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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A006632 a(n) = 3*binomial(4*n-1, n-1)/(4*n-1).

Original entry on oeis.org

1, 3, 15, 91, 612, 4389, 32890, 254475, 2017356, 16301164, 133767543, 1111731933, 9338434700, 79155435870, 676196049060, 5815796869995, 50318860986108, 437662920058980, 3824609516638444, 33563127932394060, 295655735395397520, 2613391671568320765
Offset: 1

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Comments

a(n) is the number of ordered trees (A000108) with 3n-1 edges in which every non-leaf vertex has exactly two leaf children (no restriction on non-leaf children). For example, a(2) counts the 3 trees
\/......\/......\/
.\|/...\|/....\|/ . - David Callan, Aug 22 2014
a(n) is the number of lattice paths from (0,0) to (3n,n) using only the steps (1,0) and (0,1) and which are strictly below the line y = x/3 except at the path's endpoints. - Lucas A. Brown, Aug 21 2020
This is instance k = 3 of the family {c(k, n)}A130564.%20-%20_Wolfdieter%20Lang">{n>=1} given in a comment in A130564. - _Wolfdieter Lang, Feb 04 2024

References

  • H. M. Finucan, Some decompositions of generalized Catalan numbers, pp. 275-293 of Combinatorial Mathematics IX. Proc. Ninth Australian Conference (Brisbane, August 1981). Ed. E. J. Billington, S. Oates-Williams and A. P. Street. Lecture Notes Math., 952. Springer-Verlag, 1982.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Magma
    A006632:= func< n | Binomial(4*n-2,n-1)/n >;
    [A006632(n): n in [1..40]]; // G. C. Greubel, Sep 01 2025
    
  • Maple
    A006632:=n->3*binomial(4*n-1,n-1)/(4*n-1): seq(A006632(n), n=1..30); # Wesley Ivan Hurt, Oct 23 2017
  • Mathematica
    InverseSeries[Series[y*(1-y)^3, {y, 0, 24}], x] (* then A(x)=y(x) *) (* Len Smiley, Apr 07 2000 *)
    a[ n_] := If[n<1, 0, Binomial[4 n - 2, n - 1] / n]; (* Michael Somos, Aug 22 2014 *)
  • PARI
    a(n) = 3*binomial(4*n-1, n-1)/(4*n-1) \\ Felix Fröhlich, Oct 23 2017
    
  • SageMath
    def A006632(n): return binomial(4*n-2,n-1)//n
    print([A006632(n) for n in range(1,41)]) # G. C. Greubel, Sep 01 2025

Formula

a(n) = binomial(4*n-1, n)/(4*n-1) = 3*binomial(4*n-2, n-1) - binomial(4*n-2, n). - David Callan, Sep 15 2004
G.f.: g^3 where g = 1+x*g^4 is the g.f. of A002293. - Mark van Hoeij, Nov 11 2011
a(n) = (3/4)*binomial(4*n,n)/(4*n-1). - Bruno Berselli, Jan 17 2014
From Wolfdieter Lang, Feb 06 2020: (Start)
G.f.: (3/4)*(1 - hypergeometric3F2([-1, 1, 2]/4, [1, 2]/3, (4^4/3^3)*x)).
E.g.f.: (3/4)*(1 - hypergeometric3F3([-1, 1, 2]/4, [1, 2, 3]/3, (4^4/3^3)*x)). (End)
D-finite with recurrence 3*n*(3*n-1)*(3*n-2)*a(n) -8*(4*n-5)*(4*n-3)*(2*n-1)*a(n-1)=0. - R. J. Mathar, May 07 2021
a(n) = (2n-1)*A000260(n). - F. Chapoton, Jul 15 2021
G.f. A(x) satisfies: A(x) = x / (1 - A(x))^3. - Ilya Gutkovskiy, Nov 03 2021
G.f.: x*( Sum_{n >= 0} binomial(4*n+3, n)*x^n ) / ( Sum_{n >= 0} binomial(4*n, n)*x^n ) = x*( Sum_{n >= 0} binomial(4*n+3, n)*x^n ) / ( 1 + 4*x*Sum_{n >= 0} binomial(4*n+3, n)*x^n ). - Peter Bala, Dec 13 2024
Working with a offset of 0, the g.f. A(x) = 1 + 3*x + 15*x^2 + ... is uniquely determined by the conditions A(0) = 1 and [x^n] A(x)^(-n) = -3 for all n >= 1. - Peter Bala, Jul 24 2025

A025174 a(n) = binomial(3n-1, n-1).

Original entry on oeis.org

0, 1, 5, 28, 165, 1001, 6188, 38760, 245157, 1562275, 10015005, 64512240, 417225900, 2707475148, 17620076360, 114955808528, 751616304549, 4923689695575, 32308782859535, 212327989773900, 1397281501935165, 9206478467454345, 60727722660586800, 400978991944396320
Offset: 0

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Keywords

Comments

Number of standard tableaux of shape (2n-1,n). Example: a(2)=5 because in the top row we can have 123, 124, 125, 134, or 135. - Emeric Deutsch, May 23 2004
Number of peaks in all generalized {(1,2),(1,-1)}-Dyck paths of length 3n.
Positive terms in this sequence are the numbers k such that k and 2k are consecutive terms in a row of Pascal's triangle. 1001 is the only k such that k, 2k, and 3k are consecutive terms in a row of Pascal's triangle. - J. Lowell, Mar 11 2023

Examples

			L.g.f.: L(x) = x + 5*x^2/2 + 28*x^3/3 + 165*x^4/4 + 1001*x^5/5 + 6188*x^6/6 + ...
where G(x) = exp(L(x)) satisfies G(x) = 1 + x*G(x)^3, and begins:
exp(L(x)) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + ... + A001764(n)*x^n + ...
		

References

  • B. C. Berndt, Ramanujan's Notebooks Part I, Springer-Verlag, see Entry 14, Corollary 1, p. 71.

Crossrefs

Cf. A001764 (binomial(3n,n)/(2n+1)), A117671 (C(3n+1,n+1)), A004319, A005809, A006013, A013698, A045721, A117671, A165817, A224274, A236194.

Programs

  • Magma
    [Binomial(3*n-1,n-1): n in [0..30]]; // Vincenzo Librandi, Nov 12 2014
    
  • Maple
    with(combinat):seq(numbcomp(3*i,i), i=0..20); # Zerinvary Lajos, Jun 16 2007
  • Mathematica
    Table[ GegenbauerC[ n, n, 1 ]/2, {n, 0, 24} ]
    Join[{0},Table[Binomial[3n-1,n-1],{n,20}]] (* Harvey P. Dale, Oct 19 2022 *)
    nmax=23; CoefficientList[Series[(2+HypergeometricPFQ[{1/3,2/3},{1/2,1},27x/4])/3-1,{x,0,nmax}],x]Range[0,nmax]! (* Stefano Spezia, Dec 31 2024 *)
  • PARI
    vector(30, n, n--; binomial(3*n-1, n-1)) \\ Altug Alkan, Nov 04 2015

