A054391 Number of permutations with certain forbidden subsequences.

1, 1, 2, 5, 14, 41, 123, 374, 1147, 3538, 10958, 34042, 105997, 330632, 1032781, 3229714, 10109310, 31667245, 99260192, 311294876, 976709394, 3065676758, 9625674442, 30231524869, 94972205349, 298419158008, 937861780439, 2947969125284, 9267666915326

Offset: 0

Author

N. J. A. Sloane, Elisa Pergola (elisa(AT)dsi.unifi.it), May 21 2000

Keywords

• nonn

Comments

Hankel transform is [1,1,1,...] = A000012. - Paul Barry, Jan 19 2009

The inverse Motzkin transform apparently yields 1 followed by A000930, which implies a generating function g(x)=1+z/(1-z-z^3) where z=x*A001006(x). - R. J. Mathar, Jul 07 2009

It appears that the infinite set of interpolated sequences between the Motzkin and the Catalan can be generated with a succession of INVERT transforms, given each sequence has two leading 1's. Also, the N-th sequence in the set starting with (N=1, A001006) can be generated from a production matrix of the form "M" in the formula section, such that the main diagonal is (N leading 1's, 0, 0, 0, ...). M with a diagonal of (1, 0, 0, 0, ...) generates A001006, while M with a main diagonal of all 1's is the production matrix for A000108. - Gary W. Adamson, Jul 29 2011

From Gus Wiseman, Jun 22 2019: (Start)

Conjecture: Also the number of non-capturing set partitions of {1..n}. A set partition is capturing if it has two blocks of the form {...x...y...} and {...z...t...} where x < z and y > t or x > z and y < t. This is a weaker condition than nesting, so for example {{1,3,5},{2,4}} is capturing but not nesting. The a(0) = 1 through a(4) = 14 non-capturing set partitions are:

{} {{1}} {{1,2}} {{1,2,3}} {{1,2,3,4}}

{{1},{2}} {{1},{2,3}} {{1},{2,3,4}}

{{1,2},{3}} {{1,2},{3,4}}

{{1,3},{2}} {{1,2,3},{4}}

{{1},{2},{3}} {{1,2,4},{3}}

{{1,3},{2,4}}

{{1,3,4},{2}}

{{1},{2},{3,4}}

{{1},{2,3},{4}}

{{1,2},{3},{4}}

{{1},{2,4},{3}}

{{1,3},{2},{4}}

{{1,4},{2},{3}}

{{1},{2},{3},{4}}

Cf. A000108, A000110, A058681, A326212, A326237, A326243, A326244, A326249, A326255.

(End)

The above conjecture is true: A partition is non-capturing iff its representation in canonical sequential form avoids the patterns 1221 and 2112. In the context of these partition representations, the pattern 2112 is equivalent to the pattern 12112. Partitions whose canonical sequence form avoid 1221 and 12112 are one of the classes that are handled in the Mansour/Shattuck "Pattern Avoiding Partitions,..." paper. It shows that they are counted by this sequence. - Christian Sievers, Oct 29 2024

Examples

			a(4) = 14, a(5) = 41 since the top row of M^4 = (14, 14, 9, 3, 1), with 41 = (14 + 14 + 9 + 3 + 1).
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• E. Barcucci et al., From Motzkin to Catalan Permutations, Discr. Math., 217 (2000), 33-49.
• Jean-Luc Baril and Sergey Kirgizov, Bijections from Dyck and Motzkin meanders with catastrophes to pattern avoiding Dyck paths, arXiv:2104.01186 [math.CO], 2021.
• Paul Barry, On sequences with {-1, 0, 1} Hankel transforms, arXiv preprint arXiv:1205.2565 [math.CO], 2012. - From _N. J. A. Sloane_, Oct 18 2012
• Petr Gregor, Torsten Mütze, and Namrata, Combinatorial generation via permutation languages. VI. Binary trees, arXiv:2306.08420 [cs.DM], 2023.
• Nickolas Hein and Jia Huang, Variations of the Catalan numbers from some nonassociative binary operations, arXiv:1807.04623 [math.CO], 2018.
• J. W. Layman, The Hankel Transform and Some of its Properties, J. Integer Sequences, 4 (2001), #01.1.5
• Toufik Mansour and Mark Shattuck, Pattern Avoiding Partitions, Sequence A054391 and the Kernel Method, Applications and Applied Mathematics, Vol. 6, Issue 2 (December 2011), pp. 397-411.
• T. Mansour and M. Shattuck, Restricted partitions and generalized Catalan numbers, PU. M. A., Vol. (2011), No. 2, pp. 239-251. - From _N. J. A. Sloane_, Oct 13 2012
• Eric Marberg, Crossings and nestings in colored set partitions, arXiv preprint arXiv:1203.5738 [math.CO], 2012.

