A218225 G.f. A(x) satisfies: (1 - x*A(x)) / (1 - x^2*A(x)^2)^2 = 1 - x.

1, 2, 6, 23, 101, 480, 2400, 12434, 66142, 359112, 1981904, 11085198, 62696874, 357970472, 2060459256, 11943445311, 69656978837, 408466559630, 2406825745010, 14243262687023, 84618295006269, 504485687485408, 3017344000161296, 18099717207764928

Offset: 0

Author

Paul D. Hanna, Oct 23 2012

Keywords

• nonn

Comments

Binomial transform of A001002. - Vladimir Kruchinin, Oct 03 2014

Conjecture: a(n) is the number of permutations of [1..n+1] that avoid one of the following sets of patterns: (2134, 42153, 24153), (3124, 42153, 24153), (2143, 42135, 24135). - Alexander Burstein, Dec 20 2017

From Alexander Burstein, Sep 29 2023: (Start)

Chern et al. (see links) proves the above conjecture for (3124, 42153, 24153) as well as the following conjecture of Hong and Li:

a(n) is the number of inversion sequences of length n+1 avoiding pattern 0021. (End)

Examples

			G.f.: A(x) = 1 + 2*x + 6*x^2 + 23*x^3 + 101*x^4 + 480*x^5 + 2400*x^6 + ...
The series reversion of x*A(x) begins:
x - 2*x^2 + 2*x^3 - 3*x^4 + 3*x^5 - 4*x^6 + 4*x^7 - 5*x^8 + 5*x^9 + ...
so A(1 - (1-x)/(1-x^2)^2) = x + 1/(1-x-x^2).
The logarithm of the g.f. equals the series:
log(A(x)) = (1 + (1-x)*A(x))*x +
(1 + 2^2*(1-x)*A(x) + (1-x)^2*A(x)^2)*x^2/2 +
(1 + 3^2*(1-x)*A(x) + 3^2*(1-x)^2*A(x)^2 + (1-x)^3*A(x)^3)*x^3/3 +
(1 + 4^2*(1-x)*A(x) + 6^2*(1-x)^2*A(x)^2 + 4^2*(1-x)^3*A(x)^3 + (1-x)^4*A(x)^4)*x^4/4 +
(1 + 5^2*(1-x)*A(x) + 10^2*(1-x)^2*A(x)^2 + 10^2*(1-x)^3*A(x)^3 + 5^2*(1-x)^4*A(x)^4 + (1-x)^5*A(x)^5)*x^5/5 + ...
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200
• Paul Barry, Riordan arrays, generalized Narayana triangles, and series reversion, Linear Algebra and its Applications, 491 (2016) 343-385.
• Shane Chern, Shishuo Fu, and Zhicong Lin, Burstein's permutation conjecture, Hong and Li's inversion sequence conjecture, and restricted Eulerian distributions, arXiv:2209.12137 [math.CO], 2022.
• Letong Hong and Rupert Li, Length-Four Pattern Avoidance in Inversion Sequences, arXiv:2112.15081 [math.CO], 2021.

Programs

Mathematica

Table[1/(n+1)*SeriesCoefficient[(((x-1)*(x+1)^2)/(x^2+x-1))^(n+1),{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, May 22 2013 *)
Flatten[{1,Table[FullSimplify[SeriesCoefficient[(2*(1-x)^(1/3)-2^(2/3)*(-11-16*x-3*Sqrt[-15+96*x])^(1/3)+2^(2/3)*(11+16*x-3*Sqrt[-15+96*x])^(1/3))/(6*(-1+x)^(1/3)*x),{x,0,n}]],{n,1,10}]}] (* Vaclav Kotesovec, Jul 06 2013 *)
CoefficientList[Series[(-1+Cos[2/3*(ArcCot[3*Sqrt[3/5]]-ArcCot[(3*Sqrt[3])/Sqrt[5-32*x]])]+Sqrt[15]*Sin[2/3*(ArcCot[3*Sqrt[3/5]]-ArcCot[(3*Sqrt[3])/Sqrt[5-32*x]])])/(3*x), {x, 0, 20}], x] (* Vaclav Kotesovec, Jul 06 2013 *)

PARI

{a(n)=polcoeff((1/x)*serreverse(x*(1-x-x^2)/((1-x)*(1+x)^2 +x*O(x^n))),n)}
for(n=0,30,print1(a(n),", "))

PARI

{a(n)=local(A=1);for(i=1,n,A=exp(sum(m=1,n,x^m/m*sum(k=0,m,binomial(m,k)^2*(1-x)^k*A^k)+x*O(x^n))));polcoeff(A,n)}
for(n=0,20,print1(a(n),", "))

Formula

G.f. A(x) satisfies:

(1) A(1 - (1-x)/(1-x^2)^2) = x + 1/(1-x-x^2).

(2) A(x) = (1/x) * Series_Reversion( x*(1-x-x^2)/((1-x)*(1+x)^2) ).

(3) A(x) = (1 - x*A(x)) * (1 + x*A(x))^2 / (1 - x*A(x) - x^2*A(x)^2).

(4) A(x) = exp( Sum_{n>=1} (x^n/n) * Sum_{k=0..n} binomial(n,k)^2 * (1-x)^k * A(x)^k ).

Recurrence: 5*n*(n+1)*a(n) = 21*n*(2*n-1)*a(n-1) - 3*(23*n^2-46*n+24)*a(n-2) + 16*(n-2)*(2*n-3)*a(n-3). - Vaclav Kotesovec, May 22 2013

a(n) ~ 2^(5*n+6)/(27*sqrt(Pi)*5^(n+1/2)*n^(3/2)). - Vaclav Kotesovec, May 22 2013

G.f.: (-1+cos(2/3*(arccot(3*sqrt(3/5))-arccot((3*sqrt(3))/sqrt(5-32*x))))+sqrt(15)*sin(2/3*(arccot(3*sqrt(3/5))-arccot((3*sqrt(3))/sqrt(5-32*x)))))/(3*x). - Vaclav Kotesovec, Jul 06 2013