Formula

G.f.: z*g^2/(1-3*z*g^2), where g=g(z) is given by g=1+z*g^3, g(0)=1, that is, (in Maple command) g := 2*sin(arcsin(3*sqrt(3*z)/2)/3)/sqrt(3*z). - Emeric Deutsch, May 22 2003
a(n) = Sum_{k=0..n} ((3k+1)/(2n+k+1))C(3n, 2n+k)*A001045(k). - Paul Barry, Oct 07 2005
Hankel transform of a(n+1) is A005156(n+1). - Paul Barry, Apr 14 2008
G.f.: x*B'(x)/B(x) where B(x) is the g.f. of A001764. - Vladimir Kruchinin Feb 03 2013
D-finite with recurrence: 2*n*(2*n-1)*a(n) -3*(3*n-1)*(3*n-2)*a(n-1)=0. - R. J. Mathar, Feb 05 2013
Logarithmic derivative of A001764; g.f. of A001764 satisfies G(x) = 1 + x*G(x)^3. - Paul D. Hanna, Jul 14 2013
G.f.: (2*cos((1/3)*arcsin((3/2)*sqrt(3*x)))-sqrt(4-27*x))/(3*sqrt(4-27*x)). - Emanuele Munarini, Oct 14 2014
a(n) = Sum_{k=1..n} binomial(n-1,n-k)*binomial(2*n,n-k). - Vladimir Kruchinin, Nov 12 2014
a(n) = [x^n] C(x)^n for n >= 1, where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the generating function for A000108 (Ramanujan). - Peter Bala, Jun 24 2015
From Peter Bala, Nov 04 2015: (Start)
Without the initial term 0, the o.g.f. equals f(x)*g(x)^2, where f(x) is the o.g.f. for A005809 and g(x) is the o.g.f. for A001764. g(x)^2 is the o.g..f for A006013. More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(3*n + k,n). Cf. A045721 (k = 1), A004319 (k = 3), A236194 (k = 4), A013698 (k = 5), A165817 (k = -1), A117671 (k = -2). (End)
G.f.: ( 2F1(1/3,2/3;1/2;27*x/4)-1)/3. - R. J. Mathar, Jan 27 2020
O.g.f. without the initial term 0, in the form g(x)=(2*cos(arcsin((3*sqrt(3)*sqrt(x))/2)/3)/sqrt(4-27*x)-1)/(3*x), satisfies the following algebraic equation: 1+(9*x-1)*g(x)+x*(27*x-4)*g(x)^2+x^2*(27*x-4)*g(x)^3=0. - Karol A. Penson, Oct 11 2021
O.g.f. equals f(x)/(1 - 2*f(x)), where f(x) = series reversion (x/(1 + x)^3) = x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + ... is the o.g.f. of A001764 with the initial term omitted. Cf. A224274. - Peter Bala, Feb 03 2022
Right-hand side of the identities (1/2)*Sum_{k = 0..n} (-1)^(n+k)*C(x*n,n-k)*C((x+2)*n+k-1,k) = C(3*n-1,n-1) and (1/3)*Sum_{k = 0..n} (-1)^k* C(x*n,n-k)*C((x-3)*n+k-1,k) = C(3*n-1,n-1), both valid for n >= 1 and x arbitrary. - Peter Bala, Feb 28 2022
a(n) ~ 2^(-2*n)*3^(3*n)/(2*sqrt(3*n*Pi)). - Stefano Spezia, Apr 25 2024
a(n) = Sum_{k = 0..n-1} binomial(2*n+k-1, k) = Sum_{k = 0..n-1} (-1)^(n+k+1)* binomial(3*n, k). - Peter Bala, Jul 21 2024
E.g.f.: (2 + hypergeom([1/3, 2/3], [1/2, 1], 27*x/4))/3 - 1. - Stefano Spezia, Dec 31 2024
a(n+1) = Sum_{k = 0..n} (-1)^(n+k)*binomial(4*n+3, k)*binomial(2*n-k, n-k). - Peter Bala, Sep 04 2025

A134264 Coefficients T(j, k) of a partition transform for Lagrange compositional inversion of a function or generating series in terms of the coefficients of the power series for its reciprocal. Enumeration of noncrossing partitions and primitive parking functions. T(n,k) for n >= 1 and 1 <= k <= A000041(n-1), an irregular triangle read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 1, 4, 2, 6, 1, 1, 5, 5, 10, 10, 10, 1, 1, 6, 6, 3, 15, 30, 5, 20, 30, 15, 1, 1, 7, 7, 7, 21, 42, 21, 21, 35, 105, 35, 35, 70, 21, 1, 1, 8, 8, 8, 4, 28, 56, 56, 28, 28, 56, 168, 84, 168, 14, 70, 280, 140, 56, 140, 28, 1, 1, 9, 9, 9, 9, 36, 72
Offset: 1

