cp's OEIS Frontend

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

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A001764 a(n) = binomial(3*n,n)/(2*n+1) (enumerates ternary trees and also noncrossing trees).

Original entry on oeis.org

1, 1, 3, 12, 55, 273, 1428, 7752, 43263, 246675, 1430715, 8414640, 50067108, 300830572, 1822766520, 11124755664, 68328754959, 422030545335, 2619631042665, 16332922290300, 102240109897695, 642312451217745, 4048514844039120, 25594403741131680, 162250238001816900
Offset: 0

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Author

Keywords

Comments

Smallest number of straight line crossing-free spanning trees on n points in the plane.
Number of dissections of some convex polygon by nonintersecting diagonals into polygons with an odd number of sides and having a total number of 2n+1 edges (sides and diagonals). - Emeric Deutsch, Mar 06 2002
Number of lattice paths of n East steps and 2n North steps from (0,0) to (n,2n) and lying weakly below the line y=2x. - David Callan, Mar 14 2004
With interpolated zeros, this has g.f. 2*sqrt(3)*sin(arcsin(3*sqrt(3)*x/2)/3)/(3*x) and a(n) = C(n+floor(n/2),floor(n/2))*C(floor(n/2),n-floor(n/2))/(n+1). This is the first column of the inverse of the Riordan array (1-x^2,x(1-x^2)) (essentially reversion of y-y^3). - Paul Barry, Feb 02 2005
Number of 12312-avoiding matchings on [2n].
Number of complete ternary trees with n internal nodes, or 3n edges.
Number of rooted plane trees with 2n edges, where every vertex has even outdegree ("even trees").
a(n) is the number of noncrossing partitions of [2n] with all blocks of even size. E.g.: a(2)=3 counts 12-34, 14-23, 1234. - David Callan, Mar 30 2007
Pfaff-Fuss-Catalan sequence C^{m}_n for m=3, see the Graham et al. reference, p. 347. eq. 7.66.
Also 3-Raney sequence, see the Graham et al. reference, p. 346-7.
The number of lattice paths from (0,0) to (2n,0) using an Up-step=(1,1) and a Down-step=(0,-2) and staying above the x-axis. E.g., a(2) = 3; UUUUDD, UUUDUD, UUDUUD. - Charles Moore (chamoore(AT)howard.edu), Jan 09 2008
a(n) is (conjecturally) the number of permutations of [n+1] that avoid the patterns 4-2-3-1 and 4-2-5-1-3 and end with an ascent. For example, a(4)=55 counts all 60 permutations of [5] that end with an ascent except 42315, 52314, 52413, 53412, all of which contain a 4-2-3-1 pattern and 42513. - David Callan, Jul 22 2008
Central terms of pendular triangle A167763. - Philippe Deléham, Nov 12 2009
With B(x,t)=x+t*x^3, the comp. inverse in x about 0 is A(x,t) = Sum_{j>=0} a(j) (-t)^j x^(2j+1). Let U(x,t)=(x-A(x,t))/t. Then DU(x,t)/Dt=dU/dt+U*dU/dx=0 and U(x,0)=x^3, i.e., U is a solution of the inviscid Burgers's, or Hopf, equation. Also U(x,t)=U(x-t*U(x,t),0) and dB(x,t)/dt = U(B(x,t),t) = x^3 = U(x,0). The characteristics for the Hopf equation are x(t) = x(0) + t*U(x(t),t) = x(0) + t*U(x(0),0) = x(0) + t*x(0)^3 = B(x(0),t). These results apply to all the Fuss-Catalan sequences with 3 replaced by n>0 and 2 by n-1 (e.g., A000108 with n=2 and A002293 with n=4), see also A086810, which can be generalized to A133437, for associahedra. - Tom Copeland, Feb 15 2014
Number of intervals (i.e., ordered pairs (x,y) such that x<=y) in the Kreweras lattice (noncrossing partitions ordered by refinement) of size n, see the Bernardi & Bonichon (2009) and Kreweras (1972) references. - Noam Zeilberger, Jun 01 2016
Number of sum-indecomposable (4231,42513)-avoiding permutations. Conjecturally, number of sum-indecomposable (2431,45231)-avoiding permutations. - Alexander Burstein, Oct 19 2017
a(n) is the number of topologically distinct endstates for the game Planted Brussels Sprouts on n vertices, see Ji and Propp link. - Caleb Ji, May 14 2018
Number of complete quadrillages of 2n+2-gons. See Baryshnikov p. 12. See also Nov 10 2014 comments in A134264. - Tom Copeland, Jun 04 2018
a(n) is the number of 2-regular words on the alphabet [n] that avoid the patterns 231 and 221. Equivalently, this is the number of 2-regular tortoise-sortable words on the alphabet [n] (see the Defant and Kravitz link). - Colin Defant, Sep 26 2018
a(n) is the number of Motzkin paths of length 3n with n steps of each type, with the condition that (1, 0) and (1, 1) steps alternate (starting with (1, 0)). - Helmut Prodinger, Apr 08 2019
a(n) is the number of uniquely sorted permutations of length 2n+1 that avoid the patterns 312 and 1342. - Colin Defant, Jun 08 2019
The compositional inverse o.g.f. pair in Copeland's comment above are related to a pair of quantum fields in Balduf's thesis by Theorem 4.2 on p. 92. - Tom Copeland, Dec 13 2019
The sequences of Fuss-Catalan numbers, of which this is the first after the Catalan numbers A000108 (the next is A002293), appear in articles on random matrices and quantum physics. See Banica et al., Collins et al., and Mlotkowski et al. Interpretations of these sequences in terms of the cardinality of specific sets of noncrossing partitions are provided by A134264. - Tom Copeland, Dec 21 2019
Call C(p, [alpha], g) the number of partitions of a cyclically ordered set with p elements, of cyclic type [alpha], and of genus g (the genus g Faa di Bruno coefficients of type [alpha]). This sequence counts the genus 0 partitions (non-crossing, or planar, partitions) of p = 3n into n parts of length 3: a(n) = C(3n, [3^n], 0). For genus 1 see A371250, for genus 2 see A371251. - Robert Coquereaux, Mar 16 2024
a(n) is the total number of down steps before the first up step in all 2_1-Dyck paths of length 3*n for n > 0. A 2_1-Dyck path is a lattice path with steps (1,2), (1,-1) that starts and ends at y = 0 and does not go below the line y = -1. - Sarah Selkirk, May 10 2020
a(n) is the number of pairs (A<=B) of noncrossing partitions of [n]. - Francesca Aicardi, May 28 2022
a(n) is the number of parking functions of size n avoiding the patterns 231 and 321. - Lara Pudwell, Apr 10 2023
Number of rooted polyominoes composed of n square cells of the hyperbolic regular tiling with Schläfli symbol {4,oo}. A rooted polyomino has one external edge identified, and chiral pairs are counted as two. A stereographic projection of the {4,oo} tiling on the Poincaré disk can be obtained via the Christensson link. - Robert A. Russell, Jan 27 2024
This is instance k = 3 of the family {C(k, n)}A130564.%20-%20_Wolfdieter%20Lang">{n>=0} given in a comment in A130564. - _Wolfdieter Lang, Feb 05 2024
The number of Apollonian networks (planar 3-trees) with n+3 vertices with a given base triangle. - Allan Bickle, Feb 20 2024
Number of rooted polyominoes composed of n tetrahedral cells of the hyperbolic regular tiling with Schläfli symbol {3,3,oo}. A rooted polyomino has one external face identified, and chiral pairs are counted as two. a(n) = T(n) in the second Beineke and Pippert link. - Robert A. Russell, Mar 20 2024

Examples

			a(2) = 3 because the only dissections with 5 edges are given by a square dissected by any of the two diagonals and the pentagon with no dissecting diagonal.
G.f. = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + 1428*x^6 + 7752*x^7 + 43263*x^8 + ...
		

