) for 0 <= n <= (p^k - 3).
(4) For prime p >= 5 and k >= 2, a(n + p^k) == a(n) (mod p^2) for 0 <= n <= (p^(k-1) - 3). (End)
The Hankel transform of this sequence with a(0) omitted gives the period-6 sequence [1, 0, -1, -1, 0, 1, ...] which is A010892 with its first term omitted, while the Hankel transform of the current sequence is the all-ones sequence A000012, and also it is the unique sequence with this property which is similar to the unique Hankel transform property of the Catalan numbers. - Michael Somos, Apr 17 2022
The number of terms in which the exponent of any variable x_i is not greater than 2 in the expansion of Product_{j=1..n} Sum_{i=1..j} x_i. E.g.: a(4) = 9: 3*x1^2*x2^2, 4*x1^2*x2*x3, 2*x1^2*x2*x4, x1^2*x3^2, x1^2*x3*x4, 2*x1*x2^2*x3, x1*x2^2*x4, x1*x2*x3^2, x1*x2*x3*x4. - Elif Baser, Dec 20 2024

Examples

			G.f.: 1 + x + 2*x^2 + 4*x^3 + 9*x^4 + 21*x^5 + 51*x^6 + 127*x^7 + 323*x^8 + ...
.
The 21 Motzkin-paths of length 5: UUDDF, UUDFD, UUFDD, UDUDF, UDUFD, UDFUD, UDFFF, UFUDD, UFDUD, UFDFF, UFFDF, UFFFD, FUUDD, FUDUD, FUDFF, FUFDF, FUFFD, FFUDF, FFUFD, FFFUD, FFFFF.
		

References

  • F. Bergeron, L. Favreau, and D. Krob, Conjectures on the enumeration of tableaux of bounded height, Discrete Math, vol. 139, no. 1-3 (1995), 463-468.
  • F. R. Bernhart, Catalan, Motzkin, and Riordan numbers, Discr. Math., 204 (1999) 73-112.
  • R. Bojicic and M. D. Petkovic, Orthogonal Polynomials Approach to the Hankel Transform of Sequences Based on Motzkin Numbers, Bulletin of the Malaysian Mathematical Sciences, 2015, doi:10.1007/s40840-015-0249-3.
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, pp. 24, 298, 618, 912.
  • A. J. Bu, Automated counting of restricted Motzkin paths, Enumerative Combinatorics and Applications, ECA 1:2 (2021) Article S2R12.
  • Naiomi Cameron, JE McLeod, Returns and Hills on Generalized Dyck Paths, Journal of Integer Sequences, Vol. 19, 2016, #16.6.1.
  • L. Carlitz, Solution of certain recurrences, SIAM J. Appl. Math., 17 (1969), 251-259.
  • Michael Dairyko, Samantha Tyner, Lara Pudwell, and Casey Wynn, Non-contiguous pattern avoidance in binary trees. Electron. J. Combin. 19 (2012), no. 3, Paper 22, 21 pp. MR2967227.
  • D. E. Davenport, L. W. Shapiro, and L. C. Woodson, The Double Riordan Group, The Electronic Journal of Combinatorics, 18(2) (2012), #P33.
  • E. Deutsch and L. Shapiro, A survey of the Fine numbers, Discrete Math., 241 (2001), 241-265.
  • T. Doslic, D. Svrtan, and D. Veljan, Enumerative aspects of secondary structures, Discr. Math., 285 (2004), 67-82.
  • Tomislav Doslic and Darko Veljan, Logarithmic behavior of some combinatorial sequences. Discrete Math. 308 (2008), no. 11, 2182-2212. MR2404544 (2009j:05019).
  • S. Dulucq and R. Simion, Combinatorial statistics on alternating permutations, J. Algebraic Combinatorics, 8, 1998, 169-191.
  • M. Dziemianczuk, "Enumerations of plane trees with multiple edges and Raney lattice paths." Discrete Mathematics 337 (2014): 9-24.
  • Wenjie Fang, A partial order on Motzkin paths, Discrete Math., 343 (2020), #111802.
  • I. P. Goulden and D. M. Jackson, Combinatorial Enumeration, Wiley, N.Y., 1983, (5.2.10).
  • N. S. S. Gu, N. Y. Li, and T. Mansour, 2-Binary trees: bijections and related issues, Discr. Math., 308 (2008), 1209-1221.
  • Kris Hatch, Presentation of the Motzkin Monoid, Senior Thesis, Univ. Cal. Santa Barbara, 2012; http://ccs.math.ucsb.edu/senior-thesis/Kris-Hatch.pdf.
  • V. Jelinek, Toufik Mansour, and M. Shattuck, On multiple pattern avoiding set partitions, Advances in Applied Mathematics Volume 50, Issue 2, February 2013, pp. 292-326.
  • Hana Kim and R. P. Stanley, A refined enumeration of hex trees and related polynomials, http://www-math.mit.edu/~rstan/papers/hextrees.pdf, Preprint 2015.
  • S. Kitaev, Patterns in Permutations and Words, Springer-Verlag, 2011. See p. 399 Table A.7.
  • A. Kuznetsov et al., Trees associated with the Motzkin numbers, J. Combin. Theory, A 76 (1996), 145-147.
  • T. Lengyel, On divisibility properties of some differences of Motzkin numbers, Annales Mathematicae et Informaticae, 41 (2013) pp. 121-136.
  • W. A. Lorenz, Y. Ponty, and P. Clote, Asymptotics of RNA Shapes, Journal of Computational Biology. 2008, 15(1): 31-63. doi:10.1089/cmb.2006.0153.
  • Piera Manara and Claudio Perelli Cippo, The fine structure of 4321 avoiding involutions and 321 avoiding involutions, PU. M. A. Vol. 22 (2011), 227-238; http://www.mat.unisi.it/newsito/puma/public_html/22_2/manara_perelli-cippo.pdf.
  • Toufik Mansour, Restricted 1-3-2 permutations and generalized patterns, Annals of Combin., 6 (2002), 65-76.
  • Toufik Mansour, Matthias Schork, and Mark Shattuck, Catalan numbers and pattern restricted set partitions. Discrete Math. 312(2012), no. 20, 2979-2991. MR2956089.
  • T. S. Motzkin, Relations between hypersurface cross ratios and a combinatorial formula for partitions of a polygon, for permanent preponderance and for non-associative products, Bull. Amer. Math. Soc., 54 (1948), 352-360.
  • Jocelyn Quaintance and Harris Kwong, A combinatorial interpretation of the Catalan and Bell number difference tables, Integers, 13 (2013), #A29.
  • J. Riordan, Enumeration of plane trees by branches and endpoints, J. Combin. Theory, A 23 (1975), 214-222.
  • A. Sapounakis et al., Ordered trees and the inorder transversal, Disc. Math., 306 (2006), 1732-1741.
  • A. Sapounakis, I. Tasoulas, and P. Tsikouras, Counting strings in Dyck paths, Discrete Math., 307 (2007), 2909-2924.
  • E. Schroeder, Vier combinatorische Probleme, Z. f. Math. Phys., 15 (1870), 361-376.
  • L. W. Shapiro et al., The Riordan group, Discrete Applied Math., 34 (1991), 229-239.
  • Mark Shattuck, On the zeros of some polynomials with combinatorial coefficients, Annales Mathematicae et Informaticae, 42 (2013) pp. 93-101, http://ami.ektf.hu.
  • 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).
  • Michael Z. Spivey and Laura L. Steil, The k-Binomial Transforms and the Hankel Transform, Journal of Integer Sequences, Vol. 9 (2006), Article 06.1.1.
  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 6.37. Also Problem 7.16(b), y_3(n).
  • P. R. Stein and M. S. Waterman, On some new sequences generalizing the Catalan and Motzkin numbers, Discrete Math., 26 (1979), 261-272.
  • Z.-W. Sun, Conjectures involving arithmetical sequences, Number Theory: Arithmetic in Shangri-La (eds., S. Kanemitsu, H.-Z. Li and J.-Y. Liu), Proc. the 6th China-Japan Sem. Number Theory (Shanghai, August 15-17, 2011), World Sci., Singapore, 2013, pp. 244-258; http://math.nju.edu.cn/~zwsun/142p.pdf.
  • Chenying Wang, Piotr Miska, and István Mező, "The r-derangement numbers." Discrete Mathematics 340.7 (2017): 1681-1692.
  • Ying Wang and Guoce Xin, A Classification of Motzkin Numbers Modulo 8, Electron. J. Combin., 25(1) (2018), #P1.54.
  • Wen-Jin Woan, A combinatorial proof of a recursive relation of the Motzkin sequence by lattice paths. Fibonacci Quart. 40 (2002), no. 1, 3-8.
  • Wen-jin Woan, A Recursive Relation for Weighted Motzkin Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.1.6.
  • F. Yano and H. Yoshida, Some set partition statistics in non-crossing partitions and generating functions, Discr. Math., 307 (2007), 3147-3160.