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Author

Tom Copeland, Jan 14 2008

Keywords

Comments

Coefficients are listed in Abramowitz and Stegun order (A036036).
Given an invertible function f(t) analytic about t=0 (or a formal power series) with f(0)=0 and Df(0) not equal to 0, form h(t) = t / f(t) and let h_n denote the coefficient of t^n in h(t).
Lagrange inversion gives the compositional inverse about t=0 as g(t) = Sum_{j>=1} ( t^j * (1/j) * Sum_{permutations s with s(1) + s(2) + ... + s(j) = j - 1} h_s(1) * h_s(2) * ... * h_s(j) ) = t * T(1,1) * h_0 + Sum_{j>=2} ( t^j * Sum_{k=1..(# of partitions for j-1)} T(j,k) * H(j-1,k ; h_0,h_1,...) ), where H(j-1,k ; h_0,h_1,...) is the k-th partition for h_1 through h_(j-1) corresponding to n=j-1 on page 831 of Abramowitz and Stegun (ordered as in A&S) with (h_0)^(j-m)=(h_0)^(n+1-m) appended to each partition subsumed under n and m of A&S.
Denoting h_n by (n') for brevity, to 8th order in t,
g(t) = t * (0')
+ t^2 * [ (0') (1') ]
+ t^3 * [ (0')^2 (2') + (0') (1')^2 ]
+ t^4 * [ (0')^3 (3') + 3 (0')^2 (1') (2') + (0') (1')^3 ]
+ t^5 * [ (0')^4 (4') + 4 (0')^3 (1') (3') + 2 (0')^3 (2')^2 + 6 (0')^2 (1')^2 (2') + (0') (1')^4 ]
+ t^6 * [ (0')^5 (5') + 5 (0')^4 (1') (4') + 5 (0')^4 (2') (3') + 10 (0')^3 (1')^2 (3') + 10 (0')^3 (1') (2')^2 + 10 (0')^2 (1')^3 (2') + (0') (1')^5 ]
+ t^7 * [ (0')^6 (6') + 6 (0')^5 (1') (5') + 6 (0')^5 (2') (4') + 3 (0')^5 (3')^2 + 15 (0')^4 (1')^2 (4') + 30 (0')^4 (1') (2') (3') + 5 (0')^4 (2')^3 + 20 (0')^3 (1')^3 (3') + 30 (0')^3 (1')^2 (2')^2 + 15 (0')^2 (1')^4 (2') + (0') (1')^6]
+ t^8 * [ (0')^7 (7') + 7 (0')^6 (1') (6') + 7 (0')^6 (2') (5') + 7 (0')^6 (3') (4') + 21 (0')^5 (1')^2* (5') + 42 (0')^5 (1') (2') (4') + 21 (0')^5 (1') (3')^2 + 21 (0')^5 (2')^2 (3') + 35 (0')^4 (1')^3 (4') + 105 (0)^4 (1')^2 (2') (3') + 35 (0')^4 (1') (2')^3 + 35 (0')^3 (1')^4 (3') + 70 (0')^3 (1')^3 (2')^2 + 21 (0')^2 (1')^5 (2') + (0') (1')^7 ]
+ ..., where from the formula section, for example, T(8,1',2',...,7') = 7! / ((8 - (1'+ 2' + ... + 7'))! * 1'! * 2'! * ... * 7'!) are the coefficients of the integer partitions (1')^1' (2')^2' ... (7')^7' in the t^8 term.
A125181 is an extended, reordered version of the above sequence, omitting the leading 1, with alternate interpretations.
If the coefficients of partitions with the same exponent for h_0 are summed within rows, A001263 is obtained, omitting the leading 1.
From identification of the elements of the inversion with those on page 25 of the Ardila et al. link, the coefficients of the irregular table enumerate non-crossing partitions on [n]. - Tom Copeland, Oct 13 2014
From Tom Copeland, Oct 28-29 2014: (Start)
Operating with d/d(1') = d/d(h_1) on the n-th partition polynomial Prt(n;h_0,h_1,..,h_n) in square brackets above associated with t^(n+1) generates n * Prt(n-1;h_0,h_1,..,h_(n-1)); therefore, the polynomials are an Appell sequence of polynomials in the indeterminate h_1 when h_0=1 (a special type of Sheffer sequence).
Consequently, umbrally, [Prt(.;1,x,h_2,..) + y]^n = Prt(n;1,x+y,h_2,..); that is, Sum_{k=0..n} binomial(n,k) * Prt(k;1,x,h_2,..) * y^(n-k) = Prt(n;1,x+y,h_2,..).
Or, e^(x*z) * exp[Prt(.;1,0,h_2,..) * z] = exp[Prt(.;1,x,h_2,..) * z]. Then with x = h_1 = -(1/2) * d^2[f(t)]/dt^2 evaluated at t=0, the formal Laplace transform from z to 1/t of this expression generates g(t), the comp. inverse of f(t), when h_0 = 1 = df(t)/dt eval. at t=0.
I.e., t / (1 - t*(x + Prt(.;1,0,h_2,..))) = t / (1 - t*Prt(.;1,x,h_2,..)) = g(t), interpreted umbrally, when h_0 = 1.
(End)
Connections to and between arrays associated to the Catalan (A000108 and A007317), Riordan (A005043), Fibonacci (A000045), and Fine (A000957) numbers and to lattice paths, e.g., the Motzkin, Dyck, and Łukasiewicz, can be made explicit by considering the inverse in x of the o.g.f. of A104597(x,-t), i.e., f(x) = P(Cinv(x),t-1) = Cinv(x) / (1 + (t-1)*Cinv(x)) = x*(1-x) / (1 + (t-1)*x*(1-x)) = (x-x^2) / (1 + (t-1)*(x-x^2)), where Cinv(x) = x*(1-x) is the inverse of C(x) = (1 - sqrt(1-4*x)) / 2, a shifted o.g.f. for the Catalan numbers, and P(x,t) = x / (1+t*x) with inverse Pinv(x,t) = -P(-x,t) = x / (1-t*x). Then h(x,t) = x / f(x,t) = x * (1+(t-1)Cinv(x)) / Cinv(x) = 1 + t*x + x^2 + x^3 + ..., i.e., h_1=t and all other coefficients are 1, so the inverse of f(x,t) in x, which is explicitly in closed form finv(x,t) = C(Pinv(x,t-1)), is given by A091867, whose coefficients are sums of the refined Narayana numbers above obtained by setting h_1=(1')=t in the partition polynomials and all other coefficients to one. The group generators C(x) and P(x,t) and their inverses allow associations to be easily made between these classic number arrays. - Tom Copeland, Nov 03 2014
From Tom Copeland, Nov 10 2014: (Start)
Inverting in x with t a parameter, let F(x;t,n) = x - t*x^(n+1). Then h(x) = x / F(x;t,n) = 1 / (1-t*x^n) = 1 + t*x^n + t^2*x^(2n) + t^3*x^(3n) + ..., so h_k vanishes unless k = m*n with m an integer in which case h_k = t^m.
Finv(x;t,n) = Sum_{j>=0} {binomial((n+1)*j,j) / (n*j + 1)} * t^j * x^(n*j + 1), which gives the Catalan numbers for n=1, and the Fuss-Catalan sequences for n>1 (see A001764, n=2). [Added braces to disambiguate the formula. - N. J. A. Sloane, Oct 20 2015]
This relation reveals properties of the partitions and sums of the coefficients of the array. For n=1, h_k = t^k for all k, implying that the row sums are the Catalan numbers. For n = 2, h_k for k odd vanishes, implying that there are no blocks with only even-indexed h_k on the even-numbered rows and that only the blocks containing only even-sized bins contribute to the odd-row sums giving the Fuss-Catalan numbers for n=2. And so on, for n > 2.
These relations are reflected in any combinatorial structures enumerated by this array and the partitions, such as the noncrossing partitions depicted for a five-element set (a pentagon) in Wikipedia.
(End)
From Tom Copeland, Nov 12 2014: (Start)
An Appell sequence possesses an umbral inverse sequence (cf. A249548). The partition polynomials here, Prt(n;1,h_1,...), are an Appell sequence in the indeterminate h_1=u, so have an e.g.f. exp[Prt(.;1,u,h_2...)*t] = e^(u*t) * exp[Prt(.;1,0,h2,...)*t] with umbral inverses with an e.g.f e^(-u*t) / exp[Prt(.;1,0,h2,...)*t]. This makes contact with the formalism of A133314 (cf. also A049019 and A019538) and the signed, refined face partition polynomials of the permutahedra (or their duals), which determine the reciprocal of exp[Prt(.,0,u,h2...)*t] (cf. A249548) or exp[Prt(.;1,u,h2,...)*t], forming connections among the combinatorics of permutahedra and the noncrossing partitions, Dyck paths and trees (cf. A125181), and many other important structures isomorphic to the partitions of this entry, as well as to formal cumulants through A127671 and algebraic structures of Lie algebras. (Cf. relationship of permutahedra with the Eulerians A008292.)
(End)
From Tom Copeland, Nov 24 2014: (Start)
The n-th row multiplied by n gives the number of terms in the homogeneous symmetric monomials generated by [x(1) + x(2) + ... + x(n+1)]^n under the umbral mapping x(m)^j = h_j, for any m. E.g., [a + b + c]^2 = [a^2 + b^2 + c^2] + 2 * [a*b + a*c + b*c] is mapped to [3 * h_2] + 2 * [3 * h_1^2], and 3 * A134264(3) = 3 *(1,1)= (3,3) the number of summands in the two homogeneous polynomials in the square brackets. For n=3, [a + b + c + d]^3 = [a^3 + b^3 + ...] + 3 [a*b^2 + a*c^2 + ...] + 6 [a*b*c + a*c*d + ...] maps to [4 * h_3] + 3 [12 * h_1 * h_2] + 6 [4 * (h_1)^3], and the number of terms in the brackets is given by 4 * A134264(4) = 4 * (1,3,1) = (4,12,4).
The further reduced expression is 4 h_3 + 36 h_1 h_2 + 24 (h_1)^3 = A248120(4) with h_0 = 1. The general relation is n * A134264(n) = A248120(n) / A036038(n-1) where the arithmetic is performed on the coefficients of matching partitions in each row n.
Abramowitz and Stegun give combinatorial interpretations of A036038 and relations to other number arrays.
This can also be related to repeated umbral composition of Appell sequences and topology with the Bernoulli numbers playing a special role. See the Todd class link.
(End)
These partition polynomials are dubbed the Voiculescu polynomials on page 11 of the He and Jejjala link. - Tom Copeland, Jan 16 2015
See page 5 of the Josuat-Verges et al. reference for a refinement of these partition polynomials into a noncommutative version composed of nondecreasing parking functions. - Tom Copeland, Oct 05 2016
(Per Copeland's Oct 13 2014 comment.) The number of non-crossing set partitions whose block sizes are the parts of the n-th integer partition, where the ordering of integer partitions is first by total, then by length, then lexicographically by the reversed sequence of parts. - Gus Wiseman, Feb 15 2019
With h_0 = 1 and the other h_n replaced by suitably signed partition polynomials of A263633, the refined face partition polynomials for the associahedra of normalized A133437 with a shift in indices are obtained (cf. In the Realm of Shadows). - Tom Copeland, Sep 09 2019
Number of primitive parking functions associated to each partition of n. See Lemma 3.8 on p. 28 of Rattan. - Tom Copeland, Sep 10 2019
With h_n = n + 1, the d_k (A006013) of Table 2, p. 18, of Jong et al. are obtained, counting the n-point correlation functions in a quantum field theory. - Tom Copeland, Dec 25 2019
By inspection of the diagrams on Robert Dickau's website, one can see the relationship between the monomials of this entry and the connectivity of the line segments of the noncrossing partitions. - Tom Copeland, Dec 25 2019
Speicher has examples of the first four inversion partition polynomials on pp. 22 and 23 with his k_n equivalent to h_n = (n') here with h_0 = 1. Identifying z = t, C(z) = t/f(t) = h(t), and M(z) = f^(-1)(t)/t, then statement (3), on p. 43, of Theorem 3.26, C(z M(z)) = M(z), is equivalent to substituting f^(-1)(t) for t in t/f(t), and statement (4), M(z/C(z)) = C(z), to substituting f(t) for t in f^(-1)(t)/t. - Tom Copeland, Dec 08 2021
Given a Laurent series of the form f(z) = 1/z + h_1 + h_2 z + h_3 z^2 + ..., the compositional inverse is f^(-1)(z) = 1/z + Prt(1;1,h_1)/z^2 + Prt(2;1,h_1,h_2)/z^3 + ... = 1/z + h_1/z^2 + (h_1^2 + h_2)/z^3 + (h_1^3 + 3 h_1 h_2 + h_3)/z^4 + (h_1^4 + 6 h_1^2 h_2 + 4 h_1 h_3 + 2 h_2^2 + h_4)/z^5 + ... for which the polynomials in the numerators are the partition polynomials of this entry. For example, this formula applied to the q-expansion of Klein's j-invariant / function with coefficients A000521, related to monstrous moonshine, gives the compositional inverse with the coefficients A091406 (see He and Jejjala). - Tom Copeland, Dec 18 2021
The partition polynomials of A350499 'invert' the polynomials of this entry giving the indeterminates h_n. A multinomial formula for the coefficients of the partition polynomials of this entry, equivalent to the multinomial formula presented in the first four sentences of the formula section below, is presented in the MathOverflow question referenced in A350499. - Tom Copeland, Feb 19 2022

