A228907 G.f. satisfies: A(x) = 1 + Sum_{n>=0} x^n * (1 - A(x)^(2*n))/(1 - A(x)).

1, 2, 6, 24, 114, 598, 3336, 19402, 116302, 713368, 4455650, 28240942, 181180912, 1174280146, 7677229718, 50570040088, 335289825874, 2235856077798, 14985808827416, 100900119437082, 682145490613118, 4628755102582328, 31514118237222850, 215214456560655070

Offset: 0

Author

Paul D. Hanna, Sep 08 2013

Keywords

• nonn

Comments

Conjectured to be the number of permutations of length n+1 avoiding the partially ordered pattern (POP) {1>4, 1>5, 2>4, 2>5, 5>3} of length 5. That is, conjectured to be the number of length n+1 permutations having no subsequences of length 5 in which the first and second elements are larger than the elements in positions 4 and 5, and the fifth element is larger than the element in position 3.- Sergey Kitaev, Dec 13 2020

This conjecture has been proven. It can be restated as the number of size n permutations avoiding 45123, 45132, 45213, 54123, 54132, 54213. - Christian Bean, Jul 24 2024

Examples

			G.f.: A(x) = 1 + 2*x + 6*x^2 + 24*x^3 + 114*x^4 + 598*x^5 + 3336*x^6 +...
where g.f. A = A(x) satisfies the equivalent expressions:
A = 1 + x*(1-A^2)/(1-A) + x^2*(1-A^4)/(1-A) + x^3*(1-A^6)/(1-A) +...
A = 1 + x*(1 + A) + x^2*(1 + A + A^2 + A^3) + x^3*(1 + A + A^2 + A^3 + A^4 + A^5) +...
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200
• Michael H. Albert, Christian Bean, Anders Claesson, Émile Nadeau, Jay Pantone, and Henning Ulfarsson, PermPAL database.
• Christian Bean, Émile Nadeau, Jay Pantone, and Henning Ulfarsson, Permutations avoiding bipartite partially ordered patterns have a regular insertion encoding, The Electronic Journal of Combinatorics, Volume 31, Issue 3 (2024); arXiv preprint, arXiv:2312.07716 [math.CO], 2023.
• Alice L. L. Gao and Sergey Kitaev, On partially ordered patterns of length 4 and 5 in permutations, arXiv:1903.08946 [math.CO], 2019.
• Alice L. L. Gao and Sergey Kitaev, On partially ordered patterns of length 4 and 5 in permutations, The Electronic Journal of Combinatorics 26(3) (2019), P3.26.

Crossrefs

Cf. A199548.

Programs

Mathematica

nmax=20;aa=ConstantArray[0,nmax];aa[[1]]=2;Do[AGF=1+Sum[aa[[n]]*x^n,{n,1,j-1}]+koef*x^j;sol=Solve[Coefficient[1+x*AGF*(2-AGF+AGF^2)+x^2*AGF^2*(1-AGF)-AGF,x,j]==0,koef][[1]];aa[[j]]=koef/.sol[[1]],{j,2,nmax}];Flatten[{1,aa}] (* Vaclav Kotesovec, Sep 09 2013 *)

PARI

{a(n) = my(A=1+x); for(i=1, n, A = 1 + sum(m=0, n, x^m*sum(k=0, 2*m-1, A^k) + x*O(x^n))); polcoeff(A, n)}
for(n=0,25,print1(a(n),", "))

PARI

{a(n) = my(A=1+x); for(i=1, n, A = 1+x*A*(2-A+A^2)+x^2*A^2*(1-A)+x*O(x^n)); polcoeff(A, n)}
for(n=0,25,print1(a(n),", "))

Formula

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

G.f.: (1/x)*Series_Reversion( x*(1+x+x^2 + sqrt(1-2*x-5*x^2-2*x^3+x^4)) / (2*(1+x)^2) ).

Recurrence: n*(2*n+1)*(7*n^2 - 35*n + 34)*a(n) = (126*n^4 - 742*n^3 + 1193*n^2 - 601*n + 78)*a(n-1) - (182*n^4 - 1253*n^3 + 2788*n^2 - 2293*n + 450)*a(n-2) + (56*n^4 - 434*n^3 + 1189*n^2 - 1273*n + 318)*a(n-3) + (n-4)*(2*n-5)*(7*n^2 - 21*n + 6)*a(n-4). - Vaclav Kotesovec, Sep 09 2013

a(n) ~ c*d^n/n^(3/2), where d = 8/3+14/3*cos(arctan(3*sqrt(3)/13)/3) = 7.29589694323977237... is the root of the equation 1 + 5*d - 8*d^2 + d^3 = 0 and c = sqrt(1/3 + sqrt(7)*cos((4*Pi + arccos(-1/(2*sqrt(7))))/3)/6) / sqrt(Pi) = 0.33910585091755684322274547... - Vaclav Kotesovec, Sep 09 2013, updated Mar 17 2024