Crossrefs

Bisections: A026945, A099250.
Sequences related to chords in a circle: A001006, A054726, A006533, A006561, A006600, A007569, A007678. See also entries for chord diagrams in Index file.
a(n) = A005043(n)+A005043(n+1).
A086246 is another version, although this is the main entry. Column k=3 of A182172.
Motzkin numbers A001006 read mod 2,3,4,5,6,7,8,11: A039963, A039964, A299919, A258712, A299920, A258711, A299918, A258710.
Cf. A004148, A004149, A023421, A023422, A023423, A290277 (inv. Euler Transf.).

Programs

  • Haskell
    a001006 n = a001006_list !! n
    a001006_list = zipWith (+) a005043_list $ tail a005043_list
    -- Reinhard Zumkeller, Jan 31 2012
    
  • Maple
    # Three different Maple scripts for this sequence:
    A001006 := proc(n)
        add(binomial(n,2*k)*A000108(k),k=0..floor(n/2)) ;
    end proc:
    A001006 := proc(n) option remember; local k; if n <= 1 then 1 else procname(n-1) + add(procname(k)*procname(n-k-2),k=0..n-2); end if; end proc:
    # n -> [a(0),a(1),..,a(n)]
    A001006_list := proc(n) local w, m, j, i; w := proc(i,j,n) option remember;
    if min(i,j,n) < 0 or max(i,j) > n then 0
    elif n = 0 then if i = 0 and j = 0 then 1 else 0 fi else
    w(i, j + 1, n - 1) + w(i - 1, j, n - 1) + w(i + 1, j - 1, n - 1) fi end:
    [seq( add( add( w(i, j, m), i = 0..m), j = 0..m), m = 0..n)] end:
    A001006_list(29); # Peter Luschny, May 21 2011
  • Mathematica
    a[0] = 1; a[n_Integer] := a[n] = a[n - 1] + Sum[a[k] * a[n - 2 - k], {k, 0, n - 2}]; Array[a, 30]
    (* Second program: *)
    CoefficientList[Series[(1 - x - (1 - 2x - 3x^2)^(1/2))/(2x^2), {x, 0, 29}], x] (* Jean-François Alcover, Nov 29 2011 *)
    Table[Hypergeometric2F1[(1-n)/2, -n/2, 2, 4], {n,0,29}] (* Peter Luschny, May 15 2016 *)
    Table[GegenbauerC[n,-n-1,-1/2]/(n+1),{n,0,100}] (* Emanuele Munarini, Oct 20 2016 *)
    MotzkinNumber = DifferenceRoot[Function[{y, n}, {(-3n-3)*y[n] + (-2n-5)*y[n+1] + (n+4)*y[n+2] == 0, y[0] == 1, y[1] == 1}]];
    Table[MotzkinNumber[n], {n, 0, 29}] (* Jean-François Alcover, Oct 27 2021 *)
  • Maxima
    a[0]:1$
    a[1]:1$
    a[n]:=((2*n+1)*a[n-1]+(3*n-3)*a[n-2])/(n+2)$
    makelist(a[n],n,0,12); /* Emanuele Munarini, Mar 02 2011 */
    
  • Maxima
    M(n) := coeff(expand((1+x+x^2)^(n+1)),x^n)/(n+1);
    makelist(M(n),n,0,60); /* Emanuele Munarini, Apr 04 2012 */
    
  • Maxima
    makelist(ultraspherical(n,-n-1,-1/2)/(n+1),n,0,12); /* Emanuele Munarini, Oct 20 2016 */
    
  • PARI
    {a(n) = polcoeff( ( 1 - x - sqrt((1 - x)^2 - 4 * x^2 + x^3 * O(x^n))) / (2 * x^2), n)}; /* Michael Somos, Sep 25 2003 */
    
  • PARI
    {a(n) = if( n<0, 0, n++; polcoeff( serreverse( x / (1 + x + x^2) + x * O(x^n)), n))}; /* Michael Somos, Sep 25 2003 */
    
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff( exp(x + x * O(x^n)) * besseli(1, 2 * x + x * O(x^n)), n))}; /* Michael Somos, Sep 25 2003 */
    
  • Python
    from gmpy2 import divexact
    A001006 = [1, 1]
    for n in range(2, 10**3):
        A001006.append(divexact(A001006[-1]*(2*n+1)+(3*n-3)*A001006[-2],n+2))
    # Chai Wah Wu, Sep 01 2014
    
  • Python
    def mot():
        a, b, n = 0, 1, 1
        while True:
            yield b//n
            n += 1
            a, b = b, (3*(n-1)*n*a+(2*n-1)*n*b)//((n+1)*(n-1))
    A001006 = mot()
    print([next(A001006) for n in range(30)]) # Peter Luschny, May 16 2016
    
  • Python
    # A simple generator of Motzkin-paths (see the first comment of David Callan).
    C = str.count
    def aGen(n: int):
        a = [""]
        for w in a:
            if len(w) == n:
                if C(w, "U") == C(w, "D"): yield w
            else:
                for j in "UDF":
                    u = w + j
                    if C(u, "U") >= C(u, "D"): a += [u]
        return a
    for n in range(6):
        MP = [w for w in aGen(n)];
        print(len(MP), ":", MP)  # Peter Luschny, Dec 03 2024