Examples

			1) With f(t) = t / (t-1), then h(t) = -(1-t), giving h_0 = -1, h_1 = 1 and h_n = 0 for n>1. Then g(t) = -t - t^2 - t^3 - ... = t / (t-1).
2) With f(t) = t*(1-t), then h(t) = 1 / (1-t), giving h_n = 1 for all n. The compositional inverse of this f(t) is g(t) = t*A(t) where A(t) is the o.g.f. for the Catalan numbers; therefore the sum over k of T(j,k), i.e., the row sum, is the Catalan number A000108(j-1).
3) With f(t) = (e^(-a*t)-1) / (-a), h(t) = Sum_{n>=0} Bernoulli(n) * (-a*t)^n / n! and g(t) = log(1-a*t) / (-a) = Sum_{n>=1} a^(n-1) * t^n / n. Therefore with h_n = Bernoulli(n) * (-a)^n / n!, Sum_{permutations s with s(1)+s(2)+...+s(j)=j-1} h_s(1) * h_s(2) * ... * h_s(j) = j * Sum_{k=1..(# of partitions for j-1)} T(j,k) * H(j-1,k ; h_0,h_1,...) = a^(j-1). Note, in turn, Sum_{a=1..m} a^(j-1) = (Bernoulli(j,m+1) - Bernoulli(j)) / j for the Bernoulli polynomials and numbers, for j>1.
4) With f(t,x) = t / (x-1+1/(1-t)), then h(t,x) = x-1+1/(1-t), giving (h_0)=x and (h_n)=1 for n>1. Then g(t,x) = (1-(1-x)*t-sqrt(1-2*(1+x)*t+((x-1)*t)^2)) / 2, a shifted o.g.f. in t for the Narayana polynomials in x of A001263.
5) With h(t)= o.g.f. of A075834, but with A075834(1)=2 rather than 1, which is the o.g.f. for the number of connected positroids on [n] (cf. Ardila et al., p. 25), g(t) is the o.g.f. for A000522, which is the o.g.f. for the number of positroids on [n]. (Added Oct 13 2014 by author.)
6) With f(t,x) = x / ((1-t*x)*(1-(1+t)*x)), an o.g.f. for A074909, the reverse face polynomials of the simplices, h(t,x) = (1-t*x) * (1-(1+t)*x) with h_0=1, h_1=-(1+2*t), and h_2=t*(1+t), giving as the inverse in x about 0 the o.g.f. (1+(1+2*t)*x-sqrt(1+(1+2*t)*2*x+x^2)) / (2*t*(1+t)*x) for signed A033282, the reverse face polynomials of the Stasheff polytopes, or associahedra. Cf. A248727. (Added Jan 21 2015 by author.)
7) With f(x,t) = x / ((1+x)*(1+t*x)), an o.g.f. for the polynomials (-1)^n * (1 + t + ... + t^n), h(t,x) = (1+x) * (1+t*x) with h_0=1, h_1=(1+t), and h_2=t, giving as the inverse in x about 0 the o.g.f. (1-(1+t)*x-sqrt(1-2*(1+t)*x+((t-1)*x)^2)) / (2*x*t) for the Narayana polynomials A001263. Cf. A046802. (Added Jan 24 2015 by author.)
From _Gus Wiseman_, Feb 15 2019: (Start)
Triangle begins:
   1
   1
   1   1
   1   3   1
   1   4   2   6   1
   1   5   5  10  10  10   1
   1   6   6   3  15  30   5  20  30  15   1
   1   7   7   7  21  42  21  21  35 105  35  35  70  21   1
Row 5 counts the following non-crossing set partitions:
  {{1234}}  {{1}{234}}  {{12}{34}}  {{1}{2}{34}}  {{1}{2}{3}{4}}
            {{123}{4}}  {{14}{23}}  {{1}{23}{4}}
            {{124}{3}}              {{12}{3}{4}}
            {{134}{2}}              {{1}{24}{3}}
                                    {{13}{2}{4}}
                                    {{14}{2}{3}}
(End)
		

References

  • A. Nica and R. Speicher (editors), Lectures on the Combinatorics of Free Probability, London Mathematical Society Lecture Note Series: 335, Cambridge University Press, 2006 (see in particular, Eqn. 9.14 on p. 141, enumerating noncrossing partitions).

Crossrefs

(A001263,A119900) = (reduced array, associated g(x)). See A145271 for meaning and other examples of reduced and associated.
Other orderings are A125181 and A306438.
Cf. A119900 (e.g.f. for reduced W(x) with (h_0)=t and (h_n)=1 for n>0).
Cf. A248927 and A248120, "scaled" versions of this Lagrange inversion.
Cf. A091867 and A125181, for relations to lattice paths and trees.
Cf. A249548 for use of Appell properties to generate the polynomials.
Cf. A133314, A049019, A019538, A127671, and A008292 for relations to permutahedra, Eulerians.
Cf. A006013.

Programs

  • Mathematica
    Table[Binomial[Total[y],Length[y]-1]*(Length[y]-1)!/Product[Count[y,i]!,{i,Max@@y}],{n,7},{y,Sort[Sort/@IntegerPartitions[n]]}] (* Gus Wiseman, Feb 15 2019 *)
  • PARI
    C(v)={my(n=vecsum(v), S=Set(v)); n!/((n-#v+1)!*prod(i=1, #S, my(x=S[i]); (#select(y->y==x, v))!))}
    row(n)=[C(Vec(p)) | p<-partitions(n-1)]
    { for(n=1, 7, print(row(n))) } \\ Andrew Howroyd, Feb 01 2022