References

  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 23.
  • I. M. H. Etherington, On non-associative combinations, Proc. Royal Soc. Edinburgh, 59 (Part 2, 1938-39), 153-162.
  • I. M. H. Etherington, Some problems of non-associative combinations (I), Edinburgh Math. Notes, 32 (1940), pp. i-vi. Part II is by A. Erdelyi and I. M. H. Etherington, and is on pages vii-xiv of the same issue.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, pp. 200, 347. See also the Pólya-Szegő reference.
  • W. Kuich, Languages and the enumeration of planted plane trees. Nederl. Akad. Wetensch. Proc. Ser. A 73 = Indag. Math. 32, (1970), 268-280.
  • T. V. Narayana, Lattice Path Combinatorics with Statistical Applications. Univ. Toronto Press, 1979, p. 98.
  • G. Pólya and G. Szegő, Problems and Theorems in Analysis, Springer-Verlag, New York, Heidelberg, Berlin, 2 vols., 1972, Vol. 1, problem 211, p. 146 with solution on p. 348.
  • 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

Cf. A001762, A001763, A002294 - A002296, A006013, A025174, A063548, A064017, A072247, A072248, A134264, A143603, A258708, A256311, A188687 (binomial transform), A346628 (inverse binomial transform).
A column of triangle A102537.
Bisection of A047749 and A047761.
Row sums of triangles A108410 and A108767.
Second column of triangle A062993.
Mod 3 = A113047.
2D Polyominoes: A005034 (oriented), A005036 (unoriented), A369315 (chiral), A047749 (achiral), A000108 {3,oo}, A002293 {5,oo}.
3D Polyominoes: A007173 (oriented), A027610 (unoriented), A371350 (chiral), A371351 (achiral).
Cf. A130564 (for C(k, n) cases).

Programs

  • GAP
    List([0..25],n->Binomial(3*n,n)/(2*n+1)); # Muniru A Asiru, Oct 31 2018
    
  • Haskell
    a001764 n = a001764_list !! n
    a001764_list = 1 : [a258708 (2 * n) n | n <- [1..]]
    -- Reinhard Zumkeller, Jun 23 2015
    
  • Magma
    [Binomial(3*n,n)/(2*n+1): n in [0..30]]; // Vincenzo Librandi, Sep 04 2014
    
  • Maple
    A001764 := n->binomial(3*n,n)/(2*n+1): seq(A001764(n), n=0..25);
    with(combstruct): BB:=[T,{T=Prod(Z,F),F=Sequence(B),B=Prod(F,Z,F)}, unlabeled]:seq(count(BB,size=i),i=0..22); # Zerinvary Lajos, Apr 22 2007
    with(combstruct):BB:=[S, {B = Prod(S,S,Z), S = Sequence(B)}, labelled]: seq(count(BB, size=n)/n!, n=0..21); # Zerinvary Lajos, Apr 25 2008
    n:=30:G:=series(RootOf(g = 1+x*g^3, g),x=0,n+1):seq(coeff(G,x,k),k=0..n); # Robert FERREOL, Apr 03 2015
    alias(PS=ListTools:-PartialSums): A001764List := proc(m) local A, P, n;
    A := [1,1]; P := [1]; for n from 1 to m - 2 do P := PS(PS([op(P), P[-1]]));
    A := [op(A), P[-1]] od; A end: A001764List(25); # Peter Luschny, Mar 26 2022
  • Mathematica
    InverseSeries[Series[y-y^3, {y, 0, 24}], x] (* then a(n)=y(2n+1)=ways to place non-crossing diagonals in convex (2n+4)-gon so as to create only quadrilateral tiles *) (* Len Smiley, Apr 08 2000 *)
    Table[Binomial[3n,n]/(2n+1),{n,0,25}] (* Harvey P. Dale, Jul 24 2011 *)
  • PARI
    {a(n) = if( n<0, 0, (3*n)! / n! / (2*n + 1)!)};
    
  • PARI
    {a(n) = if( n<0, 0, polcoeff( serreverse( x - x^3 + O(x^(2*n + 2))), 2*n + 1))};
    
  • PARI
    {a(n) = my(A); if( n<0, 0, A = 1 + O(x); for( m=1, n, A = 1 + x * A^3); polcoeff(A, n))};
    
  • PARI
    b=vector(22);b[1]=1;for(n=2,22,for(i=1,n-1,for(j=1,n-1,for(k=1,n-1,if((i-1)+(j-1)+(k-1)-(n-2),NULL,b[n]=b[n]+b[i]*b[j]*b[k])))));a(n)=b[n+1]; print1(a(0));for(n=1,21,print1(", ",a(n))) \\ Gerald McGarvey, Oct 08 2008
    
  • PARI
    Vec(1 + serreverse(x / (1+x)^3 + O(x^30))) \\ Gheorghe Coserea, Aug 05 2015
    
  • Python
    from math import comb
    def A001764(n): return comb(3*n,n)//(2*n+1) # Chai Wah Wu, Nov 10 2022
  • Sage
    def A001764_list(n) :
        D = [0]*(n+1); D[1] = 1
        R = []; b = false; h = 1
        for i in range(2*n) :
            for k in (1..h) : D[k] += D[k-1]
            if not b : R.append(D[h])
            else : h += 1
            b = not b
        return R
    A001764_list(22) # Peter Luschny, May 03 2012
    