Formula

G.f.: A(x) = ( 1 - x - (1-2*x-3*x^2)^(1/2) ) / (2*x^2).
G.f. A(x) satisfies A(x) = 1 + x*A(x) + x^2*A(x)^2.
G.f.: F(x)/x where F(x) is the reversion of x/(1+x+x^2). - Joerg Arndt, Oct 23 2012
a(n) = (-1/2) Sum_{i+j = n+2, i >= 0, j >= 0} (-3)^i*C(1/2, i)*C(1/2, j).
a(n) = (3/2)^(n+2) * Sum_{k >= 1} 3^(-k) * Catalan(k-1) * binomial(k, n+2-k). [Doslic et al.]
a(n) ~ 3^(n+1)*sqrt(3)*(1 + 1/(16*n))/((2*n+3)*sqrt((n+2)*Pi)). [Barcucci, Pinzani and Sprugnoli]
Limit_{n->infinity} a(n)/a(n-1) = 3. [Aigner]
a(n+2) - a(n+1) = a(0)*a(n) + a(1)*a(n-1) + ... + a(n)*a(0). [Bernhart]
a(n) = (1/(n+1)) * Sum_{i} (n+1)!/(i!*(i+1)!*(n-2*i)!). [Bernhart]
From Len Smiley: (Start)
a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*A000108(k+1), inv. Binomial Transform of A000108.
a(n) = (1/(n+1))*Sum_{k=0..ceiling((n+1)/2)} binomial(n+1, k)*binomial(n+1-k, k-1);
D-finite with recurrence: (n+2)*a(n) = (2*n+1)*a(n-1) + (3*n-3)*a(n-2). (End)
a(n) = Sum_{k=0..n} C(n, 2k)*A000108(k). - Paul Barry, Jul 18 2003
E.g.f.: exp(x)*BesselI(1, 2*x)/x. - Vladeta Jovovic, Aug 20 2003
a(n) = A005043(n) + A005043(n+1).
The Hankel transform of this sequence gives A000012 = [1, 1, 1, 1, 1, 1, ...]. E.g., Det([1, 1, 2, 4; 1, 2, 4, 9; 2, 4, 9, 21; 4, 9, 21, 51]) = 1. - Philippe Deléham, Feb 23 2004
a(m+n) = Sum_{k>=0} A064189(m, k)*A064189(n, k). - Philippe Deléham, Mar 05 2004
a(n) = (1/(n+1))*Sum_{j=0..floor(n/3)} (-1)^j*binomial(n+1, j)*binomial(2*n-3*j, n). - Emeric Deutsch, Mar 13 2004
a(n) = A086615(n) - A086615(n-1) (n >= 1). - Emeric Deutsch, Jul 12 2004
G.f.: A(x)=(1-y+y^2)/(1-y)^2 where (1+x)*(y^2-y)+x=0; A(x)=4*(1+x)/(1+x+sqrt(1-2*x-3*x^2))^2; a(n)=(3/4)*(1/2)^n*Sum_(k=0..2*n, 3^(n-k)*C(k)*C(k+1, n+1-k) ) + 0^n/4 [after Doslic et al.]. - Paul Barry, Feb 22 2005
G.f.: c(x^2/(1-x)^2)/(1-x), c(x) the g.f. of A000108. - Paul Barry, May 31 2006
Asymptotic formula: a(n) ~ sqrt(3/4/Pi)*3^(n+1)/n^(3/2). - Benoit Cloitre, Jan 25 2007
a(n) = A007971(n+2)/2. - Zerinvary Lajos, Feb 28 2007
a(n) = (1/(2*Pi))*Integral_{x=-1..3} x^n*sqrt((3-x)*(1+x)) is the moment representation. - Paul Barry, Sep 10 2007
Given an integer t >= 1 and initial values u = [a_0, a_1, ..., a_{t-1}], we may define an infinite sequence Phi(u) by setting a_n = a_{n-1} + a_0*a_{n-1} + a_1*a_{n-2} + ... + a_{n-2}*a_1 for n >= t. For example, Phi([1]) is the Catalan numbers A000108. The present sequence is Phi([0,1,1]), see the 6th formula. - Gary W. Adamson, Oct 27 2008
G.f.: 1/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/.... (continued fraction). - Paul Barry, Dec 06 2008
G.f.: 1/(1-(x+x^2)/(1-x^2/(1-(x+x^2)/(1-x^2/(1-(x+x^2)/(1-x^2/(1-.... (continued fraction). - Paul Barry, Feb 08 2009
a(n) = (-3)^(1/2)/(6*(n+2)) * (-1)^n*(3*hypergeom([1/2, n+1],[1],4/3) - hypergeom([1/2, n+2],[1],4/3)). - Mark van Hoeij, Nov 12 2009
G.f.: 1/(1-x/(1-x/(1-x^2/(1-x/(1-x/(1-x^2/(1-x/(1-x/(1-x^2/(1-... (continued fraction). - Paul Barry, Mar 02 2010
G.f.: 1/(1-x/(1-x/(1+x-x/(1-x/(1+x-x/(1-x/(1+x-x/(1-x/(1+x-x/(1-... (continued fraction). - Paul Barry, Jan 26 2011 [Adds apparently a third '1' in front. - R. J. Mathar, Jan 29 2011]
Let A(x) be the g.f., then B(x)=1+x*A(x) = 1 + 1*x + 1*x^2 + 2*x^3 + 4*x^4 + 9*x^5 + ... = 1/(1-z/(1-z/(1-z/(...)))) where z=x/(1+x) (continued fraction); more generally B(x)=C(x/(1+x)) where C(x) is the g.f. for the Catalan numbers (A000108). - Joerg Arndt, Mar 18 2011
a(n) = (2/Pi)*Integral_{x=-1..1} (1+2*x)^n*sqrt(1-x^2). - Peter Luschny, Sep 11 2011
G.f.: (1-x-sqrt(1-2*x-3*(x^2)))/(2*(x^2)) = 1/2/(x^2)-1/2/x-1/2/(x^2)*G(0); G(k) = 1+(4*k-1)*x*(2+3*x)/(4*k+2-x*(2+3*x)*(4*k+1)*(4*k+2) /(x*(2+3*x)*(4*k+1)+(4*k+4)/G(k+1))), if -1 < x < 1/3; (continued fraction). - Sergei N. Gladkovskii, Dec 01 2011
G.f.: (1-x-sqrt(1-2*x-3*(x^2)))/(2*(x^2)) = (-1 + 1/G(0))/(2*x); G(k) = 1-2*x/(1+x/(1+x/(1-2*x/(1-x/(2-x/G(k+1)))))); (continued fraction). - Sergei N. Gladkovskii, Dec 11 2011
0 = a(n) * (9*a(n+1) + 15*a(n+2) - 12*a(n+3)) + a(n+1) * ( -3*a(n+1) + 10*a(n+2) - 5*a(n+3)) + a(n+2) * (a(n+2) + a(n+3)) unless n=-2. - Michael Somos, Mar 23 2012
a(n) = (-1)^n*hypergeometric([-n,3/2],[3],4). - Peter Luschny, Aug 15 2012
Representation in terms of special values of Jacobi polynomials P(n,alpha,beta,x), in Maple notation: a(n)= 2*(-1)^n*n!*JacobiP(n,2,-3/2-n,-7)/(n+2)!, n>=0. - Karol A. Penson, Jun 24 2013
G.f.: Q(0)/x - 1/x, where Q(k) = 1 + (4*k+1)*x/((1+x)*(k+1) - x*(1+x)*(2*k+2)*(4*k+3)/(x*(8*k+6)+(2*k+3)*(1+x)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 14 2013
Catalan(n+1) = Sum_{k=0..n} binomial(n,k)*a(k). E.g.: 42 = 1*1 + 4*1 + 6*2 + 4*4 + 1*9. - Doron Zeilberger, Mar 12 2015
G.f. A(x) with offset 1 satisfies: A(x)^2 = A( x^2/(1-2*x) ). - Paul D. Hanna, Nov 08 2015
a(n) = GegenbauerPoly(n,-n-1,-1/2)/(n+1). - Emanuele Munarini, Oct 20 2016
a(n) = a(n-1) + A002026(n-1). Number of Motzkin paths that start with an F step plus number of Motzkin paths that start with an U step. - R. J. Mathar, Jul 25 2017
G.f. A(x) satisfies A(x)*A(-x) = F(x^2), where F(x) is the g.f. of A168592. - Alexander Burstein, Oct 04 2017
G.f.: A(x) = exp(int((E(x)-1)/x dx)), where E(x) is the g.f. of A002426. Equivalently, E(x) = 1 + x*A'(x)/A(x). - Alexander Burstein, Oct 05 2017
G.f. A(x) satisfies: A(x) = Sum_{j>=0} x^j * Sum_{k=0..j} binomial(j,k)*x^k*A(x)^k. - Ilya Gutkovskiy, Apr 11 2019
From Gennady Eremin, May 08 2021: (Start)
G.f.: 2/(1 - x + sqrt(1-2*x-3*x^2)).
a(n) = A107587(n) + A343386(n) = 2*A107587(n) - A343773(n) = 2*A343386(n) + A343773(n). (End)
Revert transform of A049347 (after Michael Somos). - Gennady Eremin, Jun 11 2021
Sum_{n>=0} 1/a(n) = 2.941237337631025604300320152921013604885956025483079699366681494505960039781389... - Vaclav Kotesovec, Jun 17 2021
Let a(-1) = (1 - sqrt(-3))/2 and a(n) = a(-3-n)*(-3)^(n+3/2) for all n in Z. Then a(n) satisfies my previous formula relation from Mar 23 2012 now for all n in Z. - Michael Somos, Apr 17 2022
Let b(n) = 1 for n <= 1, otherwise b(n) = Sum_{k=2..n} b(k-1) * b(n-k), then a(n) = b(n+1) (conjecture). - Joerg Arndt, Jan 16 2023
From Peter Bala, Feb 03 2024: (Start)
G.f.: A(x) = 1/(1 + x)*c(x/(1 + x))^2 = 1 + x/(1 + x)*c(x/(1 + x))^3, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108.
A(x) = 1/(1 - 3*x)*c(-x/(1 -3*x))^2.
a(n+1) = Sum_{k = 0..n} (-1)^(n-k)*binomial(n, k)*A000245(k+1).
a(n) = 3^n * Sum_{k = 0..n} (-3)^(-k)*binomial(n, k)*Catalan(k+1).
a(n) = 3^n * hypergeom([3/2, -n], [3], 4/3). (End)
G.f. A(x) satisfies A(x) = exp( x*A(x) + Integral x*A(x)/(1 - x^2*A(x)) dx ). - Paul D. Hanna, Mar 04 2024
a(n) = hypergeom([-n/2,1/2-n/2],[2],4). - Karol A. Penson, May 18 2025

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

Original entry on oeis.org

1, 1, 2, 3, 6, 10, 20, 35, 70, 126, 252, 462, 924, 1716, 3432, 6435, 12870, 24310, 48620, 92378, 184756, 352716, 705432, 1352078, 2704156, 5200300, 10400600, 20058300, 40116600, 77558760, 155117520, 300540195, 601080390, 1166803110
Offset: 0