Formula

For j>1, there are P(j,m;a...) = j! / [ (j-m)! (a_1)! (a_2)! ... (a_(j-1))! ] permutations of h_0 through h_(j-1) in which h_0 is repeated (j-m) times; h_1, repeated a_1 times; and so on with a_1 + a_2 + ... + a_(j-1) = m.
If, in addition, a_1 + 2 * a_2 + ... + (j-1) * a_(j-1) = j-1, then each distinct combination of these arrangements is correlated with a partition of j-1.
T(j,k) is [ P(j,m;a...) / j ] for the k-th partition of j-1 as described in the comments.
For example from g(t) above, T(5,4) = (5! / ((5-3)! * 2!)) / 5 = 6 for the 4th partition under n=5-1=4 with m=3 parts in A&S.
From Tom Copeland, Sep 30 2011: (Start)
Let W(x) = 1/(df(x)/dx)= 1/{d[x/h(x)]/dx}
= [(h_0)-1+:1/(1-h.*x):]^2 / {(h_0)-:[h.x/(1-h.x)]^2:}
= [(h_0)+(h_1)x+(h_2)x^2+...]^2 / [(h_0)-(h_2)x^2-2(h_3)x^3-3(h_4)x^4-...], where :" ": denotes umbral evaluation of the expression within the colons and h. is an umbral coefficient.
Then for the partition polynomials of A134264,
Poly[n;h_0,...,h_(n-1)]=(1/n!)(W(x)*d/dx)^n x, evaluated at x=0, and the compositional inverse of f(t) is g(t) = exp(t*W(x)*d/dx) x, evaluated at x=0. Also, dg(t)/dt = W(g(t)), and g(t) gives A001263 with (h_0)=u and (h_n)=1 for n>0 and A000108 with u=1.
(End)
From Tom Copeland, Oct 20 2011: (Start)
With exp(x* PS(.,t)) = exp(t*g(x)) = exp(x*W(y)d/dy) exp(t*y) eval. at y=0, the raising (creation) and lowering (annihilation) operators defined by R PS(n,t) = PS(n+1,t) and L PS(n,t) = n*PS(n-1,t) are
R = t*W(d/dt) = t*((h_0) + (h_1)d/dt + (h_2)(d/dt)^2 + ...)^2 / ((h_0) - (h_2)(d/dt)^2 - 2(h_3)(d/dt)^3 - 3(h_4)(d/dt)^4 + ...), and
L = (d/dt)/h(d/dt) = (d/dt) 1/((h_0) + (h_1)*d/dt + (h_2)*(d/dt)^2 + ...)
Then P(n,t) = (t^n/n!) dPS(n,z)/dz eval. at z=0 are the row polynomials of A134264. (Cf. A139605, A145271, and link therein to Mathemagical Forests for relation to planted trees on p. 13.)
(End)
Using the formalism of A263634, the raising operator for the partition polynomials of this array with h_0 = 1 begins as R = h_1 + h_2 D + h_3 D^2/2! + (h_4 - h_2^2) D^3/3! + (h_5 - 5 h_2 h_3) D^4/4! + (h_6 + 5 h_2^3 - 7 h_3^2 - 9 h_2 h_4) D^5/5! + (h_7 - 14 h_2 h_5 + 56 h_2^2 h_3) D^6/6! + ... with D = d/d(h_1). - Tom Copeland, Sep 09 2016
Let h(x) = x/f^{-1}(x) = 1/[1-(c_2*x+c_3*x^2+...)], with c_n all greater than zero. Then h_n are all greater than zero and h_0 = 1. Determine P_n(t) from exp[t*f^{-1}(x)] = exp[x*P.(t)] with f^{-1}(x) = x/h(x) expressed in terms of the h_n (cf. A133314 and A263633). Then P_n(b.) = 0 gives a recursion relation for the inversion polynomials of this entry a_n = b_n/n! in terms of the lower order inversion polynomials and P_j(b.)P_k(b.) = P_j(t)P_k(t)|{t^n = b_n} = d{j,k} >= 0 is the coefficient of x^j/j!*y^k/k! in the Taylor series expansion of the formal group law FGL(x,y) = f[f^{-1}(x)+f^{-1}(y)]. - Tom Copeland, Feb 09 2018
A raising operator for the partition polynomials with h_0 = 1 regarded as a Sheffer Appell sequence in h_1 is described in A249548. - Tom Copeland, Jul 03 2018

Extensions

Added explicit t^6, t^7, and t^8 polynomials and extended initial table to include the coefficients of t^8. - Tom Copeland, Sep 14 2016
Title modified by Tom Copeland, May 28 2018
More terms from Gus Wiseman, Feb 15 2019
Title modified by Tom Copeland, Sep 10 2019

A045721 a(n) = binomial(3*n+1,n).

Original entry on oeis.org

1, 4, 21, 120, 715, 4368, 27132, 170544, 1081575, 6906900, 44352165, 286097760, 1852482996, 12033222880, 78378960360, 511738760544, 3348108992991, 21945588357420, 144079707346575, 947309492837400, 6236646703759395, 41107996877935680, 271250494550621040, 1791608261879217600
Offset: 0

Views

Author

Keywords

Comments

Number of leaves in all noncrossing rooted trees on n nodes on a circle.
Number of standard tableaux of shape (n-1,1^(2n-3)). - Emeric Deutsch, May 25 2004
a(n) = number of Dyck (2n-3)-paths with exactly one descent of odd length. For example, a(3) counts all 5 Dyck 3-paths except UDUDUD. - David Callan, Jul 25 2005
a(n+2) gives row sums of A119301. - Paul Barry, May 13 2006
a(n) is the number of paths avoiding UU from (0,0) to (3n,n) and taking steps from {U,H}. - Shanzhen Gao, Apr 15 2010
Central coefficients of triangle A078812. - Vladimir Kruchinin, May 10 2012
Row sums of A252501. - L. Edson Jeffery, Dec 18 2014

Crossrefs

Programs

Formula

a(n) is asymptotic to c/sqrt(n)*(27/4)^n with c=0.73... - Benoit Cloitre, Jan 27 2003
G.f.: gz^2/(1-3zg^2), where g=g(z) is given by g=1+zg^3, g(0)=1, i.e. (in Maple command) g := 2*sin(arcsin(3*sqrt(3*z)/2)/3)/sqrt(3*z). - Emeric Deutsch, May 22 2003
a(n+2) = C(3n+1,n) = Sum_{k=0..n} C(3n-k,n-k). - Paul Barry, May 13 2006
a(n+2) = C(3n+1,2n+1) = A078812(2n,n). - Paul Barry, Nov 09 2006
G.f.: A(x)=(2*cos(asin((3^(3/2)*sqrt(x))/2)/3)* sin(asin((3^(3/2)* sqrt(x))/2)/3))/(sqrt(3)*sqrt(1-(27*x)/4)*sqrt(x)). - Vladimir Kruchinin, Jun 10 2012
From Peter Luschny, Sep 04 2012: (Start)
O.g.f.: hypergeometric2F1([2/3, 4/3], [3/2], x*27/4).
a(n) = (n+1)*hypergeometric2F1([-2*n, -n], [2], 1). (End)
D-finite with recurrence 2*n*(2*n+1)*a(n) - 3*(3*n-1)*(3*n+1)*a(n-1) = 0. - R. J. Mathar, Feb 05 2013
a(n) = Sum_{r=0..n} C(n,r) * C(2*n+1,r). - J. M. Bergot, Mar 18 2014
From Peter Bala, Nov 04 2015: (Start)
a(n) = Sum_{k = 0..n} 1/(2*k + 1)*binomial(3*n - 3*k,n - k)*binomial(3*k, k).
O.g.f. equals f(x)*g(x), where f(x) is the o.g.f. for A005809 and g(x) is the o.g.f. for A001764. More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(3*n + k,n). Cf. A025174 (k = 2), A004319 (k = 3), A236194 (k = 4), A013698 (k = 5), A165817 (k = -1), A117671 (k = -2). (End)
a(n) = [x^n] 1/(1 - x)^(2*(n+1)). - Ilya Gutkovskiy, Oct 10 2017
O.g.f.: (i/24)*((4*sqrt(4 - 27*z) + 12*i*sqrt(3)*sqrt(z))^(2/3) - (4*sqrt(4 - 27*z) - 12*i*sqrt(3)*sqrt(z))^(2/3))*sqrt(3)*8^(1/3)*sqrt(4 - 27*z)/(sqrt(z)*(-4 + 27*z)), where i = sqrt(-1). - Karol A. Penson, Dec 13 2023
a(n-1) = (1/(4*n))*binomial(2*n, n)^2 * (1 - 3*((n - 1)/(n + 1))^3 + 5*((n - 1)*(n - 2)/((n + 1)*(n + 2)))^3 - 7*((n - 1)*(n - 2)*(n - 3)/((n + 1)*(n + 2)*(n + 3)))^3 + - ...) for n >= 1. Cf. A112029. - Peter Bala, Aug 08 2024
a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(4*n+2, k)*binomial(2*n-k, n-k). - Peter Bala, Sep 04 2025
a(n) ~ 3^(3*n+3/2) / (4^(n+1) * sqrt(Pi*n)). - Amiram Eldar, Sep 05 2025

Extensions

Simpler definition from Ira M. Gessel, May 26 2007. This change means that most of the offsets in the comments will now need to be changed too.

A069271 a(n) = binomial(4*n+1,n)*2/(3*n+2).