Formula

From Karol A. Penson, Nov 08 2001: (Start)
G.f.: (2/sqrt(3*x))*sin((1/3)*arcsin(sqrt(27*x/4))).
E.g.f.: hypergeom([1/3, 2/3], [1, 3/2], 27/4*x).
Integral representation as n-th moment of a positive function on [0, 27/4]: a(n) = Integral_{x=0..27/4} (x^n*((1/12) * 3^(1/2) * 2^(1/3) * (2^(1/3)*(27 + 3 * sqrt(81 - 12*x))^(2/3) - 6 * x^(1/3))/(Pi * x^(2/3)*(27 + 3 * sqrt(81 - 12*x))^(1/3)))), n >= 0. This representation is unique. (End)
G.f. A(x) satisfies A(x) = 1+x*A(x)^3 = 1/(1-x*A(x)^2) [Cyvin (1998)]. - Ralf Stephan, Jun 30 2003
a(n) = n-th coefficient in expansion of power series P(n), where P(0) = 1, P(k+1) = 1/(1 - x*P(k)^2).
G.f. Rev(x/c(x))/x, where c(x) is the g.f. of A000108 (Rev=reversion of). - Paul Barry, Mar 26 2010
From Gary W. Adamson, Jul 07 2011: (Start)
Let M = the production matrix:
1, 1
2, 2, 1
3, 3, 2, 1
4, 4, 3, 2, 1
5, 5, 4, 3, 2, 1
...
a(n) = upper left term in M^n. Top row terms of M^n = (n+1)-th row of triangle A143603, with top row sums generating A006013: (1, 2, 7, 30, 143, 728, ...). (End)
Recurrence: a(0)=1; a(n) = Sum_{i=0..n-1, j=0..n-1-i} a(i)a(j)a(n-1-i-j) for n >= 1 (counts ternary trees by subtrees of the root). - David Callan, Nov 21 2011
G.f.: 1 + 6*x/(Q(0) - 6*x); Q(k) = 3*x*(3*k + 1)*(3*k + 2) + 2*(2*(k^2) + 5*k +3) - 6*x*(2*(k^2) + 5*k + 3)*(3*k + 4)*(3*k + 5)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 27 2011
D-finite with recurrence: 2*n*(2n+1)*a(n) - 3*(3n-1)*(3n-2)*a(n-1) = 0. - R. J. Mathar, Dec 14 2011
REVERT transform of A115140. BINOMIAL transform is A188687. SUMADJ transform of A188678. HANKEL transform is A051255. INVERT transform of A023053. INVERT transform is A098746. - Michael Somos, Apr 07 2012
(n + 1) * a(n) = A174687(n).
G.f.: F([2/3,4/3], [3/2], 27/4*x) / F([2/3,1/3], [1/2], (27/4)*x) where F() is the hypergeometric function. - Joerg Arndt, Sep 01 2012
a(n) = binomial(3*n+1, n)/(3*n+1) = A062993(n+1,1). - Robert FERREOL, Apr 03 2015
a(n) = A258708(2*n,n) for n > 0. - Reinhard Zumkeller, Jun 23 2015
0 = a(n)*(-3188646*a(n+2) + 20312856*a(n+3) - 11379609*a(n+4) + 1437501*a(n+5)) + a(n+1)*(177147*a(n+2) - 2247831*a(n+3) + 1638648*a(n+4) - 238604*a(n+5)) + a(n+2)*(243*a(n+2) + 31497*a(n+3) - 43732*a(n+4) + 8288*a(n+5)) for all integer n. - Michael Somos, Jun 03 2016
a(n) ~ 3^(3*n + 1/2)/(sqrt(Pi)*4^(n+1)*n^(3/2)). - Ilya Gutkovskiy, Nov 21 2016
Given g.f. A(x), then A(1/8) = -1 + sqrt(5), A(2/27) = (-1 + sqrt(3))*3/2, A(4/27) = 3/2, A(3/64) = -2 + 2*sqrt(7/3), A(5/64) = (-1 + sqrt(5))*2/sqrt(5), etc. A(n^2/(n+1)^3) = (n+1)/n if n > 1. - Michael Somos, Jul 17 2018
From Peter Bala, Sep 14 2021: (Start)
A(x) = exp( Sum_{n >= 1} (1/3)*binomial(3*n,n)*x^n/n ).
The sequence defined by b(n) := [x^n] A(x)^n = A224274(n) for n >= 1 and satisfies the congruence b(p) == b(1) (mod p^3) for prime p >= 3. Cf. A060941. (End)
G.f.: 1/sqrt(B(x)+(1-6*x)/(9*B(x))+1/3), with B(x):=((27*x^2-18*x+2)/54-(x*sqrt((-(4-27*x))*x))/(2*3^(3/2)))^(1/3). - Vladimir Kruchinin, Sep 28 2021
x*A'(x)/A(x) = (A(x) - 1)/(- 2*A(x) + 3) = x + 5*x^2 + 28*x^3 + 165*x^4 + ... is the o.g.f. of A025174. Cf. A002293 - A002296. - Peter Bala, Feb 04 2022
a(n) = hypergeom([1 - n, -2*n], [2], 1). Row sums of A108767. - Peter Bala, Aug 30 2023
G.f.: z*exp(3*z*hypergeom([1, 1, 4/3, 5/3], [3/2, 2, 2], (27*z)/4)) + 1.
- Karol A. Penson, Dec 19 2023
G.f.: hypergeometric([1/3, 2/3], [3/2], (3^3/2^2)*x). See the e.g.f. above. - Wolfdieter Lang, Feb 04 2024
a(n) = (3*n)! / (n!*(2*n+1)!). - Allan Bickle, Feb 20 2024
Sum_{n >= 0} a(n)*x^n/(1 + x)^(3*n+1) = 1. See A316371 and A346627. - Peter Bala, Jun 02 2024
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^5). - Seiichi Manyama, Jun 16 2025

A005807 Sum of adjacent Catalan numbers.

Original entry on oeis.org

2, 3, 7, 19, 56, 174, 561, 1859, 6292, 21658, 75582, 266798, 950912, 3417340, 12369285, 45052515, 165002460, 607283490, 2244901890, 8331383610, 31030387440, 115948830660, 434542177290, 1632963760974, 6151850548776
Offset: 0

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Author

Keywords

Comments

The aerated sequence has Hankel transform F(n+2)*F(n+3) (A001654(n+2)). - Paul Barry, Nov 04 2008

Examples

			G.f. = 2 + 3*x+ 7*x^2 + 19*x^3 + 56*x^4 + 174*x^5 + 561*x^6 + 1859*x^7 + ...
		

References

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

Crossrefs

Programs

  • Magma
    [((5*n+4)*Factorial(2*n))/(Factorial(n)*Factorial(n+2)): n in [0..30] ];  // Vincenzo Librandi, Aug 19 2011
    
  • Maple
    A005807List := proc(m) local A, P, n; A := [2,3]; P := [2,3];
    for n from 1 to m - 2 do P := ListTools:-PartialSums([op(P), P[-2]]);
    A := [op(A), P[-1]] od; A end: A005807List(25); # Peter Luschny, Mar 26 2022
  • Mathematica
    a[n_]:=Binomial[2*n, n]*(5*n+4)/(n+1)/(n+2); (* Vladimir Joseph Stephan Orlovsky, Dec 13 2008 *)
    a[ n_] := If[ n < 0, 0, CatalanNumber[n] + CatalanNumber[n + 1]]; (* Michael Somos, Jan 17 2015 *)
    Total/@Partition[CatalanNumber[Range[0,30]],2,1] (* Harvey P. Dale, Jun 21 2025 *)
  • PARI
    {a(n) = if( n<0, 0, binomial(2*n, n) * (5*n+4) / ((n+1) * (n+2)))};
    
  • Python
    from _future_ import division
    A005807_list, b = [], 2
    for n in range(10**3):
        A005807_list.append(b)
        b = b*(4*n+2)*(5*n+9)//((n+3)*(5*n+4)) # Chai Wah Wu, Jan 28 2016
  • Sage
    [catalan_number(i)+catalan_number(i+1) for i in range(0,25)] # Zerinvary Lajos, May 17 2009
    