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Comments

Sperner's theorem says that this is the maximal number of subsets of an n-set such that no one contains another.
When computed from index -1, [seq(binomial(n,floor(n/2)), n = -1..30)]; -> [1,1,1,2,3,6,10,20,35,70,126,...] and convolved with aerated Catalan numbers [seq(((n+1) mod 2)*binomial(n,n/2)/((n/2)+1), n = 0..30)]; -> [1,0,1,0,2,0,5,0,14,0,42,0,132,0,...] shifts left by one: [1,1,2,3,6,10,20,35,70,126,252,...] and if again convolved with aerated Catalan numbers, gives A037952 apart from the initial term. - Antti Karttunen, Jun 05 2001 [This is correct because the g.f.'s satisfy (1+x*g001405(x))*g126120(x) = g001405(x) and g001405(x)*g126120(x) = g037952(x)/x. - R. J. Mathar, Sep 23 2021]
Number of ordered trees with n+1 edges, having nonroot nodes of outdegree 0 or 2. - Emeric Deutsch, Aug 02 2002
Gives for n >= 1 the maximum absolute column sum norm of the inverse of the Vandermonde matrix (a_ij) i=0..n-1, j=0..n-1 with a_00=1 and a_ij=i^j for (i,j) != (0,0). - Torsten Muetze, Feb 06 2004
Image of Catalan numbers A000108 under the Riordan array (1/(1-2x),-x/(1-2x)) or A065109. - Paul Barry, Jan 27 2005
Number of left factors of Dyck paths, consisting of n steps. Example: a(4)=6 because we have UDUD, UDUU, UUDD, UUDU, UUUD and UUUU, where U=(1,1) and D=(1,-1). - Emeric Deutsch, Apr 23 2005
Number of dispersed Dyck paths of length n; they are defined as concatenations of Dyck paths and (1,0)-steps on the x-axis; equivalently, Motzkin paths with no (1,0)-steps at positive height. Example: a(4)=6 because we have HHHH, HHUD, HUDH, UDHH, UDUD, and UUDD, where U=(1,1), H=(1,0), and D=(1,-1). - Emeric Deutsch, Jun 04 2011
a(n) is odd iff n=2^k-1. - Jon Perry, May 05 2005
An inverse Chebyshev transform of binomial(1,n)=(1,1,0,0,0,...) where g(x)->(1/sqrt(1-4*x^2))*g(x*c(x^2)), with c(x) the g.f. of A000108. - Paul Barry, May 13 2005
In a random walk on the number line, starting at 0 and with 0 absorbing after the first step, number of ways of ending up at a positive integer after n steps. - Joshua Zucker, Jul 31 2005
Maximum number of sums of the form Sum_{i=1..n} e(i)*a(i) that are congruent to 0 mod q, where e_i=0 or 1 and gcd(a_i,q)=1, provided that q > ceiling(n/2). - Ralf Stephan, Apr 27 2003
Also the number of standard tableaux of height <= 2. - Mike Zabrocki, Mar 24 2007
Hankel transform of this sequence forms A000012 = [1,1,1,1,1,1,1,...]. - Philippe Deléham, Oct 24 2007
A001263 * [1, -2, 3, -4, 5, ...] = [1, -1, -2, 3, 6, -10, -20, 35, 70, -126, ...]. - Gary W. Adamson, Jan 02 2008
Equals right border of triangle A153585. - Gary W. Adamson, Dec 28 2008
Second binomial transform of A168491. - Philippe Deléham, Nov 27 2009
a(n) is also the number of distinct strings of length n, each of which is a prefix of a string of balanced parentheses; see example. - Lee A. Newberg, Apr 26 2010
Number of symmetric balanced strings of n pairs of parentheses; see example. - Joerg Arndt, Jul 25 2011
a(n) is the number of permutation patterns modulo 2. - Olivier Gérard, Feb 25 2011
For n >= 2, a(n-1) is the number of incongruent two-color bracelets of 2*n-1 beads, n of which are black (A007123), having a diameter of symmetry. - Vladimir Shevelev, May 03 2011
The number of permutations of n elements where p(k-2) < p(k) for all k. - Joerg Arndt, Jul 23 2011
Also size of the equivalence class of S_{n+1} containing the identity permutation under transformations of positionally adjacent elements of the form abc <--> cba where a < b < c, cf. A210668. - Tom Roby, May 15 2012
a(n) is the number of symmetric Dyck paths of length 2n. - Matt Watson, Sep 26 2012
a(n) is divisible by A000108(floor(n/2)) = abs(A129996(n-2)). - Paul Curtz, Oct 23 2012
a(n) is the number of permutations of length n avoiding both 213 and 231 in the classical sense which are breadth-first search reading words of increasing unary-binary trees. For more details, see the entry for permutations avoiding 231 at A245898. - Manda Riehl, Aug 05 2014
Number of symmetric standard Young tableaux of shape (n,n). - Ran Pan, Apr 10 2015
From Luciano Ancora, May 09 2015: (Start)
Also "stepped path" in the array formed by partial sums of the all 1's sequence (or a Pascal's triangle displayed as a square). Example:
[1], [1], 1, 1, 1, 1, 1, ... A000012
1, [2], [3], 4, 5, 6, 7, ...
1, 3, [6], [10], 15, 21, 28, ...
1, 4, 10, [20], [35], 56, 84, ...
1, 5, 15, 35, [70], [126], 210, ...
Sequences in second formula are the mixed diagonals shown in this array. (End)
a(n) = A265848(n,n). - Reinhard Zumkeller, Dec 24 2015
The constant Sum_{n >= 0} a(n)/n! is 1 + A130820. - Peter Bala, Jul 02 2016
Number of meanders (walks starting at the origin and ending at any altitude >= 0 that may touch but never go below the x-axis) with n steps from {-1,1}. - David Nguyen, Dec 20 2016
a(n) is also the number of paths of n steps (either up or down by 1) that end at the maximal value achieved along the path. - Winston Luo, Jun 01 2017
Number of binary n-tuples such that the number of 1's in the even positions is the same as the number of 1's in the odd positions. - Juan A. Olmos, Dec 21 2017
Equivalently, a(n) is the number of subsets of {1,...,n} containing as many even numbers as odd numbers. - Gus Wiseman, Mar 17 2018
a(n) is the number of Dyck paths with semilength = n+1, returns to the x-axis = floor((n+3)/2) and up movements in odd positions = floor((n+3)/2). Example: a(4)=6, U=up movement in odd position, u=up movement in even position, d=down movement, -=return to x-axis: Uududd-Ud-Ud-, Ud-Uudd-Uudd-, Uudd-Uudd-Ud-, Ud-Ud-Uududd-, Uudd-Ud-Uudd-, Ud-Uududd-Ud-. - Roger Ford, Dec 29 2017
Let C_n(R, H) denote the transition matrix from the ribbon basis to the homogeneous basis of the graded component of the algebra of noncommutative symmetric functions of order n. Letting I(2^(n-1)) denote the identity matrix of order 2^(n-1), it has been conjectured that the dimension of the kernel of C_n(R, H) - I(2^(n-1)) is always equal to a(n-1). - John M. Campbell, Mar 30 2018
The number of U-equivalence classes of Łukasiewicz paths. Łukasiewicz paths are U-equivalent iff the positions of pattern U are identical in these paths. - Sergey Kirgizov, Apr 08 2018
All binary self-dual codes of length 2n, for n > 0, must contain at least a(n) codewords of weight n. More to the point, there will always be at least one, perhaps unique, binary self-dual code of length 2n that will contain exactly a(n) codewords that have a hamming weight equal to half the length of the code (n). This code can be constructed by direct summing the unique binary self-dual code of length 2 (up to permutation equivalence) to itself n times. A permutation equivalent code can be constructed by augmenting two identity matrices of length n together. - Nathan J. Russell, Nov 25 2018
Closed under addition. - Torlach Rush, Apr 18 2019
The sequence starting (1, 2, 3, 6, ...) is the invert transform of A097331: (1, 1, 0, 1, 0, 2, 0, 5, 0, 14, 0, 42, ...). - Gary W. Adamson, Feb 22 2020
From Gary W. Adamson, Feb 24 2020: (Start)
The sequence is the culminating limit of an infinite set of sequences with convergents of 2*cos(Pi/N), N = (3, 5, 7, 9, ...).
The first few such sequences are:
N = 3: (1, 1, 1, 1, 1, 1, 1, 1, ...)
N = 5: (1, 1, 2, 3, 5, 8, 13, 21, ...) = A000045
N = 7: (1, 1, 2, 3, 6, 10, 19, 33, ...) = A028495, a(n)/a(n-1) tends to 1.801937...
N = 9 (1, 1, 2, 3, 6, 10, 20, 35, ...) = A061551, a(n)/a(n_1) tends to 1.879385...
...
In the limit one gets the current sequence with ratio 2. (End)
a(n) is also the number of monotone lattice paths from (0,0) to (floor(n/2),ceiling(n/2)). These are the number of Grand Dyck paths when n is even. - Nachum Dershowitz, Aug 12 2020
The maximum number of preimages that a permutation of length n+1 can have under the consecutive-132-avoiding stack-sorting map. - Colin Defant, Aug 28 2020
Counts faro permutations of length n. Faro permutations are permutations avoiding the three consecutive patterns 231, 321 and 312. They are obtained by a perfect faro shuffle of two nondecreasing words of lengths differing by at most one. - Sergey Kirgizov, Jan 12 2021
Per "Sperner's Theorem", the largest possible familes of finite sets none of which contain any other sets in the family. - Renzo Benedetti, May 26 2021
a(n-1) are the incomplete, primitive Dyck paths of n steps without a first return: paths of U and D steps starting at the origin, never touching the horizontal axis later on, and ending above the horizontal axis. n=1: {U}, n=2: {UU}, n=3: {UUU, UUD}, n=4: {UUUU, UUUD, UUDU}, n=5: {UUUUU, UUUUD, UUUDD, UUDUU, UUUDU, UUDUD}. For comparison: A037952 counts incomplete Dyck paths with n steps with any number of intermediate returns to the horizontal axis, ending above the horizontal axis. - R. J. Mathar, Sep 24 2021
a(n) is the number of noncrossing partitions of [n] whose nontrivial blocks are of type {a,b}, with a <= n/2, b > n/2. - Francesca Aicardi, May 29 2022
Maximal coefficient of (1+x)^n. - Vaclav Kotesovec, Dec 30 2022
Sums of lower-left-to-upper-right diagonals of the Catalan Triangle A001263. - Howard A. Landman, Sep 16 2024

