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A005043 Riordan numbers: a(n) = (n-1)*(2*a(n-1) + 3*a(n-2))/(n+1).
(Formerly M2587)
1, 0, 1, 1, 3, 6, 15, 36, 91, 232, 603, 1585, 4213, 11298, 30537, 83097, 227475, 625992, 1730787, 4805595, 13393689, 37458330, 105089229, 295673994, 834086421, 2358641376, 6684761125, 18985057351, 54022715451, 154000562758, 439742222071, 1257643249140 (list; graph; refs; listen; history; text; internal format)



Also called Motzkin summands or ring numbers.

The old name was "Motzkin sums", used in certain publications. The sequence has the property that Motzkin(n) = A001006(n) = a(n) + a(n+1), e.g., A001006(4) = 9 = 3 + 6 = a(4) + a(5).

Number of 'Catalan partitions', that is partitions of a set 1,2,3,...,n into parts that are not singletons and whose convex hulls are disjoint when the points are arranged on a circle (so when the parts are all pairs we get Catalan numbers). - Aart Blokhuis (aartb(AT)win.tue.nl), Jul 04 2000

Number of ordered trees with n edges and no vertices of outdegree 1. For n > 1, number of dissections of a convex polygon by nonintersecting diagonals with a total number of n+1 edges. - Emeric Deutsch, Mar 06 2002

Number of Motzkin paths of length n with no horizontal steps at level 0. - Emeric Deutsch, Nov 09 2003

Number of Dyck paths of semilength n with no peaks at odd level. Example: a(4)=3 because we have UUUUDDDD, UUDDUUDD and UUDUDUDD, where U=(1,1), D=(1,-1). Number of Dyck paths of semilength n with no ascents of length 1 (an ascent in a Dyck path is a maximal string of up steps). Example: a(4)=3 because we have UUUUDDDD, UUDDUUDD and UUDUUDDD. - Emeric Deutsch, Dec 05 2003

Arises in Schubert calculus as follows. Let P = complex projective space of dimension n+1. Take n projective subspaces of codimension 3 in P in general position. Then a(n) is the number of lines of P intersecting all these subspaces. - F. Hirzebruch, Feb 09 2004

Difference between central trinomial coefficient and its predecessor. Example: a(6) = 15 = 141 - 126 and (1 + x + x^2)^6 = ... + 126*x^5 + 141*x^6 + ... (Catalan number A000108(n) is the difference between central binomial coefficient and its predecessor.) - David Callan, Feb 07 2004

a(n) = number of 321-avoiding permutations on [n] in which each left-to-right maximum is a descent (i.e., is followed by a smaller number). For example, a(4) counts 4123, 3142, 2143. - David Callan, Jul 20 2005

The Hankel transform of this sequence give A000012 = [1, 1, 1, 1, 1, 1, 1, ...]; example: Det([1, 0, 1, 1; 0, 1, 1, 3; 1, 1, 3, 6; 1, 3, 6, 15]) = 1. - Philippe Deléham, May 28 2005

The number of projective invariants of degree 2 for n labeled points on the projective line. - Benjamin J. Howard (bhoward(AT)ima.umn.edu), Nov 24 2006

Define a random variable X=trA^2, where A is a 2 X 2 unitary symplectic matrix chosen from USp(2) with Haar measure. The n-th central moment of X is E[(X+1)^n] = a(n). - Andrew V. Sutherland, Dec 02 2007

Let V be the adjoint representation of the complex Lie algebra sl(2). The dimension of the invariant subspace of the n-th tensor power of V is a(n). - Samson Black (sblack1(AT)uoregon.edu), Aug 27 2008