Formula

a(n) = C(n)+C(n+1) = ((5*n+4)*(2*n)!)/(n!*(n+2)!).
G.f. A(x) satisfies x^2*A(x)^2 + (x-1)*A(x) + (x+2) = 0. - Michael Somos, Sep 11 2003
G.f.: (1-x - (1+x)*sqrt(1-4*x)) / (2*x^2) = (4+2*x) / (1-x + (1+x)*sqrt(1-4*x)). a(n)*(n+2)*(5*n-1) = a(n-1)*2*(2*n-1)*(5*n+4), n>0. - Michael Somos, Sep 11 2003
a(n) ~ 5*Pi^(-1/2)*n^(-3/2)*2^(2*n)*{1 - 93/40*n^-1 + 625/128*n^-2 - 10227/1024*n^-3 + 661899/32768*n^-4 ...}. - Joe Keane (jgk(AT)jgk.org), Sep 13 2002
G.f.: c(x)*(1+c(x))= (-1 +(1+x)*c(x))/x with the g.f. c(x) of A000108 (Catalan).
a(n) = binomial(2*n,n)/(n+1)*hypergeom([-1,n+1/2],[n+2],-4). - Peter Luschny, Aug 15 2012
D-finite with recurrence (n+2)*a(n) + (-3*n-2)*a(n-1) + 2*(-2*n+3)*a(n-2)=0. - R. J. Mathar, Dec 02 2012
0 = a(n)*(+16*a(n+1) + 38*a(n+2) - 18*a(n+3)) + a(n+1)*(-14*a(n+1) + 19*a(n+2) - 7*a(n+3)) + a(n+2)*(+a(n+2) + a(n+3)) for all n>=0. - Michael Somos, Jan 17 2015
0 = a(n)^2*(+368*a(n+1) - 182*a(n+2)) + a(n)*a(n+1)*(-306*a(n+1) + 317*a(n+2)) + a(n)*a(n+2)*(-77*a(n+2)) + a(n+1)^2*(-14*a(n+1) - 6*a(n+2)) + a(n+1)*a(n+2)*(+8*a(n+2)) for all n>=0. - Michael Somos, Jan 17 2015
E.g.f.: (BesselI(0,2*x) - (x - 1)*BesselI(1,2*x)/x)*exp(2*x). - Ilya Gutkovskiy, Jun 08 2016
G.f. with 1 prepended: Let E(x) = exp( Sum_{n >= 1} binomial(5*n,2*n)*x^n/n ). Then A(x) = ( x/series reversion of x*E(x) )^(1/5) = ( x/series reversion of x*D(x)^5 )^(1/5), where D(x) = 1 + 2*x + 23*x^2 + 371*x^3 + ... is the o.g.f. for A060941 .... Cf. A274052 and A274244. - Peter Bala, Jan 01 2020

Extensions

More terms from Joe Keane (jgk(AT)jgk.org), Feb 08 2000
Asymptotic series corrected and extended by Michael Somos, Sep 11 2003

A079489 a(n) = (2*4^n*binomial(2*n, n) - binomial(4*n + 1, 2*n)) / (n + 1).

Original entry on oeis.org

1, 3, 22, 211, 2306, 27230, 338444, 4362627, 57788170, 781825066, 10757497972, 150073096238, 2117778107732, 30176799215196, 433586825237912, 6274885068167651, 91383942213277530, 1338275570267001458, 19695358741104824036, 291137841642777382330, 4320734864185863437820
Offset: 0

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Author

N. J. A. Sloane, Jan 20 2003

Keywords

Comments

a(n) is the number of ordered trees on 2n-1 edges in which every subtree of the root (including its rooting edge) has an even number of edges, except for the leftmost subtree which has an odd number of edges (including its rooting edge). - David Callan, Apr 10 2012
a(n) is the number of 2 X 2n Young tableaux with a wall between the first and second row in each even column. If there is a wall between two cells, the entries may be decreasing; see [Banderier, Wallner 2021].
Example for a(1)=3:
3 4 2 4 2 3
- - -
1 2, 1 3, 1 4. - Michael Wallner, Mar 09 2022

Crossrefs

Final diagonal of triangle in A078990.

Programs

  • Maple
    a := n -> (2*4^n*binomial(2*n, n) - binomial(4*n + 1, 2*n)) / (n + 1):
    seq(a(n), n = 0..20);  # Peter Luschny, Aug 26 2024
  • Mathematica
    ((Sqrt[2] Sqrt[1 + Sqrt[1 - 16 x]] - Sqrt[1 - 16 x] - 1)/(4 x) + O[x]^20)[[3]] (* Vladimir Reshetnikov, Sep 25 2016 *)
    CoefficientList[Series[-(1 - Sqrt[1 - 4*Sqrt[x]])*(1 - Sqrt[1 + 4*Sqrt[x]])/(4*x), {x,0,50}], x] (* G. C. Greubel, Apr 13 2017 *)
  • PARI
    a(n)=if(n<0,0,polcoeff(serreverse(x*(1-x^2)/(1+x^2)^2+O(x^(2*n+3))),2*n+1))
    
  • PARI
    {a(n)=polcoeff(exp(sum(m=1,n,binomial(4*m-1,2*m)*x^m/m)+x*O(x^n)),n)} \\ Paul D. Hanna, Dec 30 2010

Formula

Series reversion of x(1-x^2)/(1+x^2)^2 expanded in odd powers of x. [Previous name.]
If x = y*(1-y^2)/(1+y^2)^2 then y = x + 3*x^3 + 22*x^5 + 211*x^7 + 2306*x^9 + ...
G.f. A(x) satisfies x*A(x^2) = (C(x) - C(-x))/(C(x) + C(-x)) where C(x) is g.f. of the Catalan numbers A000108.
a(n) = Sum_{k=0..2n} (-1)^k * A000108(2*n-k) * A000108(k). - David Callan, Aug 16 2006
a(n) = ((2^(4n+2))/Gamma(1/2)) * ((Gamma(n+1/2)/(2*Gamma(n+2))) - Gamma(2n+3/2)/Gamma(2n+3)). [David Dickson (dcmd(AT)unimelb.edu.au), Nov 10 2009]
G.f.: exp( Sum_{n>=1} C(4n-1,2n)*x^n/n ). - Paul D. Hanna, Dec 30 2010
G.f.: C(sqrt(x))*C(-sqrt(x)) where C(x) is the g.f. for the Catalan numbers A000108. - David Callan, Apr 10 2012
D-finite with recurrence n*(n+1)*(2*n+1)*a(n) -2*n*(32*n^2-32*n+11)*a(n-1) +16*(4*n-5)*(4*n-3)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Nov 29 2012
a(n) ~ (2-sqrt(2))*16^n/(sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Aug 20 2013
a(n) = 2^(2*n+1)*Catalan(n) - Catalan(2*n+1) (see Regev). It follows that the 2-adic valuations of a(n) and Catalan(n) are equal. In particular, a(n) is odd iff n is of the form 2^m - 1. - Peter Bala, Aug 02 2016
G.f.: (sqrt(2) * sqrt(1 + sqrt(1-16*x)) - sqrt(1-16*x) - 1)/(4*x). - Vladimir Reshetnikov, Sep 25 2016
G.f. A(x) satisfies A(x^2) = C(x)^2*r(-x*C(x)^2), where C(x) is g.f. of the Catalan numbers A000108, and r(x) is g.f. of the large Schröder numbers A006318. - Alexander Burstein, Nov 21 2019
From Peter Bala, Sep 14 2021: (Start)
A(x) = exp( Sum_{n >= 1} (1/2)*binomial(4*n,2*n)*x^n/n ).
1 + x*A(x) is the o.g.f. of A066357.
The sequence defined by b(n) := [x^n] A(x)^n begins [1, 3, 53, 1056, 22181, 480003, 10588508, 236720424, ...] and satisfies the congruence b(p) == b(1) (mod p^3) for prime p >= 3. See A333563. Cf. A060941. (End)
From Peter Bala, Oct 23 2024: (Start)
For integer r and positive integer s, define a sequence {u(n) : n >= 0} by setting u(n) = [x^(s*n)] A(x)^(r*n). We conjecture that the supercongruence u(n*p^k) == u(n*p^(k-1)) (mod p^(3*k)) holds for all primes p >= 5 and for all positive integers n and k.
Let B(x) = 1/x * series_reversion(x*A(x)). Define a sequence {v(n) : n >= 0} by setting v(n) = [x^(s*n)] B(x)^(r*n). We conjecture that the supercongruence v(n*p^k) == v(n*p^(k-1)) (mod p^(3*k)) holds for all primes p >= 5 and for all positive integers n and k. (End)