Original entry on oeis.org

1, 2, 9, 52, 340, 2394, 17710, 135720, 1068012, 8579560, 70068713, 580034052, 4855986044, 41043559340, 349756577100, 3001701610320, 25921837477692, 225083787458904, 1963988670706228, 17211860478150800, 151433425446423120
Offset: 0

Views

Author

Henry Bottomley, Mar 12 2002

Keywords

Comments

This sequence counts the set B_n of plane trees defined in the Poulalhon and Schaeffer link (Definition 2.2 and Section 4.2, Proposition 4). - David Callan, Aug 20 2014
a(n) is the number of lattice paths of length 4n starting and ending on the x-axis consisting of steps {(1, 1), (1, -3)} that remain on or above the line y=-1. - Sarah Selkirk, Mar 31 2020
a(n) is the number of ordered pairs of 4-ary trees with a (summed) total of n internal nodes. - Sarah Selkirk, Mar 31 2020

Examples

			a(3) = C(4*3+1,3)*2/(3*3+2) = C(13,3)*2/11 = 286*2/11 = 52.
a(3) = 52 since the top row of M^3 = (22, 22, 7, 1).
1 + 2*x + 9*x^2 + 52*x^3 + 340*x^4 + 2394*x^5 + 17710*x^6 + 135720*x^7 + ...
q + 2*q^3 + 9*q^5 + 52*q^7 + 340*q^9 + 2394*q^11 + 17710*q^13 + 135720*q^15 + ...
		

Crossrefs

Cf. A002293, A006013, A006632, A069270 for similar generalized Catalan sequences.

Programs

  • Magma
    [2*Binomial(4*n+1, n)/(3*n+2): n in [0..20]];  // Bruno Berselli, Mar 04 2011
  • Maple
    BB:=[T,{T=Prod(Z,Z,Z,F,F),F=Sequence(B),B=Prod(F,F,F,Z)}, unlabeled]: seq(count(BB,size=i),i=3..23); # Zerinvary Lajos, Apr 22 2007
  • Mathematica
    f[n_] := 2 Binomial[4 n + 1, n]/(3 n + 2); Array[f, 21, 0] (* Robert G. Wilson v *)
  • PARI
    a(n)=if(n<0,0,polcoeff(serreverse(x/(1+x^2)^2+O(x^(2*n+2))),2*n+1)) /* Ralf Stephan */
    
  • PARI
    {a(n) =  binomial(4*n + 2, n)*2 / (2*n + 1)} /* Michael Somos, Mar 28 2012 */
    
  • PARI
    {a(n) =  local(A); if( n<0, 0, A = 1 + O(x); for( k=1, n, A = (1 + x * A^2)^2); polcoeff( A, n))} /* Michael Somos, Mar 28 2012 */
    

Formula

a(n) = A069270(n+1, n) = A005810(n)*A016813(n)/A060544(n+1)
O.g.f. A(x) satisfies 2*x^2*A(x)^3 = 1-2*x*A(x)-sqrt(1-4*x*A(x)). - Vladimir Kruchinin, Feb 23 2011
a(n) is the sum of top row terms in M^n, where M is the infinite square production matrix with the triangular series in each column as follows, with the rest zeros:
1, 1, 0, 0, 0, 0, ...
3, 3, 1, 0, 0, 0, ...
6, 6, 3, 1, 0, 0, ...
10, 10, 6, 3, 1, 0, ...
15, 15, 10, 6, 3, 1, ...
... - Gary W. Adamson, Aug 11 2011
Given g.f. A(x) then B(x) = x * A(x^2) satisfies x = B(x) / (1 + B(x)^2)^2. - Michael Somos, Mar 28 2012
Given g.f. A(x) then A(x) = (1 + x * A(x)^2)^2. - Michael Somos, Mar 28 2012
a(n) / (n+1) = A000260(n). - Michael Somos, Mar 28 2012
REVERT transform is A115141. - Michael Somos, Mar 28 2012
D-finite with recurrence 3*n*(3*n+2)*(3*n+1)*a(n) - 8*(4*n+1)*(2*n-1)*(4*n-1)*a(n-1) = 0. - R. J. Mathar, Jun 07 2013
a(n) = 2*binomial(4n+1,n-1)/n for n>0, a(0)=1. - Bruno Berselli, Jan 19 2014
G.f.: hypergeom([1/2, 3/4, 5/4], [4/3, 5/3], (256/27)*x). - Robert Israel, Aug 24 2014
From Peter Bala, Oct 08 2015: (Start)
O.g.f. A(x) = (1/x) * series reversion (x/C(x)^2), where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108. Cf. A163456.
(1/2)*x*A'(x)/A(x) is the o.g.f. for A224274. (End)
E.g.f.: hypergeom([1/2, 3/4, 5/4], [1, 4/3, 5/3], (256/27)*x). - Karol A. Penson, Jun 26 2017
a(n) = binomial(4*n+2,n)/(2*n+1). - Alexander Burstein, Nov 08 2021

A118971 a(n) = binomial(5*n+3,n)/(n+1).

Original entry on oeis.org

1, 4, 26, 204, 1771, 16380, 158224, 1577532, 16112057, 167710664, 1772645420, 18974357220, 205263418941, 2240623268512, 24648785802336, 272994644359580, 3041495503591365, 34064252968167180, 383302465665133014
Offset: 0

Views

Author

Paul Barry, May 07 2006

Keywords

Comments

A quadrisection of A118968.
For n >= 1, a(n-1) is the number of lattice paths from (0,0) to (4n,n) using only the steps (1,0) and (0,1) and which stay strictly below the line y = x/4 except at the path's endpoints. - Lucas A. Brown, Aug 21 2020
This is instance k = 4 of the family {c(k, n+1)}A130564.%20-%20_Wolfdieter%20Lang">{n>=0} given in a comment in A130564. - _Wolfdieter Lang, Feb 04 2024

Crossrefs

Cf. A000108, A006013, A006632, A130564, A130565, A234466, A234513, A234573, A235340 (members of the same family).

Programs

  • Mathematica
    Table[4*Binomial[5n+3,n]/(4n+4),{n,0,30}] (* Harvey P. Dale, Apr 09 2012 *)

Formula

G.f.: If the inverse series of y*(1-y)^4 is G(x) then A(x)=G(x)/x.
D-finite with recurrence 8*(4*n+1)*(2*n+1)*(4*n+3)*(n+1)*a(n) -5*(5*n+1)*(5*n+2)*(5*n+3)*(5*n-1)*a(n-1)=0. - R. J. Mathar, Nov 26 2012
a(n) = (4/5)*binomial(5*(n+1),n+1)/(5*(n+1)-1). - Bruno Berselli, Jan 17 2014
E.g.f.: 4F4(4/5,6/5,7/5,8/5; 5/4,3/2,7/4,2; 3125*x/256). - Ilya Gutkovskiy, Jan 23 2018
G.f.: 5F4([4,5,6,7,8]/5, [5,6,7,8]/4; (5^5/4^4)*x) = (4/(5*x))*(1 - 4F3([-1,1,2,3]/5, [1,2,3]/4; (5^5/4^4)*x)). - Wolfdieter Lang, Feb 15 2024

A130564 Member k=5 of a family of generalized Catalan numbers.

Original entry on oeis.org

1, 5, 40, 385, 4095, 46376, 548340, 6690585, 83615350, 1064887395, 13770292256, 180320238280, 2386316821325, 31864803599700, 428798445360120, 5809228810425801, 79168272296871450, 1084567603590147950
Offset: 1

Views

Author

Wolfdieter Lang, Jul 13 2007

Keywords

Comments

The generalized Catalan numbers C(k,n):= binomial(k*n+1,n)/(k*n+1) become for negative k=-|k|, with |k|>=2, ((-1)^(n-1))*binomial((|k|+1)*n-2,n)/(|k|*n-1), n>=0.
The family c(k,n):=binomial((k+1)*n-2,n)/(k*n-1), n>=1, has the members A000108, A006013, A006632, A118971 for k=1,2,3,4, respectively (but the offset there is 0).
The members of the C(k,n) family for positive k are: A000012 (powers of 1), A000108, A001764, A002293, A002294, A002295, A002296, A007556, A062994, for k=1..9.