Starting with offset 3 = iterates of M * [1,1,1,...], where M = a tridiagonal matrix with [0,1,1,1,...] in the main diagonal and [1,1,1,...] in the super and subdiagonals. - Gary W. Adamson, Jan 08 2009

a(n) has the following standard-Young-tableaux (SYT) interpretation: binomial(n+1,k)*binomial(n-k-1,k-1)/(n+1)=f^(k,k,1^{n-2k}) where f^lambda equals the number of SYT of shape lambda. - Amitai Regev (amotai.regev(AT)weizmann.ac.il), Mar 02 2010

a(n) is also the sum of the numbers of standard Young tableaux of shapes (k,k,1^{n-2k}) for all 1 <= k <= floor(n/2). - Amitai Regev (amotai.regev(AT)weizmann.ac.il), Mar 10 2010

a(n) is the number of derangements of {1,2,...,n} having genus 0. The genus g(p) of a permutation p of {1,2,...,n} is defined by g(p)=(1/2)[n+1-z(p)-z(cp')], where p' is the inverse permutation of p, c = 234...n1 = (1,2,...,n), and z(q) is the number of cycles of the permutation q. Example: a(3)=1 because p=231=(123) is the only derangement of {1,2,3} with genus 0. Indeed, cp'=231*312=123=(1)(2)(3) and so g(p)= (1/2)(3+1-1-3)=0. - Emeric Deutsch, May 29 2010

Apparently: Number of Dyck 2n-paths with all ascents length 2 and no descent length 2. - David Scambler, Apr 17 2012

This is true. Proof: The mapping "insert a peak (UD) after each upstep (U)" is a bijection from all Dyck n-paths to those Dyck (2n)-paths in which each ascent is of length 2. It sends descents of length 1 in the n-path to descents of length 2 in the (2n)-path. But Dyck n-paths with no descents of length 1 are equinumerous with Riordan n-paths (Motzkin n-paths with no flatsteps at ground level) as follows. Given a Dyck n-path with no descents of length 1, split it into consecutive step pairs, then replace UU with U, DD with D, UD with a blue flatstep (F), DU with a red flatstep, and concatenate the new steps to get a colored Motzkin path. Each red F will be (immediately) preceded by a blue F or a D. In the latter case, transfer the red F so that it precedes the matching U of the D. Finally, erase colors to get the required Riordan path. For example, with lowercase f denoting a red flatstep, U^5 D^2 U D^4 U^4 D^3 U D^2 -> (U^2, U^2, UD, DU, D^2, D^2, U^2, U^2 D^2, DU, D^2) -> UUFfDDUUDfD -> UUFFDDUFUDD. - David Callan, Apr 25 2012

From Nolan Wallach, Aug 20 2014: (Start)

Let ch[part1, part2] be the value of the character of the symmetric group on n letters corresponding to the partition part1 of n on the conjucgacy class given by part2. Let A[n] be the set of (n+1) partitions of 2n with parts 1 or 2. Then deleting the first term of the sequence one has a(n) = Sum_{k=1..n+1} binomial(n,k-1)*ch[[n,n], A[n][[k]]])/2^n. This via the Frobenius Character Formula can be interpreted as the dimension of the SL(n,C) invariants in tensor^n (wedge^2 C^n).

Explanation: Let p_j denote sum (x_i)^j the sum in k variables. Then the Frobenius formula says then (p_1)^j_1 (p_2)^j_2 ... (p_r)^j_r is equal to sum(lambda, ch[lambda, 1^j_12^j_2 ... r^j_r] S_lambda) with S_lambda the Shur function corresponding to lambda. This formula implies that the coefficient of S([n,n]) in (((p_1)^1+p_2)/2)^n in its expansion in terms of Shur functions is the right hand side of our formula. If we specialize the number of variables to 2 then S[n,n](x,y)=(xy)^n. Which when restricted to y=x^(-1) is 1. That is it is 1 on SL(2).