Extensions

New name by Peter Luschny, Aug 26 2024

A100982 Number of admissible sequences of order j; related to 3x+1 problem and Wagon's constant.

Original entry on oeis.org

1, 1, 2, 3, 7, 12, 30, 85, 173, 476, 961, 2652, 8045, 17637, 51033, 108950, 312455, 663535, 1900470, 5936673, 13472296, 39993895, 87986917, 257978502, 820236724, 1899474678, 5723030586, 12809477536, 38036848410, 84141805077, 248369601964
Offset: 1

Views

Author

Steven Finch, Jan 13 2005

Keywords

Comments

Eric Roosendaal counted all admissible sequences up to order j=1000 (2005). Note: there is a typo in both Wagon and Chamberland in the definition of Wagon's constant 9.477955... The expression floor(1 + 2*i + i*log_2(3)) should be replaced by floor(1 + i + i*log_2(3)).
The length of all admissible sequences of order j is A020914(j). - T. D. Noe, Sep 11 2006
Conjecture: a(n) is given for each n > 3 by a formula using a(2)..a(n-1). This allows us to create an iterative algorithm which generates a(n) for each n > 6. This has been proved for each n <= 53. For higher values of n the algorithm must be slightly modified. - Mike Winkler, Jan 03 2018
Theorem 1: a(k) is given for each k > 1 by a formula using a(1)..a(k-1). Namely, a(1)=1 and a(k+1) = Sum_{m=1..k} (-1)^(m-1)*binomial(floor((k-m+1)*(log(3)/log(2))) + m - 1, m)*a(k-m+1) for k >= 1. - Vladimir M. Zarubin, Sep 25 2015
Theorem 2: a(n) can be generated for each n > 2 algorithmically in a Pascal's triangle-like manner from the two starting values 0 and 1. This result is based on the fact that the Collatz residues (mod 2^k) can be evolved according to a binary tree. There is a direct connection with A076227, A056576 and A022921. - Mike Winkler, Sep 12 2017
A177789 shows another theorem and algorithm for generating a(n). - Mike Winkler, Sep 12 2017

Examples

			The unique admissible sequence of order 1 is 3/2, 1/2.
The unique admissible sequence of order 2 is 3/2, 3/2, 1/2, 1/2.
The two admissible sequences of order 3 are 3/2, 3/2, 3/2, 1/2, 1/2 and 3/2, 3/2, 1/2, 3/2, 1/2.
a(13) = 8045 = binomial(floor(5*(13-2)/3), 13-2)
- Sum_{i=2..6} binomial(floor((3*(13-i)+0)/2), 13-i)*a(i)
- Sum_{i=7..11} binomial(floor((3*(13-i)-1)/2), 13-i)*a(i)
- Sum_{i=12..12} binomial(floor((3*(13-i)-2)/2), 13-i)*a(i)
= 31824 - 4368*1 - 3003*2 - 715*3 - 495*7 - 120*12 - 28*30 - 21*85 - 5*173 - 4*476 - 1*961 - 0*2652. (Conjecture)
From _Mike Winkler_, Sep 12 2017: (Start)
The next table shows how Theorem 2 works. No entry is equal to zero.
n =       3  4  5   6   7   8   9  10  11   12 .. |A076227(k)=
--------------------------------------------------|
k =  2 |  1                                       |     1
k =  3 |  1  1                                    |     2
k =  4 |     2  1                                 |     3
k =  5 |        3   1                             |     4
k =  6 |        3   4   1                         |     8
k =  7 |            7   5   1                     |    13
k =  8 |               12   6   1                 |    19
k =  9 |               12  18   7   1             |    38
k = 10 |                   30  25   8   1         |    64
k = 11 |                   30  55  33   9    1    |   128
:      |                        :   :   :    : .. |    :
--------------------------------------------------|---------
a(n) =    2  3  7  12  30  85 173 476 961 2652 .. |
The entries (k,n) in this table are generated by the rule (k+1,n) = (k,n) + (k,n-1). The last value of (k+1,n) is given by k+1 = A056576(n-1), or the highest value in column n is given twice only if A022921(n-2) = 2. Then a(n) is equal to the sum of the entries in column n. For n = 7 there is 1 = 0 + 1, 5 = 1 + 4, 12 = 5 + 7, 12 = 12 + 0. Therefore a(7) = 1 + 5 + 12 + 12 = 30. The sum of row k is equal to A076227(k). (End)
From _Ruud H.G. van Tol_, Dec 04 2023: (Start)
A tree view.
n-tree--A098294--ids-----paths-----------------a(n)
0 ._          0  0       0                       -
1 |_          1  1       10                      1
2 |_._        2  2       1100                    1
3 |_|_        2  3-4     11010     -   11100     2
4 |_|_._      3  5-7     1101100   -  1111000    3
5 |_|_|_      3  8-14    11011010  - 11111000    7
6 |_|_|_._    4  15-26   1101101100-1111110000  12
7 |_|_|_|_._  5  27-56   ...                    30
8 |_|_|_|_|_  5  57-141  ...                    85
...
For n>=1, the endpoints are at A098294(n) to the right.
(End)
		

Crossrefs

Cf. A122790 (Wagon's constant), A076227, A056576, A022921, A098294, A177789.