Examples

			For n = 4, the a(4) = 6 distinct strings of length 4, each of which is a prefix of a string of balanced parentheses, are ((((, (((), (()(, ()((, ()(), and (()). - _Lee A. Newberg_, Apr 26 2010
There are a(5)=10 symmetric balanced strings of 5 pairs of parentheses:
[ 1] ((((()))))
[ 2] (((()())))
[ 3] ((()()()))
[ 4] ((())(()))
[ 5] (()()()())
[ 6] (()(())())
[ 7] (())()(())
[ 8] ()()()()()
[ 9] ()((()))()
[10] ()(()())() - _Joerg Arndt_, Jul 25 2011
G.f. = 1 + x + 2*x^2 + 3*x^3 + 6*x^4 + 10*x^5 + 20*x^6 + 35*x^7 + 70*x^8 + ...
The a(4)=6 binary 4-tuples such that the number of 1's in the even positions is the same as the number of 1's in the odd positions are 0000, 1100, 1001, 0110, 0011, 1111. - _Juan A. Olmos_, Dec 21 2017
		

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.
  • M. Aigner and G. M. Ziegler, Proofs from The Book, Springer-Verlag, Berlin, 1999; see p. 135.
  • K. Engel, Sperner Theory, Camb. Univ. Press, 1997; Theorem 1.1.1.
  • P. Frankl, Extremal sets systems, Chap. 24 of R. L. Graham et al., eds, Handbook of Combinatorics, North-Holland.
  • 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).
  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 7.16(b), p. 452.

Crossrefs

Row sums of Catalan triangle A053121 and of symmetric Dyck paths A088855.
Enumerates the structures encoded by A061854 and A061855.
First differences are in A037952.
Apparently a(n) = lim_{k->infinity} A094718(k, n).
Partial sums are in A036256. Column k=2 of A182172. Column k=1 of A335570.
Bisections: A000984 (even part), A001700 (odd part).
Cf. A097331.
Cf. A107373, A340567, A340568, A340569 (popularity of certain patterns in faro permutations).

Programs

  • GAP
    List([0..40],n->Binomial(n,Int(n/2))); # Muniru A Asiru, Apr 08 2018
    
  • Haskell
    a001405 n = a007318_row n !! (n `div` 2) -- Reinhard Zumkeller, Nov 09 2011
    
  • Magma
    [Binomial(n, Floor(n/2)): n in [0..40]]; // Vincenzo Librandi, Nov 16 2014
    
  • Maple
    A001405 := n->binomial(n, floor(n/2)): seq(A001405(n), n=0..33);
  • Mathematica
    Table[Binomial[n, Floor[n/2]], {n, 0, 40}] (* Stefan Steinerberger, Apr 08 2006 *)
    Table[DifferenceRoot[Function[{a,n},{-4 n a[n]-2 a[1+n]+(2+n) a[2+n] == 0,a[1] == 1,a[2] == 1}]][n], {n, 30}] (* Luciano Ancora, Jul 08 2015 *)
    Array[Binomial[#,Floor[#/2]]&,40,0] (* Harvey P. Dale, Mar 05 2018 *)
  • Maxima
    A001405(n):=binomial(n,floor(n/2))$
    makelist(A001405(n),n,0,30); /* Martin Ettl, Nov 01 2012 */
    
  • PARI
    a(n) = binomial(n, n\2);
    
  • PARI
    first(n) = x='x+O('x^n); Vec((-1+2*x+sqrt(1-4*x^2))/(2*x-4*x^2)) \\ Iain Fox, Dec 20 2017 (edited by Iain Fox, May 07 2018)
    
  • Python
    from math import comb
    def A001405(n): return comb(n,n//2) # Chai Wah Wu, Jun 07 2022

Formula

a(n) = max_{k=0..n} binomial(n, k).
a(2*n) = A000984(n), a(2*n+1) = A001700(n).
By symmetry, a(n) = binomial(n, ceiling(n/2)). - Labos Elemer, Mar 20 2003
P-recursive with recurrence: a(0) = 1, a(1) = 1, and for n >= 2, (n+1)*a(n) = 2*a(n-1) + 4*(n-1)*a(n-2). - Peter Bala, Feb 28 2011
G.f.: (1+x*c(x^2))/sqrt(1-4*x^2) = 1/(1 - x - x^2*c(x^2)); where c(x) = g.f. for Catalan numbers A000108.
G.f.: (-1 + 2*x + sqrt(1-4*x^2))/(2*x - 4*x^2). - Lee A. Newberg, Apr 26 2010
G.f.: 1/(1 - x - x^2/(1 - x^2/(1 - x^2/(1 - x^2/(1 - ... (continued fraction). - Paul Barry, Aug 12 2009
a(0) = 1; a(2*m+2) = 2*a(2*m+1); a(2*m+1) = Sum_{k = 0..2*m} (-1)^k*a(k)*a(2*m-k). - Len Smiley, Dec 09 2001
G.f.: (sqrt((1+2*x)/(1-2*x)) - 1)/(2*x). - Vladeta Jovovic, Apr 28 2003
The o.g.f. A(x) satisfies A(x) + x*A^2(x) = 1/(1-2*x). - Peter Bala, Feb 28 2011
E.g.f.: BesselI(0, 2*x) + BesselI(1, 2*x). - Vladeta Jovovic, Apr 28 2003
a(0) = 1; a(2*m+2) = 2*a(2*m+1); a(2*m+1) = 2*a(2*m) - c(m), where c(m)=A000108(m) are the Catalan numbers. - Christopher Hanusa (chanusa(AT)washington.edu), Nov 25 2003
a(n) = Sum_{k=0..n} (-1)^k*2^(n-k)*binomial(n, k)*A000108(k). - Paul Barry, Jan 27 2005
a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)*binomial(1, n-2*k). - Paul Barry, May 13 2005
From Paul Barry, Nov 02 2004: (Start)
a(n) = Sum_{k=0..floor((n+1)/2)} (binomial(n+1, k)*(cos((n-2*k+1)*Pi/2) + sin((n-2*k+1)*Pi/2))).
a(n) = Sum_{k=0..n+1}, (binomial(n+1, (n-k+1)/2)*(1-(-1)^(n-k))*(cos(k*Pi/2) + sin(k*Pi))/2). (End)
a(n) = Sum_{k=floor(n/2)..n} (binomial(n,n-k) - binomial(n,n-k-1)). - Paul Barry, Sep 06 2007
Inverse binomial transform of A005773 starting (1, 2, 5, 13, 35, 96, ...) and double inverse binomial transform of A001700. Row sums of triangle A132815. - Gary W. Adamson, Aug 31 2007
a(n) = Sum_{k=0..n} A120730(n,k). - Philippe Deléham, Oct 16 2008
a(n) = Sum_{k = 0..floor(n/2)} (binomial(n,k) - binomial(n,k-1)). - Nishant Doshi (doshinikki2004(AT)gmail.com), Apr 06 2009
Sum_{n>=0} a(n)/10^(n+1) = 0.1123724... = (sqrt(3)-sqrt(2))/(2*sqrt(2)); Sum_{n>=0} a(n)/100^(n+1) = 0.0101020306102035... = (sqrt(51)-sqrt(49))/(2*sqrt(49)). - Mark Dols, Jul 15 2010
Conjectured: a(n) = 2^n*2F1(1/2,-n;2;2), useful for number of paths in 1-d for which the coordinate is never negative. - Benjamin Phillabaum, Feb 20 2011
a(2*m+1) = (2*m+1)*a(2*m)/(m+1), e.g., a(7) = (7/4)*a(6) = (7/4)*20 = 35. - Jon Perry, Jan 20 2011
From Peter Bala, Feb 28 2011: (Start)
Let F(x) be the logarithmic derivative of the o.g.f. A(x). Then 1+x*F(x) is the o.g.f. for A027306.
Let G(x) be the logarithmic derivative of 1+x*A(x). Then x*G(x) is the o.g.f. for A058622. (End)
Let M = an infinite tridiagonal matrix with 1's in the super and subdiagonals and [1,0,0,0,...] in the main diagonal; and V = the vector [1,0,0,0,...]. a(n) = M^n*V, leftmost term. - Gary W. Adamson, Jun 13 2011
Let M = an infinite tridiagonal matrix with 1's in the super and subdiagonals and [1,0,0,0,...] in the main diagonal. a(n) = M^n_{1,1}. - Corrected by Gary W. Adamson, Jan 30 2012
a(n) = A007318(n, floor(n/2)). - Reinhard Zumkeller, Nov 09 2011
a(n+1) = Sum_{k=0..n} a(n-k)*A097331(k) = a(n) + Sum_{k=0..(n-1)/2} A000108(k)*a(n-2*k-1). - Philippe Deléham, Nov 27 2011
a(n) = A214282(n) - A214283(n), for n > 0. - Reinhard Zumkeller, Jul 14 2012
a(n) = Sum_{k=0..n} A168511(n,k)*(-1)^(n-k). - Philippe Deléham, Mar 19 2013
a(n+2*p-2) = Sum_{k=0..floor(n/2)} A009766(n-k+p-1, k+p-1) + binomial(n+2*p-2, p-2), for p >= 1. - Johannes W. Meijer, Aug 02 2013
O.g.f.: (1-x*c(x^2))/(1-2*x), with the o.g.f. c(x) of Catalan numbers A000108. See the rewritten formula given by Lee A. Newberg above. This is the o.g.f. for the row sums the Riordan triangle A053121. - Wolfdieter Lang, Sep 22 2013
a(n) ~ 2^n / sqrt(Pi * n/2). - Charles R Greathouse IV, Oct 23 2015
a(n) = 2^n*hypergeom([1/2,-n], [2], 2). - Vladimir Reshetnikov, Nov 02 2015
a(2*k) = Sum_{i=0..k} binomial(k, i)*binomial(k, i), a(2*k+1) = Sum_{i=0..k} binomial(k+1, i)*binomial(k, i). - Juan A. Olmos, Dec 21 2017
a(0) = 1, a(n) = 2 * a(n-1) for even n, a(n) = (2*n/(n+1)) * a(n-1) for odd n. - James East, Sep 25 2019
a(n) = A037952(n) + A000108(n/2) where A(.)=0 for non-integer argument. - R. J. Mathar, Sep 23 2021
From Amiram Eldar, Mar 10 2022: (Start)
Sum_{n>=0} 1/a(n) = 2*Pi/(3*sqrt(3)) + 2.
Sum_{n>=0} (-1)^n/a(n) = 2/3 - 2*Pi/(9*sqrt(3)). (End)
For k>2, Sum_{n>=0} a(n)/k^n = (sqrt((k+2)/(k-2)) - 1)*k/2. - Vaclav Kotesovec, May 13 2022
From Peter Bala, Mar 24 2023: (Start)
a(n) = Sum_{k = 0..n+1} (-1)^(k+binomial(n+2,2)) * k/(n+1) * binomial(n+1,k)^2.
(n + 1)*(2*n - 1)*a(n) = (-1)^(n+1)*2*a(n-1) + 4*(n - 1)*(2*n + 1)*a(n-2) with a(0) = a(1) = 1. (End)
a(n) = Integral_{x=-2..2} x^n*W(x)*dx, n>=0, where W(x) = sqrt((2+x)/(2-x))/(2*Pi) is a positive function on x=(-2,2) and is singular at x = 2. Therefore a(n) is a positive definite sequence. - Karol A. Penson, May 12 2025