On the other hand ((p_1)^2+p_2)/2 is the complete homogeneous symmetric function of degree 2 that is tr(S^2(X)). Thus our formula for a(n) is the same as that of Samson Black above since his V is the same as S^2(C^2) as a representation of SL(2). On the other hand, if we multiply ch(lambda) by sgn you get ch(Transpose(lambda)). So ch([n,n]) becomes ch([2,...,2]) (here there are n 2's). The formula for a(n) is now (1/2^n)*Sum_{j=0..n} ch([2,..,2], 1^(2n-2j) 2^j])*(-1)^j)*binomial(n,j), which calculates the coefficient of S_(2,...,2) in (((p_1)^2-p_2)/2)^n. But ((p_1)^2-p_2)/2 in n variables is the second elementary symmetric function which is the character of wedge^2 C^n and S_(2,...,2) is 1 on SL(n).


a(n) = number of noncrossing partitions (A000108) of [n] that contain no singletons, also number of nonnesting partitions (A000108) of [n] that contain no singletons. - David Callan, Aug 27 2014

From Tom Copeland, Nov 02 2014: (Start)

Let P(x) = x/(1+x) with comp. inverse Pinv(x) = x/(1-x) = -P[-x], and C(x)= [1-sqrt(1-4x)]/2, an o.g.f. for the shifted Catalan numbers A000108, with inverse Cinv(x) = x * (1-x).

Fin(x) = P[C(x)] = C(x)/[1 + C(x)] is an o.g.f. for the Fine numbers, A000957 with inverse Fin^(-1)(x) = Cinv[Pinv(x)] = Cinv[-P(-x)].

Mot(x) = C[P(x)] = C[-Pinv(-x)] gives an o.g.f. for shifted A005043, the Motzkin or Riordan numbers with comp. inverse Mot^(-1)(x) = Pinv[Cinv(x)] = (x - x^2) / (1 - x + x^2) (cf. A057078).

BTC(x) = C[Pinv(x)] gives A007317, a binomial transform of the Catalan numbers, with BTC^(-1)(x) = P[Cinv(x)].

Fib(x) = -Fin[Cinv(Cinv(-x))] = -P[Cinv(-x)] = x + 2 x^2 + 3 x^3 + 5 x^4 + ... = (x+x^2)/[1-x-x^2] is an o.g.f. for the shifted Fibonacci sequence A000045, so the comp. inverse is Fib^(-1)(x) = -C[Pinv(-x)] = -BTC(-x) and Fib(x) = -BTC^(-1)(-x).

Various relations among the o.g.f.s may be easily constructed, such as Fib[-Mot(-x)] = -P[P(-x)] = x/(1-2*x) a generating fct for 2^n.

Generalizing to P(x,t) = x /(1 + t*x) and Pinv(x,t) = x /(1 - t*x) = -P(-x,t) gives other relations to lattice paths, such as the o.g.f. for A091867, C[P[x,1-t]], and that for A104597, Pinv[Cinv(x),t+1]. (End)

Consistent with David Callan's comment above, A249548, provides a refinement of the Motzkin sums into the individual numbers for the non-crossing partitions he describes. - Tom Copeland, Nov 09 2014

The number of lattice paths from (0,0) to (n,0) that do not cross below the x-axis and use up-step=(1,1) and down-steps=(1,-k) where k is a positive integer. For example, a(4) = 3: [(1,1)(1,1)(1,-1)(1,-1)], [(1,1)(1,-1)(1,1)(1,-1)] and [(1,1)(1,1)(1,1)(1,-3)]. - Nicholas Ham, Aug 19 2015

A series created using 2*(a(n) + a(n+1)) + (a(n+1) + a(n+2)) has Hankel transform of F(2n), offset 3, F being a Fibonacci number, A001906 (Empirical observation). - Tony Foster III, Jul 30 2016

The series a(n) + A001006(n) has Hankel transform F(2n+1), offset n=1, F being the Fibonacci bisection A001519 (empirical observation). - Tony Foster III, Sep 05 2016

