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A263777
Number of inversion sequences avoiding pattern 201 (or 210).
35
1, 1, 2, 6, 24, 118, 674, 4306, 29990, 223668, 1763468, 14558588, 124938648, 1108243002, 10115202962, 94652608690, 905339525594, 8829466579404, 87618933380020, 883153699606024, 9028070631668540, 93478132393544988, 979246950529815364, 10368459385853924212
OFFSET
0,3
LINKS
Lars Blomberg and Gheorghe Coserea, Table of n, a(n) for n = 0..777, terms 1..100 from Lars Blomberg.
Juan S. Auli and Sergi Elizalde, Wilf equivalences between vincular patterns in inversion sequences, arXiv:2003.11533 [math.CO], 2020.
Sylvie Corteel, Megan A. Martinez, Carla D. Savage, and Michael Weselcouch, Patterns in Inversion Sequences I, arXiv:1510.05434 [math.CO], 2015. See equations (4,5).
Megan A. Martinez and Carla D. Savage, Patterns in Inversion Sequences II: Inversion Sequences Avoiding Triples of Relations, arXiv:1609.08106 [math.CO], 2016.
Jay Pantone, The enumeration of inversion sequences avoiding the patterns 201 and 210, arXiv:2310.19632 [math.CO], 2023.
Benjamin Testart, Completing the enumeration of inversion sequences avoiding one or two patterns of length 3, arXiv:2407.07701 [math.CO], 2024. See p. 2.
FORMULA
a(n) = Sum_{k=0..n-1} Sum_{p=-1,k-1} T(n,k,p), where T(n,k,p) = Sum_{i=-1..p} T(n-1,k,i) + Sum_{j=p+1..k} T(n-1,j,p) with initial conditions T(n,k,p) = 0 if k >= n and T(n,k,-1) = (n-k)/n * binomial(n-1+k,k). (eqn. (4) and (5) in Corteel link) - Gheorghe Coserea, Sep 21 2017
a(n) ~ c * (27/2)^n / n^alfa, where alfa = 5.7667921227... and c = 9.973... - Vaclav Kotesovec, Oct 16 2021
MATHEMATICA
T[n_, k_, _] /; k >= n = 0; T[n_, k_, -1] := (n-k)/n*Binomial[n+k-1, k];
T[n_, k_, p_] := T[n, k, p] = Sum[T[n-1, k, i], {i, -1, p}] + Sum[T[n-1, j, p], {j, p+1, k}];
a[0] = 1; a[n_] := Sum[T[n, k, p], {k, 0, n-1}, {p, -1, k-1}];
Table[a[n], {n, 0, 23}] (* Jean-François Alcover, Aug 10 2018 *)
PROG
(PARI)
seq(N) = {
my(a=vector(N), t=vector(2, k, matrix(N, N)), s=matrix(N+1, N+1),
C=(n, k)->(n-k)/n*binomial(n-1+k, k));
for (n=1, N, for (k=1, n, for(p=1, k-1,
s[k+1, p+1] = s[k+1, p] + t[1+n%2][k, p];
s[p+1, k+1] = s[p+1, k] + t[1+n%2][k, p];
t[1+(n+1)%2][k, p] = s[k+1, p+1] + s[p+1, k+1] + C(n-1, k-1)));
a[n] = sum(k=1, n, sum(p=1, k-1, t[1+(n+1)%2][k, p])) + C(n+1, n));
a;
};
concat(1, seq(23)) \\ Gheorghe Coserea, Nov 20 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Marcus, Oct 26 2015
EXTENSIONS
a(0)=1 prepended by Alois P. Heinz, Dec 15 2016
More terms from Lars Blomberg, Jan 18 2017
STATUS
approved