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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Comments

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

A005043 Riordan numbers: a(n) = (n-1)*(2*a(n-1) + 3*a(n-2))/(n+1).

Original entry on oeis.org

1, 0, 1, 1, 3, 6, 15, 36, 91, 232, 603, 1585, 4213, 11298, 30537, 83097, 227475, 625992, 1730787, 4805595, 13393689, 37458330, 105089229, 295673994, 834086421, 2358641376, 6684761125, 18985057351, 54022715451, 154000562758, 439742222071, 1257643249140
Offset: 0

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Keywords

Comments

Also called Motzkin summands or ring numbers.
The old name was "Motzkin sums", used in certain publications. The sequence has the property that Motzkin(n) = A001006(n) = a(n) + a(n+1), e.g., A001006(4) = 9 = 3 + 6 = a(4) + a(5).
Number of 'Catalan partitions', that is partitions of a set 1,2,3,...,n into parts that are not singletons and whose convex hulls are disjoint when the points are arranged on a circle (so when the parts are all pairs we get Catalan numbers). - Aart Blokhuis (aartb(AT)win.tue.nl), Jul 04 2000
Number of ordered trees with n edges and no vertices of outdegree 1. For n > 1, number of dissections of a convex polygon by nonintersecting diagonals with a total number of n+1 edges. - Emeric Deutsch, Mar 06 2002
Number of Motzkin paths of length n with no horizontal steps at level 0. - Emeric Deutsch, Nov 09 2003
Number of Dyck paths of semilength n with no peaks at odd level. Example: a(4)=3 because we have UUUUDDDD, UUDDUUDD and UUDUDUDD, where U=(1,1), D=(1,-1). Number of Dyck paths of semilength n with no ascents of length 1 (an ascent in a Dyck path is a maximal string of up steps). Example: a(4)=3 because we have UUUUDDDD, UUDDUUDD and UUDUUDDD. - Emeric Deutsch, Dec 05 2003
Arises in Schubert calculus as follows. Let P = complex projective space of dimension n+1. Take n projective subspaces of codimension 3 in P in general position. Then a(n) is the number of lines of P intersecting all these subspaces. - F. Hirzebruch, Feb 09 2004
Difference between central trinomial coefficient and its predecessor. Example: a(6) = 15 = 141 - 126 and (1 + x + x^2)^6 = ... + 126*x^5 + 141*x^6 + ... (Catalan number A000108(n) is the difference between central binomial coefficient and its predecessor.) - David Callan, Feb 07 2004
a(n) = number of 321-avoiding permutations on [n] in which each left-to-right maximum is a descent (i.e., is followed by a smaller number). For example, a(4) counts 4123, 3142, 2143. - David Callan, Jul 20 2005
The Hankel transform of this sequence give A000012 = [1, 1, 1, 1, 1, 1, 1, ...]; example: Det([1, 0, 1, 1; 0, 1, 1, 3; 1, 1, 3, 6; 1, 3, 6, 15]) = 1. - Philippe Deléham, May 28 2005
The number of projective invariants of degree 2 for n labeled points on the projective line. - Benjamin J. Howard (bhoward(AT)ima.umn.edu), Nov 24 2006
Define a random variable X=trA^2, where A is a 2 X 2 unitary symplectic matrix chosen from USp(2) with Haar measure. The n-th central moment of X is E[(X+1)^n] = a(n). - Andrew V. Sutherland, Dec 02 2007
Let V be the adjoint representation of the complex Lie algebra sl(2). The dimension of the invariant subspace of the n-th tensor power of V is a(n). - Samson Black (sblack1(AT)uoregon.edu), Aug 27 2008
Starting with offset 3 = iterates of M * [1,1,1,...], where M = a tridiagonal matrix with [0,1,1,1,...] in the main diagonal and [1,1,1,...] in the super and subdiagonals. - Gary W. Adamson, Jan 08 2009
a(n) has the following standard-Young-tableaux (SYT) interpretation: binomial(n+1,k)*binomial(n-k-1,k-1)/(n+1)=f^(k,k,1^{n-2k}) where f^lambda equals the number of SYT of shape lambda. - Amitai Regev (amotai.regev(AT)weizmann.ac.il), Mar 02 2010
a(n) is also the sum of the numbers of standard Young tableaux of shapes (k,k,1^{n-2k}) for all 1 <= k <= floor(n/2). - Amitai Regev (amotai.regev(AT)weizmann.ac.il), Mar 10 2010
a(n) is the number of derangements of {1,2,...,n} having genus 0. The genus g(p) of a permutation p of {1,2,...,n} is defined by g(p)=(1/2)[n+1-z(p)-z(cp')], where p' is the inverse permutation of p, c = 234...n1 = (1,2,...,n), and z(q) is the number of cycles of the permutation q. Example: a(3)=1 because p=231=(123) is the only derangement of {1,2,3} with genus 0. Indeed, cp'=231*312=123=(1)(2)(3) and so g(p) = (1/2)(3+1-1-3)=0. - Emeric Deutsch, May 29 2010
Apparently: Number of Dyck 2n-paths with all ascents length 2 and no descent length 2. - David Scambler, Apr 17 2012
This is true. Proof: The mapping "insert a peak (UD) after each upstep (U)" is a bijection from all Dyck n-paths to those Dyck (2n)-paths in which each ascent is of length 2. It sends descents of length 1 in the n-path to descents of length 2 in the (2n)-path. But Dyck n-paths with no descents of length 1 are equinumerous with Riordan n-paths (Motzkin n-paths with no flatsteps at ground level) as follows. Given a Dyck n-path with no descents of length 1, split it into consecutive step pairs, then replace UU with U, DD with D, UD with a blue flatstep (F), DU with a red flatstep, and concatenate the new steps to get a colored Motzkin path. Each red F will be (immediately) preceded by a blue F or a D. In the latter case, transfer the red F so that it precedes the matching U of the D. Finally, erase colors to get the required Riordan path. For example, with lowercase f denoting a red flatstep, U^5 D^2 U D^4 U^4 D^3 U D^2 -> (U^2, U^2, UD, DU, D^2, D^2, U^2, U^2 D^2, DU, D^2) -> UUFfDDUUDfD -> UUFFDDUFUDD. - David Callan, Apr 25 2012
From Nolan Wallach, Aug 20 2014: (Start)
Let ch[part1, part2] be the value of the character of the symmetric group on n letters corresponding to the partition part1 of n on the conjucgacy class given by part2. Let A[n] be the set of (n+1) partitions of 2n with parts 1 or 2. Then deleting the first term of the sequence one has a(n) = Sum_{k=1..n+1} binomial(n,k-1)*ch[[n,n], A[n][[k]]])/2^n. This via the Frobenius Character Formula can be interpreted as the dimension of the SL(n,C) invariants in tensor^n (wedge^2 C^n).