References

  • Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Mathematics, Addison-Wesley, Reading, MA, 2nd ed. 1994, pp. 200, 363.

Crossrefs

Programs

  • Mathematica
    Rest@ CoefficientList[InverseSeries[Series[y (1 - y)^5, {y, 0, 18}], x], x] (* Michael De Vlieger, Oct 13 2019 *)

Formula

a(n) = binomial((k+1)*n-2,n)/(k*n-1), with k=5.
G.f.: inverse series of y*(1-y)^5.
a(n) = (5/6)*binomial(6*n,n)/(6*n-1). [Bruno Berselli, Jan 17 2014]
From Wolfdieter Lang, Feb 06 2020: (Start)
G.f.: (5/6)*(1 - hypergeom([-1, 1, 2, 3, 4]/6, [1, 2, 3, 4]/5,(6^6/5^5)*x)).
E.g.f.: (5/6)*(1 - hypergeom([-1, 1, 2, 3, 4]/6, [1, 2, 3, 4, 5]/5,(6^6/5^5)*x)). (End)
D-finite with recurrence 5*n*(5*n-4)*(5*n-3)*(5*n-2)*(5*n-1)*a(n) -72*(6*n-7)*(3*n-1)*(2*n-1)*(3*n-2)*(6*n-5)*a(n-1)=0. - R. J. Mathar, May 07 2021

A165817 Number of compositions (= ordered integer partitions) of n into 2n parts.

Original entry on oeis.org

1, 2, 10, 56, 330, 2002, 12376, 77520, 490314, 3124550, 20030010, 129024480, 834451800, 5414950296, 35240152720, 229911617056, 1503232609098, 9847379391150, 64617565719070, 424655979547800, 2794563003870330, 18412956934908690, 121455445321173600
Offset: 0

Views

Author

Thomas Wieder, Sep 29 2009

Keywords

Comments

Number of ways to put n indistinguishable balls into 2*n distinguishable boxes.
Number of rankings of n unlabeled elements for 2*n levels.

Examples

			Let [1,1,1], [1,2] and [3] be the integer partitions of n=3.
Then [0,0,0,1,1,1], [0,0,0,0,1,2] and [0,0,0,0,0,3] are the corresponding partitions occupying 2*n = 6 positions.
We have to take into account the multiplicities of the parts including the multiplicities of the zeros.
Then
  [0,0,0,1,1,1] --> 6!/(3!*3!) = 20
  [0,0,0,0,1,2] --> 6!/(4!*1!*1!) = 30
  [0,0,0,0,0,3] --> 6!/(5!*1!) = 6
and thus a(3) = 20+30+6=56.
a(2)=10, since we have 10 ordered partitions of n=2 where the parts are distributed over 2*n=4 boxes:
  [0, 0, 0, 2]
  [0, 0, 1, 1]
  [0, 0, 2, 0]
  [0, 1, 0, 1]
  [0, 1, 1, 0]
  [0, 2, 0, 0]
  [1, 0, 0, 1]
  [1, 0, 1, 0]
  [1, 1, 0, 0]
  [2, 0, 0, 0].
		

Crossrefs

Programs

  • Magma
    [Binomial(3*n-1, n): n in [0..30]]; // Vincenzo Librandi, Aug 07 2014
    
  • Maple
    for n from 0 to 16 do
    a[n] := 9*sqrt(3)*GAMMA(n+5/3)*GAMMA(n+4/3)*27^n/(Pi*GAMMA(2*n+3))
    end do;
  • Mathematica
    Table[Binomial[3 n - 1, n], {n, 0, 20}] (* Vincenzo Librandi, Aug 07 2014 *)
  • PARI
    vector(30, n, n--; binomial(3*n-1, n)) \\ Altug Alkan, Nov 04 2015
    
  • Python
    from math import comb
    def A165817(n): return comb(3*n-1,n) if n else 1 # Chai Wah Wu, Oct 11 2023
  • Sage
    def A165817(n):
        return rising_factorial(2*n,n)/falling_factorial(n,n)
    [A165817(n) for n in (0..22)]   # Peter Luschny, Nov 21 2012
    

Formula

a(n) = 9*sqrt(3)*GAMMA(n+5/3)*GAMMA(n+4/3)*27^n/(Pi*GAMMA(2*n+3)).
a(n) = binomial(3*n-1, n);
Let denote P(n) = the number of integer partitions of n,
p(i) = the number of parts of the i-th partition of n,
d(i) = the number of different parts of the i-th partition of n,
m(i,j) = multiplicity of the j-th part of the i-th partition of n.
Then one has:
a(n) = Sum_{i=1..P(n)} (2*n)!/((2*n-p(i))!*(Prod_{j=1..d(i)} m(i,j)!)).
a(n) = rf(2*n,n)/ff(n,n), where rf is the rising factorial and ff the falling factorial. - Peter Luschny, Nov 21 2012
G.f.: A(x) = x*B'(x)/B(x), where B(x) satisfies B(x)^3-2*B(x)^2+B(x)=x, B(x)=A006013(x). - Vladimir Kruchinin, Feb 06 2013
G.f.: A(x) = sqrt(3*x)*cot(asin((3^(3/2)*sqrt(x))/2)/3)/(sqrt(4-27*x)). - Vladimir Kruchinin, Mar 18 2015
a(n) = Sum_{k=0..n} binomial(n-1,n-k)*binomial(2*n,k). - Vladimir Kruchinin, Oct 06 2015
From Peter Bala, Nov 04 2015: (Start)
The o.g.f. equals f(x)/g(x), where f(x) is the o.g.f. for A005809 and g(x) is the o.g.f. for A001764. More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(3*n + k,n). Cf. A045721 (k = 1), A025174 (k = 2), A004319 (k = 3), A236194 (k = 4), A013698 (k = 5) and A117671 (k = -2). (End)
a(n) = [x^n] 1/(1 - x)^(2*n). - Ilya Gutkovskiy, Oct 03 2017
a(n) = A059481(2n,n). - Alois P. Heinz, Oct 17 2022
From Peter Bala, Feb 14 2024: (Start)
a(n) = (-1)^n * binomial(-2*n, n).
a(n) = hypergeom([1 - 2*n, -n], [1], 1).
The g.f. A(x) satisfies A(x/(1 + x)^3) = 1/(1 - 2*x).
Sum_{n >= 0} a(n)/9^n = (1 + 4*cos(Pi/9))/3.
Sum_{n >= 0} a(n)/27^n = (3 + 4*sqrt(3)*cos(Pi/18))/9.
Sum_{n >= 0} a(n)*(2/27)^n = (2 + sqrt(3))/3. (End)
From Peter Bala, Sep 16 2024: (Start)
a(n) = Sum_{k = 0..n} binomial(n+k-1, k)*binomial(2*n-k-1, n-k).
More generally, a(n) = Sum_{k = 0..n} (-1)^k*binomial(x*n, k)*binomial((x+3)*n-k-1, n-k) for arbitrary x.
a(n) = (2/3) * Sum_{k = 0..n} (-1)^k*binomial(x*n+k-1, k)*binomial((x+3)*n, n-k) for n >= 1 and arbitrary x. (End)
G.f.: 1/(3-2*g) where g = 1+x*g^3 is the g.f. of A001764. - Seiichi Manyama, Aug 17 2025

Extensions

a(0) prepended and more terms from Alois P. Heinz, Apr 04 2012

A007226 a(n) = 2*det(M(n; -1))/det(M(n; 0)), where M(n; m) is the n X n matrix with (i,j)-th element equal to 1/binomial(n + i + j + m, n).