The Rubey and Stump reference proves a refinement of a conjecture of René Marczinzik, which they state as: "The number of 2-Gorenstein algebras which are Nakayama algebras with n simple modules and have an oriented line as associated quiver equals the number of Motzkin paths of length n. Moreover, the number of such algebras having the double centraliser property with respect to a minimal faithful projective-injective module equals the number of Riordan paths, that is, Motzkin paths without level-steps at height zero, of length n." - Eric M. Schmidt, Dec 16 2017

A connection to the Thue-Morse sequence: (-1)^a(n) = (-1)^A010060(n) * (-1)^A010060(n+1) = A106400(n) * A106400(n+1). - Vladimir Reshetnikov, Jul 21 2019

Named by Bernhart (1999) after the American mathematician John Riordan (1903-1988). - Amiram Eldar, Apr 15 2021


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a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*A000108(k). a(n) = (1/(n+1)) * Sum_{k=0..ceiling(n/2)} binomial(n+1, k)*binomial(n-k-1, k-1), for n > 1. - Len Smiley. [Comment from Amitai Regev (amitai.regev(AT)weizmann.ac.il), Mar 02 2010: the latter sum should be over the range k=1..floor(n/2).]

G.f.: (1 + x - sqrt(1-2*x-3*x^2))/(2*x*(1+x)).

G.f.: 2/(1+x+sqrt(1-2*x-3*x^2)). - Paul Peart (ppeart(AT)fac.howard.edu), May 27 2000

a(n+1) + (-1)^n = a(0)*a(n) + a(1)*a(n-1) + ... + a(n)*a(0). - Bernhart

a(n) = (1/(n+1)) * Sum_{i} (-1)^i*binomial(n+1, i)*binomial(2*n-2*i, n-i). - Bernhart

G.f. A(x) satisfies A = 1/(1+x) + x*A^2.

E.g.f.: exp(x)*(BesselI(0, 2*x) - BesselI(1, 2*x)). - Vladeta Jovovic, Apr 28 2003

a(n) = A001006(n-1) - a(n-1).

a(n+1) = Sum_{k=0..n} (-1)^k*A026300(n, k), where A026300 is the Motzkin triangle.

a(n) = Sum_{k=0..n} (-1)^k*binomial(n, k)*binomial(k, floor(k/2)). - Paul Barry, Jan 27 2005

a(n) = Sum_{k>=0} A086810(n-k, k). - Philippe Deléham, May 30 2005

a(n+2) = Sum_{k>=0} A064189(n-k, k). - Philippe Deléham, May 31 2005

Moment representation: a(n) = (1/(2*Pi))*Int(x^n*sqrt((1+x)(3-x))/(1+x),x,-1,3). - Paul Barry, Jul 09 2006

Inverse binomial transform of A000108 (Catalan numbers). - Philippe Deléham, Oct 20 2006

a(n) = (2/Pi)* Integral_{t_0..Pi} (4*cos^2(x)-1)^n*sin^2(x) dx. - Andrew V. Sutherland, Dec 02 2007

G.f.: 1/(1-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-.....(continued fraction). - Paul Barry, Jan 22 2009

G.f.: 1/(1+x-x/(1-x/(1+x-x/(1-x/(1+x-x/(1-... (continued fraction). - Paul Barry, May 16 2009

G.f.: 1/(1-x^2/(1-x/(1-x/(1-x^2/(1-x/(1-x/(1-x^2/(1-x/(1-... (continued fraction). - Paul Barry, Mar 02 2010

a(n) = -(-1)^n * hypergeom([1/2, n+2],[2],4/3) / sqrt(-3). - Mark van Hoeij, Jul 02 2010

a(n) = (-1)^n*hypergeometric([-n,1/2],[2],4). - Peter Luschny, Aug 15 2012

Let A(x) be the g.f., then x*A(x) is the reversion of x/(1 + x^2*Sum_{k>=0} x^k); see A215340 for the correspondence to Dyck paths without length-1 ascents. - Joerg Arndt, Aug 19 2012 and Apr 16 2013

a(n) ~ 3^(n+3/2)/(8*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 02 2012