Explanation: Let p_j denote sum (x_i)^j the sum in k variables. Then the Frobenius formula says then (p_1)^j_1 (p_2)^j_2 ... (p_r)^j_r is equal to sum(lambda, ch[lambda, 1^j_12^j_2 ... r^j_r] S_lambda) with S_lambda the Schur function corresponding to lambda. This formula implies that the coefficient of S([n,n]) in (((p_1)^1+p_2)/2)^n in its expansion in terms of Schur functions is the right hand side of our formula. If we specialize the number of variables to 2 then S[n,n](x,y)=(xy)^n. Which when restricted to y=x^(-1) is 1. That is it is 1 on SL(2).
On the other hand ((p_1)^2+p_2)/2 is the complete homogeneous symmetric function of degree 2 that is tr(S^2(X)). Thus our formula for a(n) is the same as that of Samson Black above since his V is the same as S^2(C^2) as a representation of SL(2). On the other hand, if we multiply ch(lambda) by sgn you get ch(Transpose(lambda)). So ch([n,n]) becomes ch([2,...,2]) (here there are n 2's). The formula for a(n) is now (1/2^n)*Sum_{j=0..n} ch([2,..,2], 1^(2n-2j) 2^j])*(-1)^j)*binomial(n,j), which calculates the coefficient of S_(2,...,2) in (((p_1)^2-p_2)/2)^n. But ((p_1)^2-p_2)/2 in n variables is the second elementary symmetric function which is the character of wedge^2 C^n and S_(2,...,2) is 1 on SL(n).
(End)
a(n) = number of noncrossing partitions (A000108) of [n] that contain no singletons, also number of nonnesting partitions (A000108) of [n] that contain no singletons. - David Callan, Aug 27 2014
From Tom Copeland, Nov 02 2014: (Start)
Let P(x) = x/(1+x) with comp. inverse Pinv(x) = x/(1-x) = -P[-x], and C(x)= [1-sqrt(1-4x)]/2, an o.g.f. for the shifted Catalan numbers A000108, with inverse Cinv(x) = x * (1-x).
Fin(x) = P[C(x)] = C(x)/[1 + C(x)] is an o.g.f. for the Fine numbers, A000957 with inverse Fin^(-1)(x) = Cinv[Pinv(x)] = Cinv[-P(-x)].
Mot(x) = C[P(x)] = C[-Pinv(-x)] gives an o.g.f. for shifted A005043, the Motzkin or Riordan numbers with comp. inverse Mot^(-1)(x) = Pinv[Cinv(x)] = (x - x^2) / (1 - x + x^2) (cf. A057078).
BTC(x) = C[Pinv(x)] gives A007317, a binomial transform of the Catalan numbers, with BTC^(-1)(x) = P[Cinv(x)].
Fib(x) = -Fin[Cinv(Cinv(-x))] = -P[Cinv(-x)] = x + 2 x^2 + 3 x^3 + 5 x^4 + ... = (x+x^2)/[1-x-x^2] is an o.g.f. for the shifted Fibonacci sequence A000045, so the comp. inverse is Fib^(-1)(x) = -C[Pinv(-x)] = -BTC(-x) and Fib(x) = -BTC^(-1)(-x).
Various relations among the o.g.f.s may be easily constructed, such as Fib[-Mot(-x)] = -P[P(-x)] = x/(1-2*x) a generating fct for 2^n.
Generalizing to P(x,t) = x /(1 + t*x) and Pinv(x,t) = x /(1 - t*x) = -P(-x,t) gives other relations to lattice paths, such as the o.g.f. for A091867, C[P[x,1-t]], and that for A104597, Pinv[Cinv(x),t+1]. (End)
Consistent with David Callan's comment above, A249548, provides a refinement of the Motzkin sums into the individual numbers for the non-crossing partitions he describes. - Tom Copeland, Nov 09 2014
The number of lattice paths from (0,0) to (n,0) that do not cross below the x-axis and use up-step=(1,1) and down-steps=(1,-k) where k is a positive integer. For example, a(4) = 3: [(1,1)(1,1)(1,-1)(1,-1)], [(1,1)(1,-1)(1,1)(1,-1)] and [(1,1)(1,1)(1,1)(1,-3)]. - Nicholas Ham, Aug 19 2015
A series created using 2*(a(n) + a(n+1)) + (a(n+1) + a(n+2)) has Hankel transform of F(2n), offset 3, F being a Fibonacci number, A001906 (Empirical observation). - Tony Foster III, Jul 30 2016
The series a(n) + A001006(n) has Hankel transform F(2n+1), offset n=1, F being the Fibonacci bisection A001519 (empirical observation). - Tony Foster III, Sep 05 2016
The Rubey and Stump reference proves a refinement of a conjecture of René Marczinzik, which they state as: "The number of 2-Gorenstein algebras which are Nakayama algebras with n simple modules and have an oriented line as associated quiver equals the number of Motzkin paths of length n. Moreover, the number of such algebras having the double centraliser property with respect to a minimal faithful projective-injective module equals the number of Riordan paths, that is, Motzkin paths without level-steps at height zero, of length n." - Eric M. Schmidt, Dec 16 2017
A connection to the Thue-Morse sequence: (-1)^a(n) = (-1)^A010060(n) * (-1)^A010060(n+1) = A106400(n) * A106400(n+1). - Vladimir Reshetnikov, Jul 21 2019
Named by Bernhart (1999) after the American mathematician John Riordan (1903-1988). - Amiram Eldar, Apr 15 2021

Examples

			a(5)=6 because the only dissections of a polygon with a total number of 6 edges are: five pentagons with one of the five diagonals and the hexagon with no diagonals.
G.f. = 1 + x^2 + x^3 + 3*x^4 + 6*x^5 + 15*x^6 + 36*x^7 + 91*x^8 + 232*x^9 + ...
From _Gus Wiseman_, Nov 15 2022: (Start)
The a(0) = 1 through a(6) = 15 lone-child-avoiding (no vertices of outdegree 1) ordered rooted trees with n + 1 vertices (ranked by A358376):
  o  .  (oo)  (ooo)  (oooo)   (ooooo)   (oooooo)
                     ((oo)o)  ((oo)oo)  ((oo)ooo)
                     (o(oo))  ((ooo)o)  ((ooo)oo)
                              (o(oo)o)  ((oooo)o)
                              (o(ooo))  (o(oo)oo)
                              (oo(oo))  (o(ooo)o)
                                        (o(oooo))
                                        (oo(oo)o)
                                        (oo(ooo))
                                        (ooo(oo))
                                        (((oo)o)o)
                                        ((o(oo))o)
                                        ((oo)(oo))
                                        (o((oo)o))
                                        (o(o(oo)))
(End)
		

References

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

Crossrefs

Row sums of triangle A020474, first differences of A082395.
First diagonal of triangular array in A059346.
Binomial transform of A126930. - Philippe Deléham, Nov 26 2009
The Hankel transform of a(n+1) is A128834. The Hankel transform of a(n+2) is floor((2*n+4)/3) = A004523(n+2). - Paul Barry, Mar 08 2011
The Kn11 triangle sums of triangle A175136 lead to A005043(n+2), while the Kn12(n) = A005043(n+4)-2^(n+1), Kn13(n) = A005043(n+6)-(n^2+9*n+56)*2^(n-2) and the Kn4(n) = A005043(2*n+2) = A099251(n+1) triangle sums are related to the sequence given above. For the definitions of these triangle sums see A180662. - Johannes W. Meijer, May 06 2011
Cf. A187306 (self-convolution), A348210 (column 1).
Bisections: A099251, A099252.