Crossrefs

Interpolates between Motzkin numbers (A001006) and Catalan numbers (A000108). Cf. A005773, A054392, ...

Binomial transform of A224747.

Programs

Maple

c := x->(1-sqrt(1-4*x))/(2*x); a := (x,j)->(x)/((1-4*x)*(c(x))^2*(1-c(x))^(j))*(-x^2*(c(x))^2*(1-c(x))*(x^2*(c(x))^4)^(j)-(1-3*x-2*x^2)*(c(x))^2*(x*(c(x))^2)^(j)+x);
b := (x,j)->1+(1)/((1-4*x)*c(x)*(1-c(x))^(j))*(-2*x^3*(c(x))^2*(x^2*(c(x))^4)^(j)+(1-3*x-2*x^2)*c(x)*(x*(c(x))^2)^(j)-2*x^2);
co := (x,j)->(1)/((1-4*x)*(1-c(x))^(j))*(x^2*(x^2*(c(x))^4)^(j)-(1-3*x-2*x^2)*(x*(c(x))^2)^(j)+x^2);
s := (x,j)->(1-b(x,j)+(-1)^j*sqrt((1-b(x,j))^2-4*a(x,j)*co(x,j)))/(2*a(x,j)); j := 3; series(s(x,j),x=0..60); od; # j=1,2,3,... inf gives A001006, A005773, A054391, A054392, ..., A000108

Mathematica

CoefficientList[Series[1 - 2*x^2/(2*x^2 - 3*x + 1 - Sqrt[1 - 2*x - 3*x^2]), {x, 0, 50}], x] (* G. C. Greubel, Apr 27 2017 *)

Maxima

a(n):=sum((sum((binomial(k,l)*l*sum(binomial(j,1-n-2*l+k+2*j)*binomial(n-1+l-k,j),j,0,n+l-k-1))/(n+l-k-1),l,1,k)),k,1,n-1)+1; /* Vladimir Kruchinin, Oct 31 2011 */

PARI

x='x+O('x^66); gf=1-2*x^2/(2*x^2-3*x+1-sqrt(1-2*x-3*x^2)); Vec(gf) \\ Joerg Arndt, Jun 29 2013

Formula

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

G.f.: 1/(1-x-x^2/(1-2x-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-... (continued fraction) (conjecture). - Paul Barry, Jan 19 2009

From Gary W. Adamson, Jul 29 2011: (Start)

a(n) = upper left term of M^n, a(n+1) = sum of top row terms of M^n; M = an infinite square production matrix as follows with a main diagonal of (1, 1, 1, 0, 0, 0, ...):

1, 1, 0, 0, 0, 0, ...

1, 1, 1, 0, 0, 0, ...

1, 1, 1, 1, 0, 0, ...

1, 1, 1, 0, 1, 0, ...

1, 1, 1, 1, 0, 1, ...

1, 1, 1, 1, 1, 0, ...

... (End)

a(n) = Sum_{k=1..n-1} (sum(l=1..k, (binomial(k,l)*l*sum(j=0..n+l-k-1, binomial(j,1-n-2*l+k+2*j)*binomial(n-1+l-k,j)))/(n+l-k-1))) + 1. - Vladimir Kruchinin, Oct 31 2011

D-finite with recurrence (-n+1)*a(n) + 3*(2*n-3)*a(n-1) + (-8*n+11)*a(n-2) + (-5*n+32)*a(n-3) + (7*n-31)*a(n-4) + 3*(-n+4)*a(n-5)= 0. - R. J. Mathar, Nov 26 2012

G.f.: 1 - x*(2*x^2-3*x+1 + 1/G(0))/(2*(x^3-3*x^2+4*x-1)), 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, Jun 29 2013