A368881 a(n) = binomial(n+3, 4) + binomial(n+1, 3) + 1.

1, 2, 7, 20, 46, 91, 162, 267, 415, 616, 881, 1222, 1652, 2185, 2836, 3621, 4557, 5662, 6955, 8456, 10186, 12167, 14422, 16975, 19851, 23076, 26677, 30682, 35120, 40021, 45416, 51337, 57817, 64890, 72591, 80956, 90022, 99827, 110410, 121811, 134071

Offset

0,2

Comments

The number of bigrassmannian permutations in the type B hyperoctahedral group of order 2^n*n!, i.e., those with a unique left and right type B descent or the identity. This can be characterized by avoiding 18 signed permutation patterns.

Links

Harvey P. Dale, Table of n, a(n) for n = 0..1000

Joshua Swanson, Bigrassmannians and pattern avoidance notes.

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Formula

a(n) = (1/24)*(n^4 + 10*n^3 + 11*n^2 + 2*n + 24).

G.f.: (x^4 - 5x^3 + 7x^2 - 3x + 1)/(1-x)^5.

E.g.f.: exp(x)*(24 + 24*x + 48*x^2 + 16*x^3 + x^4)/24. - Stefano Spezia, Jan 09 2024

Example

For n=2, all eight 2 X 2 signed permutation matrices are bigrassmannian except the negative of the identity matrix, or equivalently the one with window notation [-1 -2], so a(2) = 7.

Mathematica

Table[Binomial[n + 3, 4] + Binomial[n + 1, 3] + 1, {n, 0, 20}]
LinearRecurrence[{5, -10, 10, -5, 1}, {1, 2, 7, 20, 46}, 50] (* Harvey P. Dale, Jan 21 2025 *)

Prog

(Python)
def A368881(n): return 1+(n*(n*(n*(n + 10) + 11) + 2))//24 # Chai Wah Wu, Jan 27 2024

Crossrefs

Cf. A050407.

It appears that this is equal to {A005712}+1, also ({A212039}+2)/3 .

Sequence in context: A259144 A090145 A244307 * A270109 A360421 A123203

Adjacent sequences: A368878 A368879 A368880 * A368882 A368883 A368884

Keyword

nonn,easy

Author

Joshua Swanson, Jan 08 2024

Status

approved