Programs

  • Mathematica
    (* based on Eric Roosendaal's algorithm *) nn=100; Clear[x,y]; Do[x[i]=0, {i,0,nn+1}]; x[1]=1; t=Table[Do[y[cnt]=x[cnt]+x[cnt-1], {cnt,p+1}]; Do[x[cnt]=y[cnt], {cnt,p+1}]; admis=0; Do[If[(p+1-cnt)*Log[3]T. D. Noe, Sep 11 2006 *)
  • PARI
    /* translation of the above code from T. D. Noe */
    {limit=100; n=1; x=y=vector(limit+1); x[1]=1; for(b=2, limit, for(c=2, b+1, y[c]=x[c]+x[c-1]); for(c=2, b+1, x[c]=y[c]); a_n=0; for(c=1, b+1, if((b+1-c)*log(3)Mike Winkler, Feb 28 2015
    
  • PARI
    /* algorithm for the Conjecture */
    {limit=53; zn=vector(limit); zn[2]=1; zn[3]=2; zn[4]=3; zn[5]=7; zn[6]=12; f=1; e1=-1; e2=-2; for(n=7, limit, m=floor((n-1)*log(3)/log(2))-(n-1); j=(m+n-2)!/(m!*(n-2)!); if(n>6*f, if(frac(n/2)==0, e=e1, e=e2)); if(frac((n-6 )/12)==0, f++; e1=e1+2); if(frac((n-12)/12)==0, f++; e2=e2+2); Sum=a=b=0; c=1; d=5; until(c>=n-1, for(i=2+a*5+b, 1+d+a*5, if(i>11 && frac((i+2)/6)==0, b++); delta=e-a; Sum=Sum+binomial(floor((3*(n-i)+delta)/2),n-i)*zn[i]; c++); a++; for(k=3, 50, if(n>=k*6 && a==k-1, d=k+3))); zn[n]=j-Sum; print(n" "zn[n]))} \\ Mike Winkler, Jan 03 2018
    
  • PARI
    /* cf. code for Theorem 2 */
    {limit=100; /*or limit>100*/ p=q=vector(limit); c=2; w=log(3)/log(2); for(n=3, limit, p[1]=Sum=1; for(i=2, c, p[i]=p[i-1]+q[i]; Sum=Sum+p[i]); a_n=Sum; print(n" "a_n); for(i=1, c, q[i]=p[i]); d=floor(n*w)-floor((n-1)*w); if(d==2, c++)); } \\ Mike Winkler, Apr 14 2015
    
  • PARI
    /* algorithm for Theorem 1 */
    n=20; a=vector(n); log32=log(3)/log(2);
    {a[1]=1; for ( k=1, n-1, a[k+1]=sum( m=1,k,(-1)^(m-1)*binomial( floor( (k-m+1)*log32)+m-1,m)*a[k-m+1] ); print(k" "a[k]) );
    } \\ Vladimir M. Zarubin, Sep 25 2015
    
  • PARI
    /* algorithm for Theorem 2 */
    {limit=30; /*or limit>30*/ R=matrix(limit,limit); R[2,1]=0; R[2,2]=1; for(n=2, limit, print; print1("For n="n" in column n: "); Kappa_n=floor(n*log(3)/log(2)); a_n=0; for(k=n, Kappa_n, R[k+1,n]=R[k,n]+R[k,n-1]; print1(R[k+1,n]", "); a_n=a_n+R[k+1,n]); print; print(" and the sum is a(n)="a_n))} \\ Mike Winkler, Sep 12 2017

Formula

A sequence s(k), where k=1, 2, ..., n, is *admissible* if it satisfies s(k)=3/2 exactly j times, s(k)=1/2 exactly n-j times, s(1)*s(2)*...*s(n) < 1 but s(1)*s(2)*...*s(m) > 1 for all 1 < m < n.
a(n) = (m+n-2)!/(m!*(n-2)!) - Sum_{i=2..n-1} binomial(floor((3*(n-i)+b)/2), n-i)*a(i), where m = floor((n-1)*log_2(3))-(n-1) and b assumes different integer values within the sum at intervals of 5 or 6 terms. (Conjecture)
a(n) = Sum_{k=n-1..A056576(n-1)} (k,n). (Theorem 2, cf. example)
a(k) = 2*A076227(A020914(k)-1) - A076227(A020914(k)), for k > 0. - Vladimir M. Zarubin, Sep 29 2019
a(1)=1, a(n) = Sum_{k=0..A020914(n-1)-n-2} A325904(k)*binomial(A020914(n-1)-k-2, n-2) for n>1. - Benjamin Lombardo, Oct 18 2019

Extensions

Two more terms from Jules Renucci (jules.renucci(AT)wanadoo.fr), Nov 02 2005
More terms from T. D. Noe, Sep 11 2006

A001450 a(n) = binomial(5*n,2*n).

Original entry on oeis.org

1, 10, 210, 5005, 125970, 3268760, 86493225, 2319959400, 62852101650, 1715884494940, 47129212243960, 1300853625660225, 36052387482172425, 1002596421878664480, 27963143931814663880, 781879430625942976880, 21910242651571684460050, 615167304833936727234180
Offset: 0

Views

Author

Keywords

Crossrefs

Programs

  • Magma
    [Binomial(5*n, 2*n): n in [0..20]]; // Vincenzo Librandi, Aug 07 2014
    
  • Maple
    f := n->(5*n)!/((3*n)!*(2*n)!);
  • Mathematica
    Table[Hypergeometric2F1[-3n,-2n,1,1],{n,0,60}] (* John M. Campbell, Jul 15 2011 *)
    Table[Binomial[5n,2n],{n,0,20}] (* Harvey P. Dale, Nov 09 2011 *)
  • PARI
    a(n) = binomial(5*n,2*n) \\ Altug Alkan, Oct 06 2015

Formula

a(n) = (5*n)!/((3*n)!*(2*n)!).
a(n) = 2F1[-3n,-2n,1,1] (see Mathematica code below). - John M. Campbell, Jul 15 2011
G.f.: hypergeom([1/5, 2/5, 3/5, 4/5], [1/3, 1/2, 2/3], (3125/108)*x). - Robert Israel, Aug 07 2014
From Peter Bala, Oct 05 2015: (Start)
a(n) = [x^n] ( (1 + x)*C(x) )^(5*n), where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108.
a(n) = 5*A259550(n) for n >= 1.
exp( (1/5) * Sum_{n >= 1} a(n)*x^n/n ) = 1 + 2*x + 23*x^2 + 377*x^3 + ... is the o.g.f. for the sequence of Duchon numbers A060941. (End)
a(n) = [x^(2*n)] 1/(1 - x)^(3*n+1). - Ilya Gutkovskiy, Oct 10 2017
D-finite with recurrence 6*n*(3*n-1)*(2*n-1)*(3*n-2)*a(n) -5*(5*n-4)*(5*n-3)*(5*n-2)*(5*n-1)*a(n-1)=0. - R. J. Mathar, Feb 08 2021
a(n) = Sum_{k = 0..2*n} binomial(3*n+k-1, k). Cf. A066802. - Peter Bala, Jun 04 2024
Right-hand side of the identity Sum_{k = 0..2*n} (-1)^k*binomial(-n, k)* binomial(4*n-k, 2*n-k) = binomial(5*n, 2*n). Compare with the identity Sum_{k = 0..n} (-1)^k*binomial(n, k)*binomial(4*n-k, 2*n-k) = binomial(3*n, n). - Peter Bala, Jun 05 2024
From Karol A. Penson, May 07 2025: (Start)
G.f. denoted by h(x) satisfies the following algebraic equation of order 10:
8 - 3125*x + 20*(-13 + 3125*x)*h(x) - 45*(-74 + 9375*x)*h(x)^2 + 5*(-4023 + 175000*x)*h(x)^2 + 5*(-4023 + 175000*x)*h(x)^3 + 25*(1809 + 53125*x)*h(x)^4 + (34375*x - 738)*(3125*x - 108)*h(x)^5 + 15*(3125*x + 297)*(3125*x - 108)*h(x)^6 + 5*(3125*x - 108)^2*h(x)^7 + 135*(3125*x - 108)^2*h(x)^8 + (3125*x - 108)^3*h(x)^10=0.
a(n) = Integral_{x=0..3125/108} x^n*W(x)*dx, n>=0, where W(x) = W1(x)+W2(x)+W3(x)+W4(x) can be expressed with four generalized hypergeometric functions of type 4F3:
W1(x) = sqrt(5)*csc((2*Pi)/5)*sin((3*Pi)/10)*hypergeom([1/5, 8/15, 7/10, 13/15], [2/5, 3/5, 4/5], (108*x)/3125)/(10*Pi*x^(4/5)),
W2(x) = sqrt(5)*sec((3*Pi)/10)*sin(Pi/10)*hypergeom([2/5, 11/15, 9/10, 16/15], [3/5, 4/5, 6/5], (108*x)/3125)/(50*Pi*x^(3/5)),
W3(x) = sqrt(5)*sec((3*Pi)/10)*sin(Pi/10)*hypergeom([3/5, 14/15, 11/10, 19/15], [4/5, 6/5, 7/5], (108*x)/3125)/(125*Pi*x^(2/5)), and
W4(x) = (7*sqrt(5)*csc((2*Pi)/5)*sin((3*Pi)/10)*hypergeom([4/5, 17/15, 13/10, 22/15], [6/5, 7/5, 8/5], (108*x)/3125))/(1250*Pi*x^(1/5)).
Using the formula for a(n) only, W(x) can be shown to be a positive function. It is singular at x=0 and at x=3125/108. This integral representation is unique since W(x) is the solution of the Hausdorff moment problem. (End)
From Peter Bala, Jun 21 2025: (Start)
a(n) = [x^(3*n)] 1/(1 - x)^(2*n+1).
a(n) = Sum_{k = 0..3*n} binomial(2*n+k-1, k). (End)