G.f.: 2/(1+x+1/G(0)), where G(k)= 1 + x*(2+3*x)*(4*k+1)/( 4*k+2 - x*(2+3*x)*(4*k+2)*(4*k+3)/(x*(2+3*x)*(4*k+3) + 4*(k+1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 05 2013

D-finite (an alternative): (n+1)*a(n) = 3*(n-2)*a(n-3) + (5*n-7)*a(n-2) + (n-2)*a(n-1), n >= 3. - Fung Lam, Mar 22 2014

Asymptotics: a(n) = 3^(n+2)/sqrt(3*n*Pi)/(8*n)*(1-21/(16*n) + O(1/n^2)) (with contribution by Vaclav Kotesovec). - Fung Lam, Mar 22 2014

a(n) = T(2*n-1,n)/n, where T(n,k) = triangle of A180177. - Vladimir Kruchinin, Sep 23 2014

a(n) = (-1)^n*JacobiP(n,1,-n-3/2,-7)/(n+1). - Peter Luschny, Sep 23 2014

a(n) = Sum_{k=0..n} C(n,k)*(C(k,n-k)-C(k,n-k-1)). - Peter Luschny, Oct 01 2014

a(n) = A002426(n) - A005717(n), n > 0. - Mikhail Kurkov, Feb 24 2019

a(n) = A309303(n) + A309303(n+1). - Vladimir Reshetnikov, Jul 22 2019

From Peter Bala, Feb 11 2022: (Start)

a(n) = A005773(n+1) - 2*A005717(n) for n >= 1.

Conjectures: for n >= 1, n divides a(2*n+1) and 2*n-1 divides a(2*n). (End)


a(5)=6 because the only dissections of a polygon with a total number of 6 edges are: five pentagons with one of the five diagonals and the hexagon with no diagonals.

G.f. = 1 + x^2 + x^3 + 3*x^4 + 6*x^5 + 15*x^6 + 36*x^7 + 91*x^8 + 232*x^9 + ...

From Gus Wiseman, Nov 15 2022: (Start)

The a(0) = 1 through a(6) = 15 lone-child-avoiding (no vertices of outdegree 1) ordered rooted trees with n + 1 vertices (ranked by A358376):

o . (oo) (ooo) (oooo) (ooooo) (oooooo)

((oo)o) ((oo)oo) ((oo)ooo)

(o(oo)) ((ooo)o) ((ooo)oo)

(o(oo)o) ((oooo)o)

(o(ooo)) (o(oo)oo)

(oo(oo)) (o(ooo)o)












A005043 := proc(n) option remember; if n <= 1 then 1-n else (n-1)*(2*A005043(n-1)+3*A005043(n-2))/(n+1); fi; end;

Order := 20: solve(series((x-x^2)/(1-x+x^2), x)=y, x); # outputs g.f.


a[0]=1; a[1]=0; a[n_]:= a[n] = (n-1)*(2*a[n-1] + 3*a[n-2])/(n+1); Table[ a[n], {n, 0, 30}] (* Robert G. Wilson v, Jun 14 2005 *)

Table[(-3)^(1/2)/6 * (-1)^n*(3*Hypergeometric2F1[1/2, n+1, 1, 4/3]+ Hypergeometric2F1[1/2, n+2, 1, 4/3]), {n, 0, 32}] (* cf. Mark van Hoeij in A001006 *) (* Wouter Meeussen, Jan 23 2010 *)

RecurrenceTable[{a[0]==1, a[1]==0, a[n]==(n-1) (2a[n-1]+3a[n-2])/(n+1)}, a, {n, 30}] (* Harvey P. Dale, Sep 27 2013 *)

a[ n_]:= SeriesCoefficient[2/(1+x +Sqrt[1-2x-3x^2]), {x, 0, n}]; (* Michael Somos, Aug 21 2014 *)

a[ n_]:= If[n<0, 0, 3^(n+3/2) Hypergeometric2F1[3/2, n+2, 2, 4]/I]; (* Michael Somos, Aug 21 2014 *)

