0,5

I'm not sure "Motzkin sums" is the best name for this, but it has been used in publications, and provides a quick way to refer to the sequence. 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 by U, DD by D, UD by a blue flatstep (F), DU by 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).

(End)

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

From Tom Copeland, Nov 02 2014: (Start)

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

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

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

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

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

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

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

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

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

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

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

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D. N. Verma, Towards Classifying Finite Point-Set Configurations, preprint, 1997. [Apparently unpublished]

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D. Callan, Pattern avoidance in "flattened" partitions, Discrete Math., 309 (2009), 4187-4191.

David Callan, Bijections for Dyck paths with all peak heights of the same parity, arXiv:1702.06150 [math.CO], 2017.

Xiang-Ke Chang, X.-B. Hu, H. Lei, Y.-N. Yeh, Combinatorial proofs of addition formulas, The Electronic Journal of Combinatorics, 23(1) (2016), #P1.8.

W. Y. C. Chen, E. Y. P. Deng and L. L. M. Yang, Riordan paths and derangements, arXiv:math/0602298 [math.CO], 2006.

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J. Menashe,  Bijections on Riordan objects [From Tom Copeland, Nov 07 2014]

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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

Recurrence (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(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 + ...

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 - 2 x - 3 x^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 *)

(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$

a[1]:0$

a[n]:=(n-1)*(2*a[n-1]+3*a[n-2])/(n+1)$

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

(Haskell)

a005043 n = a005043_list !! n

a005043_list = 1 : 0 : zipWith div

   (zipWith (*) [1..] (zipWith (+)

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

-- Reinhard Zumkeller, Jan 31 2012

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

(Sage)

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

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

# Peter Luschny, Sep 23 2014

(Sage)

def ms():

    a, b, c, d, n = 0, 1, 1, -1, 1

    yield 1

    while True:

        yield -b + (-1)^n*d

        n += 1

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

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

A005043 = ms()

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

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

The self-convolution of A005043 gives A187306. - Philippe Deléham, Jan 28 2014

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

Bisections: A099251, A099252.

Sequence in context: A033192 A174297 * A099323 A058534 A063778 A279374

Adjacent sequences:  A005040 A005041 A005042 * A005044 A005045 A005046

nonn,easy,nice,changed

N. J. A. Sloane

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

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