A300386 The number of paths of length 7*n from the origin to the line y = 2*x/5 with unit East and North steps that stay below the line or touch it.

Original entry on oeis.org

1, 3, 76, 2803, 121637, 5782513, 291437249, 15297882929, 827402061954, 45790180469312, 2580588279994441, 147592910517101281, 8544927937132306600, 499811636639428519226, 29491983283370728013309, 1753398440591481772556798, 104933899400256659634374549, 6316334518803437568442071134
Offset: 0

Views

Author

Bryan T. Ek, Mar 04 2018

Keywords

Comments

Equivalent to nonnegative walks from (0,0) to (7*n,0) with step set [1,2], [1,-5].

Examples

			For n=1, the possible walks are EEEEENN, EEEENEN, EEENEEN.
		

Crossrefs

Programs

  • Mathematica
    terms = 18; f[_] = 0;
    Do[f[t_] = f[t]^21 t^3 + 2 f[t]^16 t^2 - f[t]^15 t^2 + 3 f[t]^14 t^2 + f[t]^11 t - f[t]^10 t + 2 f[t]^9 t - 2 f[t]^8 t + 3 f[t]^7 t + 1 + O[t]^terms, {terms}];
    CoefficientList[f[t], t] (* Jean-François Alcover, Dec 04 2018 *)
    nmax = 20; CoefficientList[Series[Exp[Sum[Binomial[7*k, 2*k]*x^k/(7*k), {k, 1, nmax}]], {x, 0, nmax}], x] (* Vaclav Kotesovec, Sep 16 2021 *)

Formula

G.f. satisfies: f=f^21*t^3+2*f^16*t^2-f^15*t^2+3*f^14*t^2+f^11*t-f^10*t+2*f^9*t-2*f^8*t+3*f^7*t+1.
From Peter Bala, Jan 02 2019: (Start)
O.g.f.: A(x) = exp( Sum_{n >= 1} (1/7)*binomial(7*n, 2*n)*x^n/n ) - Bizley. Cf. A274052.
Recurrence: a(0) = 1 and a(n) = (1/n) * Sum_{k = 0..n-1} (1/7)*binomial(7*n-7*k, 2*n-2*k)*a(k) for n >= 1. (End)
The sequence defined by b(n) := [x^n] A(x)^n begins [1, 3, 161, 9804, 630401, 41789278, 2824792568, 193553976353, ...] and conjecturally satisfies the congruence b(p) == b(1) (mod p^3) for prime p >= 11 (checked up to p = 101). - Peter Bala, Sep 14 2021
a(n) ~ c * 7^(7*n) / (n^(3/2) * 2^(2*n) * 5^(5*n)), where c = 0.0538519123304380623474844037127876191519207214308040151922885271364215631... = s*sqrt((3 - 2*s + 2*s^2 - s^3 + s^4 + 6*r*s^7 - 2*r*s^8 + 4*r*s^9 + 3*r^2*s^14) / (63 - 56*s + 72*s^2 - 45*s^3 + 55*s^4 + 273*r*s^7 - 105*r*s^8 + 240*r*s^9 + 210*r^2*s^14)) / (2*sqrt(Pi)), where r = 12500/823543 and s = 1.129379978325... is the root of the equation -16807 + 24010*s - 13720*s^2 + 7350*s^3 - 3500*s^4 + 1250*s^5 = 0. - Vaclav Kotesovec, Sep 16 2021

A300389 The number of paths of length 13*n from the origin to the line y = 2*x/11 with unit East and North steps that stay below the line or touch it.

Original entry on oeis.org

1, 6, 593, 87143, 15149546, 2891511017, 585739005066, 123655688922720, 26908765569970320, 5993187329634638043, 1359541058523676017369, 313029501692713279534165, 72965556751635426636633639, 17184586991024424745328563477, 4083065013894860643162116395527
Offset: 0

Views

Author

Bryan T. Ek, Mar 04 2018

Keywords

Comments

Equivalent to nonnegative walks from (0,0) to (13*n,0) with step set [1,2], [1,-11].

Examples

			For n=1, the possible walks are EEEEEEEEEEENN, EEEEEEEEEENEN, EEEEEEEEENEEN, EEEEEEEENEEEN, EEEEEEEENEEEEN, EEEEEEENEEEEN.
		

Crossrefs

Programs

  • Mathematica
    m = 15;
    Exp[Sum[(1/13) Binomial[13n, 2n] x^n/n, {n, 1, m}]] + O[x]^m // CoefficientList[#, x]& (* Jean-François Alcover, Feb 26 2020, after Peter Bala *)

Formula

G.f. satisfies: f = f^78*t^6 + 5*f^67*t^5 - f^66*t^5 + 6*f^65*t^5 + 10*f^56*t^4 - 4*f^55*t^4 + 20*f^54*t^4 - 5*f^53*t^4 + 15*f^52*t^4 + 10*f^45*t^3 - 6*f^44*t^3 + 24*f^43*t^3 - 12*f^42*t^3 + 30*f^41*t^3 - 10*f^40*t^3 + 20*f^39*t^3 + 5*f^34*t^2 - 4*f^33*t^2 + 12*f^32*t^2 - 9*f^31*t^2 + 18*f^30*t^2 - 12*f^29*t^2 + 20*f^28*t^2 - 10*f^27*t^2 + 15*f^26*t^2 + f^23*t - f^22*t + 2*f^21*t - 2*f^20*t + 3*f^19*t - 3*f^18*t + 4*f^17*t - 4*f^16*t + 5*f^15*t - 5*f^14*t + 6*f^13*t + 1.
From Peter Bala, Jan 03 2019: (Start)
O.g.f.: A(x) = exp( Sum_{n >= 1} (1/13)*binomial(13*n, 2*n)*x^n/n ) - Bizley.
Recurrence: a(0) = 1 and a(n) = (1/n) * Sum_{k = 0..n-1} (1/13)*binomial(13*n-13*k, 2*n-2*k)*a(k) for n >= 1. (End)
The sequence defined by b(n) := [x^n] A(x)^n begins [1, 6, 1222, 282993, 69239846, 17468997381, 4494716943847, 1172353182893367, ...] and conjecturally satisfies the congruence b(p) == b(1) (mod p^3) for prime p >= 5 except for p = 11 and p = 13 (checked up to p = 101). - Peter Bala, Sep 14 2021
a(n) ~ c * 13^(13*n) / (n^(3/2) * 2^(2*n) * 11^(11*n)), where c = 0.0250562444901910770802983936320823301923793538303930752981380507191770309... - Vaclav Kotesovec, Sep 16 2021

A182960 G.f.: exp( Sum_{n>=1} C(6n-1,2n-1)*x^n/n ).