Table[3^(n+3/2) CatalanNumber[n] (4(5+2n)Hypergeometric2F1[3/2, 3/2, 1/2-n, 1/4] -9 Hypergeometric2F1[3/2, 5/2, 1/2 -n, 1/4])/(4^(n+3) (n+1)), {n, 0, 31}] (* Vladimir Reshetnikov, Jul 21 2019 *)

Table[Sqrt[27]/8 (3/4)^n CatalanNumber[n] Hypergeometric2F1[1/2, 3/2, 1/2 - n, 1/4], {n, 0, 31}] (* Jan Mangaldan, Sep 12 2021 *)


(PARI) {a(n) = if( n<0, 0, n++; polcoeff( serreverse( (x - x^3) / (1 + x^3) + x * O(x^n)), n))}; /* Michael Somos, May 31 2005 */

(Maxima) a[0]:1$



makelist(a[n], n, 0, 12); /* Emanuele Munarini, Mar 02 2011 */


a005043 n = a005043_list !! n

a005043_list = 1 : 0 : zipWith div

(zipWith (*) [1..] (zipWith (+)

(map (* 2) $ tail a005043_list) (map (* 3) a005043_list))) [3..]

-- Reinhard Zumkeller, Jan 31 2012

(PARI) N=66; Vec(serreverse(x/(1+x*sum(k=1, N, x^k))+O(x^N))) \\ Joerg Arndt, Aug 19 2012


A005043 = lambda n: (-1)^n*jacobi_P(n, 1, -n-3/2, -7)/(n+1)

[simplify(A005043(n)) for n in (0..29)]

# Peter Luschny, Sep 23 2014


def ms():

a, b, c, d, n = 0, 1, 1, -1, 1

yield 1

while True:

yield -b + (-1)^n*d

n += 1

a, b = b, (3*(n-1)*n*a+(2*n-1)*n*b)/((n+1)*(n-1))

c, d = d, (3*(n-1)*c-(2*n-1)*d)/n

A005043 = ms()

print([next(A005043) for _ in range(32)]) # Peter Luschny, May 16 2016


from functools import cache


def A005043(n: int) -> int:

if n <= 1: return 1 - n

return (n - 1) * (2 * A005043(n - 1) + 3 * A005043(n - 2)) // (n + 1)

print([A005043(n) for n in range(32)]) # Peter Luschny, Nov 20 2022


Row sums of triangle A020474, first differences of A082395.

First diagonal of triangular array in A059346.

Binomial transform of A126930. - Philippe Deléham, Nov 26 2009

The Hankel transform of a(n+1) is A128834. The Hankel transform of a(n+2) is floor((2*n+4)/3) = A004523(n+2). - Paul Barry, Mar 08 2011

The Kn11 triangle sums of triangle A175136 lead to A005043(n+2), while the Kn12(n) = A005043(n+4)-2^(n+1), Kn13(n) = A005043(n+6)-(n^2+9*n+56)*2^(n-2) and the Kn4(n) = A005043(2*n+2) = A099251(n+1) triangle sums are related to the sequence given above. For the definitions of these triangle sums see A180662. - Johannes W. Meijer, May 06 2011

Cf. A187306 (self-convolution), A348210 (column 1).

Cf. A000045, A000108, A000957, A001006, A005717, A005773, A007317, A057078, A091867, A104597, A126120, A178514, A249548, A309303.

Bisections: A099251, A099252.

Sequence in context: A052827 A033192 A174297 * A099323 A342912 A058534

Adjacent sequences: A005040 A005041 A005042 * A005044 A005045 A005046




N. J. A. Sloane


Thanks to Laura L. M. Yang (yanglm(AT)hotmail.com) for a correction, Aug 29 2004

Name changed to Riordan numbers following a suggestion from Ira M. Gessel. - N. J. A. Sloane, Jul 24 2020



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