A047749 If n = 2*m then a(n) = binomial(3*m, m)/(2*m+1), if n=2*m+1 then a(n) = binomial(3*m+1, m+1)/(2*m+1).
1, 1, 1, 2, 3, 7, 12, 30, 55, 143, 273, 728, 1428, 3876, 7752, 21318, 43263, 120175, 246675, 690690, 1430715, 4032015, 8414640, 23841480, 50067108, 142498692, 300830572, 859515920, 1822766520, 5225264024, 11124755664, 31983672534, 68328754959, 196947587823, 422030545335
Offset: 0
Keywords
Examples
G.f. = 1 + x + x^2 + 2*x^3 + 3*x^4 + 7*x^5 + 12*x^6 + 30*x^7 + 55*x^8 + ...
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
- Per Alexandersson, Frether Getachew Kebede, Samuel Asefa Fufa, and Dun Qiu, Pattern-Avoidance and Fuss-Catalan Numbers, J. Int. Seq. (2023) Vol. 26, Art. 23.4.2.
- Nikos Apostolakis, Non-crossing trees, quadrangular dissections, ternary trees, and duality preserving bijections arXiv:1804.01214 [math.CO], 2018.
- Jean-Luc Baril, Alexander Burstein, and Sergey Kirgizov, Pattern statistics in faro words and permutations, arXiv:2010.06270 [math.CO], 2020. See Theorem 4.4.
- Paul Barry, Jacobsthal Decompositions of Pascal's Triangle, Ternary Trees, and Alternating Sign Matrices, Journal of Integer Sequences, 19, 2016, #16.3.5.
- Paul Barry, Centered polygon numbers, heptagons and nonagons, and the Robbins numbers, arXiv:2104.01644 [math.CO], 2021.
- Paul Barry, Extensions of Riordan Arrays and Their Applications, Mathematics (2025) Vol. 13, No. 2, 242. See pp. 23-24.
- L. W. Beineke and R. E. Pippert, Enumerating dissectable polyhedra by their automorphism groups, Can. J. Math., 26 (1974), 50-67.
- M. Bousquet and C. Lamathe, Enumeration of solid trees according to edge number and edge degree distribution, Discr. Math., 298 (2005), 115-141.
- Michel Bousquet and Cédric Lamathe, On symmetric structures of order two, Discrete Math. Theor. Comput. Sci. 10 (2008), 153-176.
- Hassen Cheriha, Yousra Gati, and Vladimir Petrov Kostov, Descartes' rule of signs, Rolle's theorem and sequences of admissible pairs, arXiv:1805.04261 [math.CA], 2018.
- Malin Christensson, Make hyperbolic tilings of images, web page, 2019.
- S. J. Cyvin, Jianji Wang, J. Brunvoll, Shiming Cao, Ying Li, B. N. Cyvin, and Yugang Wang, Staggered conformers of alkanes: complete solution of the enumeration problem, J. Molec. Struct. 413-414 (1997), 227-239.
- S. J. Cyvin et al., Enumeration of staggered conformers of alkanes and monocyclic cycloalkanes, J. Molec. Struct., 445 (1998), 127-137.
- Alexander Burstein, Sergi Elizalde, and Toufik Mansour, Restricted Dumont permutations, Dyck paths and noncrossing partitions, arXiv:math/0610234 [math.CO], 2006. [Theorem 3.5]
- Emeric Deutsch, Another Path to Generalized Catalan Numbers:Problem 10751, Amer. Math. Monthly, 108 (Nov., 2001), 872-873.
- Emeric Deutsch, S. Feretic and M. Noy, Diagonally convex directed polyominoes and even trees: a bijection and related issues, Discrete Math., 256 (2002), 645-654.
- F. Hering et al., The enumeration of stack polytopes and simplicial clusters, Discrete Math., 40 (1982), 203-217.
- Clark Kimberling, Matrix Transformations of Integer Sequences, J. Integer Seqs., Vol. 6, 2003.
- Anthony Zaleski and Doron Zeilberger, On the Intriguing Problem of Counting (n+1,n+2)-Core Partitions into Odd Parts, arXiv:1712.10072 [math.CO], 2017.
