A279565 Number of length n inversion sequences avoiding the patterns 100, 110, 120, 201, and 210.
1, 1, 2, 6, 21, 81, 332, 1420, 6266, 28318, 130412, 609808, 2887582, 13818590, 66726628, 324713196, 1590853485, 7840315329, 38843186366, 193342353214, 966409013021, 4848846341569, 24412146213116, 123290812268404, 624448756434476, 3171046361310556
Offset: 0
Keywords
Examples
The length 4 inversion sequences avoiding (100, 110, 120, 201, 210) are 0000, 0001, 0002, 0003, 0010, 0011, 0012, 0013, 0020, 0021, 0022, 0023, 0101, 0102, 0103, 0111, 0112, 0113, 0121, 0122, 0123.
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..1372
- Megan A. Martinez and Carla D. Savage, Patterns in Inversion Sequences II: Inversion Sequences Avoiding Triples of Relations, arXiv:1609.08106 [math.CO], 2016.
Crossrefs
Programs
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Magma
I:=[6, 21, 81]; [1,1,2] cat [n le 3 select I[n] else ( (n+1)*(17*n+6)*Self(n-1) +(49*n^2+11*n+22)*Self(n-2) +3*(3*n-1)*(3*n-2)*Self(n-3) )/(5*(n+2)*(n+1)) : n in [1..30]]; // G. C. Greubel, Mar 29 2019
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Maple
a:= proc(n) option remember; `if`(n<3, n!, ((n-1)*(17*n-28)*a(n-1) +(49*n^2-185*n+196)*a(n-2) +(3*(3*n-7))*(3*n-8)*a(n-3)) / (5*n*(n-1))) end: seq(a(n), n=0..30); # Alois P. Heinz, Feb 22 2017
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Mathematica
a[n_] := a[n] = If[n < 3, n!, (((n - 1)*(17*n - 28)*a[n-1] + (49*n^2 - 185*n + 196)*a[n-2] + (3*(3*n - 7))*(3*n - 8)*a[n-3]) / (5*n*(n - 1)))]; Array[a, 30, 0] (* Jean-François Alcover, Nov 06 2017, after Alois P. Heinz *) Join[{1}, Table[(1/n)*Sum[m*Sum[Binomial[k, n-m-k]*Binomial[n+k-1, k], {k, 0, n-m}], {m, 1, n}], {n, 1, 30}]] (* G. C. Greubel, Mar 29 2019 *)
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Maxima
a(n):=if n=0 then 1 else sum(m*sum(binomial(k,n-m-k)*binomial(n+k-1,k),k,0,n-m),m,1,n)/n; /* Vladimir Kruchinin, Mar 26 2019 */
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PARI
my(x='x+O('x^30)); Vec(round(3/(4-4*sin(asin((27*x+11)/16)/3)))) \\ G. C. Greubel, Mar 29 2019
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Sage
[1] +[(1/n)*(sum(sum(k*binomial(j,n-k-j)*binomial(n+j-1,j) for j in (0..n-k)) for k in (1..n))) for n in (1..30)] # G. C. Greubel, Mar 29 2019
Formula
G.f.: 3/(4-4*sin(asin((27*x+11)/16)/3)). - Vladimir Kruchinin, Mar 25 2019
a(n) = (1/n)*Sum_{m=1..n} m*Sum_{k=0..n-m} C(k,n-m-k)*C(n+k-1,k), n>0, a(0)=1. - Vladimir Kruchinin, Mar 26 2019
a(n) ~ 3^(3*n + 1/2) / (2^(7/2) * sqrt(Pi) * n^(3/2) * 5^(n - 1/2)). - Vaclav Kotesovec, Oct 07 2021
Conjecture: a(n) = (v_n + v_{n+1})/2 for n > 0 with a(0) = 1 where we start with vector v of fixed length m with elements v_i = 1 and for i=1..m-2, for j=i+2..m apply v_j := Sum_{k=0..2} v_{j-k}. - Mikhail Kurkov, Sep 03 2024
Extensions
a(10)-a(25) from Alois P. Heinz, Feb 22 2017
Comments