Original entry on oeis.org

1, 5, 95, 2496, 76063, 2524161, 88534548, 3228482908, 121171012431, 4649906785719, 181614908182551, 7196014051078368, 288537887780406468, 11686156771344086156, 477379538839242423528, 19645977861506470428324
Offset: 0

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Author

Paul D. Hanna, Dec 31 2010

Keywords

Comments

The logarithmic derivative of this sequence is a bisection of the logarithmic derivative of A001764 (ternary trees).
To see this, compare the g.f. of this sequence with g.f. of A001764:
exp(Sum_{n>=1} C(3n-1,n-1)*x^n/n) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 +...

Examples

			G.f.: A(x) = 1 + 5*x + 95*x^2 + 2496*x^3 + 76063*x^4 + 2524161*x^5 +...
log(A(x)) = 5*x + 165*x^2 + 6188*x^3 + 245157*x^4 + 10015005*x^5 +...+ A025174(2n)*x^n/n +...
G.f. satisfies: A(x) = G(x*A(x)^3) where G(x) begins:
G(x) = 1 + 5*x + 20*x^2 + 96*x^3 + 528*x^4 + 3136*x^5 + 19584*x^6 +...
		

Crossrefs

Programs

  • Mathematica
    f[n_] := Exp[ Sum[ Binomial[6 i - 1, 2 i - 1]*x^i/i, {i, n}]]; CoefficientList[ Series[f@ 15, {x, 0, 15}], x]  (* Robert G. Wilson v, Dec 31 2010 *)
  • PARI
    {a(n)=polcoeff(exp(sum(m=1,n,binomial(6*m-1,2*m-1)*x^m/m)+x*O(x^n)),n)}
    
  • PARI
    {a(n)=polcoeff((serreverse(x*(1-2*x+sqrt(1-8*x+x*O(x^n)))^3/(8*(1+x)^6))/x)^(1/3),n)}

Formula

G.f. A(x) satisfies: A(x^2) = F(x)*F(-x) where F(x) = 1 + x*F(x)^3 = Sum_{n>=0} C(3n,n)*x^n/(2n+1) is the g.f. of A001764.
Let G(x) = 2*(1+x)^2/(1-2*x+sqrt(1-8*x)), then g.f. A(x) satisfies:
* A(x) = G(x*A(x)^3) and A(x/G(x)^3) = G(x);
* A(x) = [Series_Reversion( x/G(x)^3 ) / x]^(1/3).
The sequence defined by b(n) := [x^n] A(x)^n begins [1, 5, 215, 10463, 537287, 28435880, 1534398353, 83920389642, ...] and conjecturally satisfies the congruence b(p) == b(1) (mod p^3) for prime p >= 5 (checked up to p = 101). - Peter Bala, Sep 14 2021
From Vaclav Kotesovec, Sep 16 2021: (Start)
Recurrence: 16*n*(2*n + 1)*(4*n - 1)*(4*n + 1)*(27*n - 31)*a(n) = 18*(69984*n^5 - 220320*n^4 + 267705*n^3 - 156510*n^2 + 42021*n - 3680)*a(n-1) - 59049*(n-1)*(2*n - 3)*(3*n - 5)*(3*n - 4)*(27*n - 4)*a(n-2).
a(n) ~ sqrt((sqrt(2) - 1)^(2/3) + (sqrt(2) + 1)^(2/3) - 2) * 3^(6*n + 3/2) / (sqrt(Pi) * n^(3/2) * 2^(4*n + 7/2)). (End)

A274244 Number of factor-free Dyck words with slope 7/2 and length 9n.

Original entry on oeis.org

1, 4, 34, 494, 8615, 165550, 3380923, 71999763, 1580990725, 35537491360, 813691565184, 18911247654404, 444978958424224, 10579389908116344, 253756528273411250, 6133110915783398175, 149219383150626519874, 3651756292682801022384, 89830021324956206790496, 2219945238901447637080235, 55088272581138888326634644
Offset: 0

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Author

Michael D. Weiner, Jun 15 2016

Keywords

Comments

a(n) is the number of lattice paths (allowing only north and east steps) starting at (0,0) and ending at (2n,7n) that stay below the line y=7/2x and also do not contain a proper subpath of smaller size.

Examples

			a(2) = 34 since there are 34 lattice paths (allowing only north and east steps) starting at (0,0) and ending at (4,14) that stay below the line y=7/2x and also do not contain a proper subpath of small size; e.g., EEENNNNENNNNNNNNNN is a factor-free Dyck word but EEENNENNNNNNNNNNNN contains the factor ENNENNNNN.
		

Crossrefs

Factor-free Dyck words: A005807 (slope 3/2), A274052 (slope 5/2), A274256 (slope 9/2), A274257 (slope 4/3), A274259 (slope 7/3).
Cf. A060941.

Formula

Conjectural o.g.f.: Let E(x) = exp( Sum_{n >= 1} binomial(9*n, 2*n)*x^n/n ). Then A(x) = ( x/series reversion of x*E(x) )^(1/9) = 1 + 4*x + 34*x^2 + 494*x^3 + ... . Equivalently, [x^n]( A(x)^(9*n) ) = binomial(9*n, 2*n) for n = 0,1,2,... . - Peter Bala, Jan 01 2020

A381773 Expansion of ( (1/x) * Series_Reversion( x/((1+x) * C(x))^3 ) )^(1/3), where C(x) is the g.f. of A000108.

Original entry on oeis.org

1, 2, 15, 157, 1913, 25427, 357546, 5229980, 78765793, 1213181593, 19021747383, 302595975502, 4871780511910, 79232327379407, 1299767617080662, 21481625997258747, 357350097625089497, 5978708468143961925, 100537111802285439375, 1698302173359384479307
Offset: 0

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Author

Seiichi Manyama, Mar 07 2025

Keywords

Crossrefs

Programs

  • PARI
    my(N=30, x='x+O('x^N)); Vec((serreverse(x/((1+x)*(1-sqrt(1-4*x))/(2*x))^3)/x)^(1/3))

Formula

G.f. A(x) satisfies A(x) = (1 + x*A(x)^3) * C(x*A(x)^3).
a(n) = Sum_{k=0..n} binomial(3*n+2*k+1,k) * binomial(3*n+1,n-k)/(3*n+2*k+1).
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