A290140 The number of maximal subsemigroups of the Jones monoid on the set [1..n].

1, 2, 5, 9, 13, 19, 27, 39, 57, 85, 129, 199, 311, 491, 781, 1249, 2005, 3227, 5203, 8399, 13569, 21933, 35465, 57359, 92783, 150099, 242837, 392889, 635677, 1028515, 1664139, 2692599, 4356681, 7049221, 11405841, 18454999, 29860775, 48315707, 78176413

Offset: 1

Author

James Mitchell and Wilf A. Wilson, Jul 21 2017

Keywords

• nonn
• easy

Comments

a(2n) is the number of maximal subsemigroups of the planar partition monoid of degree n.

Links

• Wilf A. Wilson, Table of n, a(n) for n = 1..1000
• James East, Jitender Kumar, James D. Mitchell, and Wilf A. Wilson, Maximal subsemigroups of finite transformation and partition monoids, arXiv:1706.04967 [math.GR], 2017.
• Index entries for linear recurrences with constant coefficients, signature (3,-2,-1,1).

Crossrefs

Cf. A000045.

Programs

Mathematica

{1, 2}~Join~Table[2 Fibonacci[n - 1] + 2 n - 3, {n, 3, 39}] (* Michael De Vlieger, Jul 21 2017 *)

PARI

Vec(x*(1 + x)*(1 - 2*x + 3*x^2 - 4*x^3 + x^4) / ((1 - x)^2*(1 - x - x^2)) + O(x^50)) \\ Colin Barker, Jul 21 2017

Formula

a(n) = 2 * A000045(n - 1) + 2n - 3, n > 2.

From Colin Barker, Jul 21 2017: (Start)

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

a(n) = -5 + (2^(-n)*((1-sqrt(5))^n*(1+sqrt(5)) + (-1+sqrt(5))*(1+sqrt(5))^n)) / sqrt(5) + 2*(1+n) for n>2.

a(n) = 3*a(n-1) - 2*a(n-2) - a(n-3) + a(n-4) for n>6.

(End)