Original entry on oeis.org

2, 3, 10, 42, 198, 1001, 5304, 29070, 163438, 937365, 5462730, 32256120, 192565800, 1160346492, 7048030544, 43108428198, 265276342782, 1641229898525, 10202773534590, 63698396932170, 399223286267190, 2510857763851185, 15842014607109600
Offset: 0

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Author

Keywords

Comments

For n >= 1, a(n) is the number of distinct perforation patterns for deriving (v,b) = (n+1,n) punctured convolutional codes from (3,1). [Edited by Petros Hadjicostas, Jul 27 2020]
Apparently Bégin's (1992) paper was presented at a poster session at the conference and was never published.
a(n) is the total number of down steps between the first and second up steps in all 2-Dyck paths of length 3*(n+1). A 2-Dyck path is a nonnegative lattice path with steps (1,2), (1,-1) that starts and ends at y = 0. - Sarah Selkirk, May 07 2020
From Petros Hadjicostas, Jul 27 2020: (Start)
"A punctured convolutional code is a high-rate code obtained by the periodic elimination (i.e., puncturing) of specific code symbols from the output of a low-rate encoder. The resulting high-rate code depends on both the low-rate code, called the original code, and the number and specific positions of the punctured symbols." (The quote is from Haccoun and Bégin (1989).)
A high-rate code (v,b) (written as R = b/v) can be constructed from a low-rate code (v0,1) (written as R = 1/v0) by deleting from every v0*b code symbols a number of v0*b - v symbols (so that the resulting rate is R = b/v).
Even though my formulas below do not appear in the two published papers in the IEEE Transactions on Communications, from the theory in those two papers, it makes sense to replace "k|b" with "k|v0*b" (and "k|gcd(v,b)" with "k|gcd(v,v0*b)"). Pab Ter, however, uses "k|b" in the Maple programs in the related sequences A007223, A007224, A007225, A007227, and A007229. (End)
Conjecture: a(n) is odd iff n = A022341(k) for some k. - Peter Bala, Mar 13 2023

References

  • Guy Bégin, On the enumeration of perforation patterns for punctured convolutional codes, Séries Formelles et Combinatoire Algébrique, 4th colloquium, 15-19 Juin 1992, Montréal, Université du Québec à Montréal, pp. 1-10.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Magma
    [Binomial(3*n,n)/(2*n+1)+Binomial(3*n+1,n)/(n+1): n in [0..25]]; // Vincenzo Librandi, Aug 10 2014
    
  • Maple
    A007226:=n->2*binomial(3*n,n)-binomial(3*n,n+1): seq(A007226(n), n=0..30); # Wesley Ivan Hurt, Aug 11 2014
  • Mathematica
    Table[2*Binomial[3n,n]-Binomial[3n,n+1], {n,0,20}] (* Harvey P. Dale, Aug 10 2014 *)
  • PARI
    a(n) = {my(M1=matrix(n,n)); my(M0=matrix(n,n)); for(i=1, n, for(j=1, n, M1[i,j] = 1/binomial(n+i+j-1,n); M0[i,j] = 1/binomial(n+i+j,n);)); 2*matdet(M1)/matdet(M0);} \\ Petros Hadjicostas, Jul 27 2020

Formula

a(n) = (2/(n + 1))*binomial(3*n, n).
a(n) = (2n+1) * A000139(n). - F. Chapoton, Feb 23 2024
a(n) = 2*C(3*n, n) - C(3*n, n+1) for n >= 0. - David Callan, Oct 25 2004
a(n) = C(3*n, n)/(2*n + 1) + C(3*n + 1, n)/(n + 1) = C(3*n, n)/(2*n + 1) + 2*C(3*n + 1, n)/(2*n + 2) for n >= 0. - Paul Barry, Nov 05 2006
G.f.: g*(2 - g)/x, where g*(1 - g)^2 = x. - Mark van Hoeij, Nov 08 2011 [Thus, g = (4/3)*sin((1/3)*arcsin(sqrt(27*x/4)))^2. - Petros Hadjicostas, Jul 27 2020]
Recurrence: 2*(n+1)*(2*n-1)*a(n) - 3*(3*n-1)*(3*n-2)*a(n-1) = 0 for n >= 1. - R. J. Mathar, Nov 26 2012
G.f.: (1 - 1/B(x))/x, where B(x) is the g.f. of A006013. [Vladimir Kruchinin, Mar 05 2013]
G.f.: ( -16 * sin(asin((3^(3/2) * sqrt(x))/2)/3)^4 + 24 * sin(asin((3^(3/2) * sqrt(x))/2)/3)^2 ) / (9*x). [Vladimir Kruchinin, Nov 16 2013]
From Petros Hadjicostas, Jul 27 2020: (Start)
The number of perforation patterns to derive high-rate convolutional code (v,b) (written as R = b/v) from a given low-rate convolutional code (v0, 1) (written as R = 1/v0) is (1/b)*Sum_{k|gcd(v,b)} phi(k)*binomial(v0*b/k, v/k).
According to Pab Ter's Maple code in the related sequences (see above), this is the coefficient of z^v in the polynomial (1/b)*Sum_{k|b} phi(k)*(1 + z^k)^(v0*b/k).
Here (v,b) = (n+1,n) and (v0,1) = (3,1), so for n >= 1,
a(n) = (1/n)*Sum_{k|gcd(n+1,n)} phi(k)*binomial(3*n/k, (n+1)/k).
This simplifies to
a(n) = (1/n)*binomial(3*n, n+1) for n >= 1. (End)
G.f.: (-(p+r)*(4*r-12*p)^(1/3)+(p-r)*(4*r+12*p)^(1/3)+8)/(12*z), where p = i*sqrt(3*z), r = sqrt(4-27*z), and i = sqrt(-1) is the imaginary unit. - Karol A. Penson, Mar 20 2025

Extensions

Edited following a suggestion of Ralf Stephan, Feb 07 2004
Offset changed to 0 and all formulas checked by Petros Hadjicostas, Jul 27 2020

A006629 Self-convolution 4th power of A001764, which enumerates ternary trees.

Original entry on oeis.org

1, 4, 18, 88, 455, 2448, 13566, 76912, 444015, 2601300, 15426840, 92431584, 558685348, 3402497504, 20858916870, 128618832864, 797168807855, 4963511449260, 31032552351570, 194743066471800, 1226232861415695
Offset: 0

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Author

Keywords

Comments

Sum of root degrees of all noncrossing trees on nodes on a circle. - Emeric Deutsch

References

  • H. M. Finucan, Some decompositions of generalized Catalan numbers, pp. 275-293 of Combinatorial Mathematics IX. Proc. Ninth Australian Conference (Brisbane, August 1981). Ed. E. J. Billington, S. Oates-Williams and A. P. Street. Lecture Notes Math., 952. Springer-Verlag, 1982.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Magma
    A006629:= func< n | 2*Binomial(3*n+3,n)/(n+2) >;
    [A006629(n): n in [0..40]]; // G. C. Greubel, Aug 29 2025
    
  • Mathematica
    Table[2*Binomial[3*n+3,n]/(n+2), {n,0,40}] (* G. C. Greubel, Aug 29 2025 *)
  • PARI
    a(n)=my(m=4);binomial(3*n+m-1,n)*m/(2*n+m) /* 4th power of A001764 with offset n=0 */ \\ Paul D. Hanna, May 10 2008
    
  • SageMath
    def A006629(n): return 2*binomial(3*n+3,n)//(n+2)
    print([A006629(n) for n in range(41)]) # G. C. Greubel, Aug 29 2025

Formula

a(n) = 2*binomial(3*n+3,n)/(n+2). - Emeric Deutsch
a(n) = (n+1) * A000139(n+1). - F. Chapoton, Feb 23 2024
G.f.: hypergeom( [ 2, 5/3, 4/3 ]; [ 3, 5/2 ]; 27*x/4 ).
G.f.: A(x) = G(x)^4 where G(x) = 1 + x*G(x)^3 = g.f. of A001764 giving a(n)=C(3n+m-1,n)*m/(2n+m) at power m=4 with offset n=0. - Paul D. Hanna, May 10 2008
G.f.: (((4*sin(arcsin((3*sqrt(3*x))/2)/3))/(sqrt(3*x))-1)^2-1)/(4*x). - Vladimir Kruchinin, Feb 17 2023
E.g.f.: hypergeom([4/3, 5/3, 2]; [1, 5/2, 3]; 27*x/4). - G. C. Greubel, Aug 29 2025

Extensions

More precise definition from Paul D. Hanna, May 10 2008
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