Crossrefs
Programs
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Magma
G:=Gamma; [Round((1+(-1)^n)*G(3*n/2+1)/(G(n/2+1)*Factorial(n+1)) + (1-(-1)^n)*G((3*n+1)/2)/(G((n+3)/2)*Factorial(n)))/2: n in [0..35]]; // G. C. Greubel, Jul 07 2019
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Maple
A047749 := proc(m) if m mod 2 = 1 then x := (m-1)/2; RETURN((3*x+1)!/((x+1)!*(2*x+1)!)); fi; x := m/2; RETURN((3*x)!/(x!*(2*x+1)!)); end; A047749 := proc(m) local x; if m mod 2 = 1 then x := (m-1)/2; RETURN((3*x+1)!/((x+1)!*(2*x+1)!)); fi; RETURN(A001764(m/2)); end;
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Mathematica
a[ n_] := If[ n < 1, Boole[n == 0], SeriesCoefficient[ InverseSeries[ Series[ (x + 2 x^2) / (1 + x)^3, {x, 0, n}]], {x, 0, n}]]; (* Michael Somos, Oct 29 2014 *) Table[If[OddQ[n],2Binomial[(3n-1)/2,(n-1)/2],Binomial[3n/2,n/2]]/(n+1),{n,0,40}] (* Robert A. Russell, Jan 19 2024 *)
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PARI
{a(n)=local(A=1+x);for(i=1,n,A=1+x*A^2*subst(A,x,-x+x*O(x^n)));polcoeff(A,n)} \\ Paul D. Hanna, Sep 20 2009
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PARI
x='x+O('x^66); C(x)=serreverse(x-x^3); /* =x+x^3+3*x^5+12*x^7+55*x^9 +..., cf. A001764 */ s=1/(1-C(x)); /* g.f. */ Vec(s) /* Joerg Arndt, Apr 16 2011 */
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PARI
apr(n, p, r) = r*binomial(n*p+r, n)/(n*p+r); a(n) = apr(n\2, 3, n%2+1); \\ Seiichi Manyama, Jul 20 2025
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Python
from math import comb def A047749(n): return comb(n+(a:=n>>1),a+(b:=n&1))//(n+1-b) # Chai Wah Wu, Jul 30 2022
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Sage
def A047749_list(n) : D = [0]*n; D[1] = 1 R = []; b = False; h = 1 for i in range(n) : for k in (1..h) : D[k] = D[k] + D[k-1] R.append(D[h]) if b : h += 1 b = not b return R A047749_list(35) # Peter Luschny, May 03 2012
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Sage
[1]+[((1+(-1)^n)*binomial(3*n/2,n/2)/(n+1) + (1-(-1)^n)* binomial((3*n-1)/2, (n+1)/2)/n)/2 for n in (1..35)] # G. C. Greubel, Jul 07 2019
Formula
G.f. is 1+Z, where Z satisfies x*Z^3 + (3*x-2)*Z^2 + (3*x-1)*Z + x = 0. Equivalently, the g.f. Y satisfies x*Y^3 - 2*Y^2 + 3*Y - 1 = 0. - Vladeta Jovovic, Dec 06 2002
Reversion of g.f. (x-2*x^2)/(1-x)^3 (ignoring signs). - Ralf Stephan, Mar 22 2004
G.f.: (4/(3*x))*(sin((1/3)*asin(sqrt(27*x^2/4))))^2 +(2/sqrt(3*x^2))*sin((1/3)*asin(sqrt(27*x^2/4))). - Paul Barry, Nov 08 2006
G.f.: 1/(1-2*sin(asin(3*sqrt(3)*x/2)/3)/sqrt(3)). - Paul Barry, Apr 16 2008
From Paul D. Hanna, Sep 20 2009: (Start)
G.f. satisfies: A(x) = 1 + x*A(x)^2*A(-x);
also, A(x)*A(-x) = B(x^2) where B(x) = 1 + x*B(x)^3 = g.f. of A001764. (End)
G.f.: 1/(1-C(x)) where C(x) = Reverse(x-x^3) = x + x^3 + 3*x^5 + 12*x^7 + 55*x^9 + ... (cf. A001764). - Joerg Arndt, Apr 16 2011
G.f.: G(z^2)+z*G(z^2)^2, where G(z) = 1 + z*G(z)^3, the generating function for A001764. - Robert A. Russell, Jan 26 2024
From Gary W. Adamson, Jul 14 2011: (Start)
a(n) is the upper left term in M^n, M = the infinite square production matrix:
1, 1, 0, 0, 0, 0, ...
0, 0, 1, 0, 0, 0, ...
1, 1, 0, 1, 0, 0, ...
0, 0, 1, 0, 1, 0, ...
1, 1, 0, 1, 0, 1, ...
... (End)
Conjecture D-finite with recurrence: 8*n*(n+1)*a(n) + 36*n*(n-2)*a(n-1) - 6*(9*n^2-18*n+14)*a(n-2) - 27*(3*n-7)*(3*n-8)*a(n-3) = 0. - R. J. Mathar, Dec 19 2011
0 = a(n)*(+7308954*a(n+2) - 16659999*a(n+3) - 4854519*a(n+4) + 6201838*a(n+5)) + a(n+1)*(-406053*a(n+2) - 1627560*a(n+3) + 1683538*a(n+4) - 245747*a(n+5)) + a(n+2)*(+45117*a(n+2) + 235870*a(n+3) + 173953*a(n+4) - 484295*a(n+5)) + a(n+3)*(-41820*a(n+3) - 50184*a(n+4) + 22304*a(n+5)) for all n in Z if a(-1) = -2/3. - Michael Somos, Oct 29 2014
a(0) = 1; a(n) = Sum_{i=0..n-1} Sum_{j=0..n-i-1} (-1)^i * a(i) * a(j) * a(n-i-j-1). - Ilya Gutkovskiy, Jul 28 2021
a(n) = binomial(A032766(n),floor((n+1)/2))/(2*floor(n/2)+1). - Miko Labalan, Nov 28 2023
a(n) = 2*A005036(n) - A005034(n) = A005034(n) - 2*A369315(n) = A005036(n) - A369315(n). - Robert A. Russell, Jan 20 2024
From Robert A. Russell, Mar 20 2024: (Start)
a(n) = U(n) in the Beineke and Pippert link.
G.f.: E(1)(t*E(3)(t^2)) (second entry in Table 1), where E(d)(t) is defined in formula 3 of Hering link. (End)
From Robert A. Russell, Jul 15 2024: (Start)
a(2m) = A001764(m) ~ (3^3/2^2)^m*sqrt(3/(2*Pi*(2*m)^3)).
a(n+2)/a(n) ~ 27/4; a(2m+1)/a(2m) ~ 3; a(2m)/a(2m-1) ~ 9/4. (End)
a(n) ~ 3^((6n+3)/4)/(sqrt(Pi)*2^((2n-1)/2)*(2n+1)^(3/2)). - Miko Labalan, Dec 05 2024
a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/2)} a(2*k) * a(n-1-2*k). - Seiichi Manyama, Jul 07 2025
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