Programs

  • Haskell
    a005043 n = a005043_list !! n
    a005043_list = 1 : 0 : zipWith div
       (zipWith (*) [1..] (zipWith (+)
           (map (* 2) $ tail a005043_list) (map (* 3) a005043_list))) [3..]
    -- Reinhard Zumkeller, Jan 31 2012
    
  • Maple
    A005043 := proc(n) option remember; if n <= 1 then 1-n else (n-1)*(2*A005043(n-1)+3*A005043(n-2))/(n+1); fi; end;
    Order := 20: solve(series((x-x^2)/(1-x+x^2),x)=y,x); # outputs g.f.
  • Mathematica
    a[0]=1; a[1]=0; a[n_]:= a[n] = (n-1)*(2*a[n-1] + 3*a[n-2])/(n+1); Table[ a[n], {n, 0, 30}] (* Robert G. Wilson v, Jun 14 2005 *)
    Table[(-3)^(1/2)/6 * (-1)^n*(3*Hypergeometric2F1[1/2,n+1,1,4/3]+ Hypergeometric2F1[1/2,n+2,1,4/3]), {n,0,32}] (* cf. Mark van Hoeij in A001006 *) (* Wouter Meeussen, Jan 23 2010 *)
    RecurrenceTable[{a[0]==1,a[1]==0,a[n]==(n-1) (2a[n-1]+3a[n-2])/(n+1)},a,{n,30}] (* Harvey P. Dale, Sep 27 2013 *)
    a[ n_]:= SeriesCoefficient[2/(1+x +Sqrt[1-2x-3x^2]), {x, 0, n}]; (* Michael Somos, Aug 21 2014 *)
    a[ n_]:= If[n<0, 0, 3^(n+3/2) Hypergeometric2F1[3/2, n+2, 2, 4]/I]; (* Michael Somos, Aug 21 2014 *)
    Table[3^(n+3/2) CatalanNumber[n] (4(5+2n)Hypergeometric2F1[3/2, 3/2, 1/2-n, 1/4] -9 Hypergeometric2F1[3/2, 5/2, 1/2 -n, 1/4])/(4^(n+3) (n+1)), {n, 0, 31}] (* Vladimir Reshetnikov, Jul 21 2019 *)
    Table[Sqrt[27]/8 (3/4)^n CatalanNumber[n] Hypergeometric2F1[1/2, 3/2, 1/2 - n, 1/4], {n, 0, 31}] (* Jan Mangaldan, Sep 12 2021 *)
  • Maxima
    a[0]:1$
    a[1]:0$
    a[n]:=(n-1)*(2*a[n-1]+3*a[n-2])/(n+1)$
    makelist(a[n],n,0,12); /* Emanuele Munarini, Mar 02 2011 */
    
  • PARI
    {a(n) = if( n<0, 0, n++; polcoeff( serreverse( (x - x^3) / (1 + x^3) + x * O(x^n)), n))}; /* Michael Somos, May 31 2005 */
    
  • PARI
    my(N=66); Vec(serreverse(x/(1+x*sum(k=1,N,x^k))+O(x^N))) \\ Joerg Arndt, Aug 19 2012
    
  • Python
    from functools import cache
    @cache
    def A005043(n: int) -> int:
        if n <= 1: return 1 - n
        return (n - 1) * (2 * A005043(n - 1) + 3 * A005043(n - 2)) // (n + 1)
    print([A005043(n) for n in range(32)]) # Peter Luschny, Nov 20 2022
  • Sage
    A005043 = lambda n: (-1)^n*jacobi_P(n,1,-n-3/2,-7)/(n+1)
    [simplify(A005043(n)) for n in (0..29)]
    # Peter Luschny, Sep 23 2014
    
  • Sage
    def ms():
        a, b, c, d, n = 0, 1, 1, -1, 1
        yield 1
        while True:
            yield -b + (-1)^n*d
            n += 1
            a, b = b, (3*(n-1)*n*a+(2*n-1)*n*b)/((n+1)*(n-1))
            c, d = d, (3*(n-1)*c-(2*n-1)*d)/n
    A005043 = ms()
    print([next(A005043) for  in range(32)]) # _Peter Luschny, May 16 2016
    

Formula

a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*A000108(k). a(n) = (1/(n+1)) * Sum_{k=0..ceiling(n/2)} binomial(n+1, k)*binomial(n-k-1, k-1), for n > 1. - Len Smiley. [Comment from Amitai Regev (amitai.regev(AT)weizmann.ac.il), Mar 02 2010: the latter sum should be over the range k=1..floor(n/2).]
G.f.: (1 + x - sqrt(1-2*x-3*x^2))/(2*x*(1+x)).
G.f.: 2/(1+x+sqrt(1-2*x-3*x^2)). - Paul Peart (ppeart(AT)fac.howard.edu), May 27 2000
a(n+1) + (-1)^n = a(0)*a(n) + a(1)*a(n-1) + ... + a(n)*a(0). - Bernhart
a(n) = (1/(n+1)) * Sum_{i} (-1)^i*binomial(n+1, i)*binomial(2*n-2*i, n-i). - Bernhart
G.f. A(x) satisfies A = 1/(1+x) + x*A^2.
E.g.f.: exp(x)*(BesselI(0, 2*x) - BesselI(1, 2*x)). - Vladeta Jovovic, Apr 28 2003
a(n) = A001006(n-1) - a(n-1).
a(n+1) = Sum_{k=0..n} (-1)^k*A026300(n, k), where A026300 is the Motzkin triangle.
a(n) = Sum_{k=0..n} (-1)^k*binomial(n, k)*binomial(k, floor(k/2)). - Paul Barry, Jan 27 2005
a(n) = Sum_{k>=0} A086810(n-k, k). - Philippe Deléham, May 30 2005
a(n+2) = Sum_{k>=0} A064189(n-k, k). - Philippe Deléham, May 31 2005
Moment representation: a(n) = (1/(2*Pi))*Int(x^n*sqrt((1+x)(3-x))/(1+x),x,-1,3). - Paul Barry, Jul 09 2006
Inverse binomial transform of A000108 (Catalan numbers). - Philippe Deléham, Oct 20 2006
a(n) = (2/Pi)* Integral_{x=0..Pi} (4*cos(x)^2-1)^n*sin(x)^2 dx. - Andrew V. Sutherland, Dec 02 2007
G.f.: 1/(1-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-... (continued fraction). - Paul Barry, Jan 22 2009
G.f.: 1/(1+x-x/(1-x/(1+x-x/(1-x/(1+x-x/(1-... (continued fraction). - Paul Barry, May 16 2009
G.f.: 1/(1-x^2/(1-x/(1-x/(1-x^2/(1-x/(1-x/(1-x^2/(1-x/(1-... (continued fraction). - Paul Barry, Mar 02 2010
a(n) = -(-1)^n * hypergeom([1/2, n+2],[2],4/3) / sqrt(-3). - Mark van Hoeij, Jul 02 2010
a(n) = (-1)^n*hypergeometric([-n,1/2],[2],4). - Peter Luschny, Aug 15 2012
Let A(x) be the g.f., then x*A(x) is the reversion of x/(1 + x^2*Sum_{k>=0} x^k); see A215340 for the correspondence to Dyck paths without length-1 ascents. - Joerg Arndt, Aug 19 2012 and Apr 16 2013
a(n) ~ 3^(n+3/2)/(8*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 02 2012
G.f.: 2/(1+x+1/G(0)), where G(k) = 1 + x*(2+3*x)*(4*k+1)/( 4*k+2 - x*(2+3*x)*(4*k+2)*(4*k+3)/(x*(2+3*x)*(4*k+3) + 4*(k+1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 05 2013
D-finite (an alternative): (n+1)*a(n) = 3*(n-2)*a(n-3) + (5*n-7)*a(n-2) + (n-2)*a(n-1), n >= 3. - Fung Lam, Mar 22 2014
Asymptotics: a(n) = (3^(n+2)/(sqrt(3*n*Pi)*(8*n)))*(1-21/(16*n) + O(1/n^2)) (with contribution by Vaclav Kotesovec). - Fung Lam, Mar 22 2014
a(n) = T(2*n-1,n)/n, where T(n,k) = triangle of A180177. - Vladimir Kruchinin, Sep 23 2014
a(n) = (-1)^n*JacobiP(n,1,-n-3/2,-7)/(n+1). - Peter Luschny, Sep 23 2014
a(n) = Sum_{k=0..n} C(n,k)*(C(k,n-k)-C(k,n-k-1)). - Peter Luschny, Oct 01 2014
Conjecture: a(n) = A002426(n) - A005717(n), n > 0. - Mikhail Kurkov, Feb 24 2019 [The conjecture is true. - Amiram Eldar, May 17 2024]
a(n) = A309303(n) + A309303(n+1). - Vladimir Reshetnikov, Jul 22 2019
From Peter Bala, Feb 11 2022: (Start)
a(n) = A005773(n+1) - 2*A005717(n) for n >= 1.
Conjectures: for n >= 1, n divides a(2*n+1) and 2*n-1 divides a(2*n). (End)

Extensions

Thanks to Laura L. M. Yang (yanglm(AT)hotmail.com) for a correction, Aug 29 2004
Name changed to Riordan numbers following a suggestion from Ira M. Gessel. - N. J. A. Sloane, Jul 24 2020
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