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a(n) = binomial(2*n,n)*(2*n+1)/2+n*binomial(2*n-2,n)+(n-1)*binomial(2*n-2,n+1).
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%I #23 Jan 17 2024 10:34:55

%S 3,17,84,393,1778,7866,34254,147433,628914,2663934,11219728,47033322,

%T 196393044,817338580,3391858530,14040986985,57998364690,239112756630,

%U 984126777480,4044255577230,16597080112860,68027923573740

%N a(n) = binomial(2*n,n)*(2*n+1)/2+n*binomial(2*n-2,n)+(n-1)*binomial(2*n-2,n+1).

%C Conjecture: These are the number of linear intervals in the Cambrian lattices of type B_n. An interval is linear if it is isomorphic to a total order. The conjecture has been checked up to the term 34254 for n = 7.

%H Michael De Vlieger, <a href="/A344228/b344228.txt">Table of n, a(n) for n = 1..1658</a>

%H Clément Chenevière, <a href="https://theses.hal.science/tel-04255439">Enumerative study of intervals in lattices of Tamari type</a>, Ph. D. thesis, Univ. Strasbourg (France), Ruhr-Univ. Bochum (Germany), HAL tel-04255439 [math.CO], 2024. See p. 151.

%F From _Peter Luschny_, May 12 2021: (Start)

%F a(n) = 3*(2*n^3 + n - 1)*2^(2*n - 2)*binomial(n - 3/2, -1/2)/((n + 1)*n).

%F a(n) = [x^n] (15*x - 24*x^2 + 8*x^3 - 2 + (1 - 4*x)^(3/2)*(2 - 3*x))/(2*(1 - 4*x)^(3/2)*x).

%F a(n) ~ 4^(n-2)*(24*n - 15)/sqrt(Pi*n). (End)

%F a(n) = a(n-1)*2*(2*n - 3)*(2*n^3 + n - 1)/((n + 1)*(2*n^3 - 6*n^2 + 7*n - 4)) for n > 1. - _Chai Wah Wu_, May 13 2021

%e For B_2, among the 18 intervals in the hexagon-shaped lattice, only one is not linear.

%p a := n -> 3*(2*n^3 + n - 1)*2^(2*n - 2)*binomial(n - 3/2, -1/2)/((n + 1)*n):

%p seq(a(n), n = 1..22); # _Peter Luschny_, May 12 2021

%t Array[3 (2 #^3 + # - 1)*2^(2 # - 2)*Binomial[# - 3/2, -1/2]/(# (# + 1)) &, 22] (* _Michael De Vlieger_, Jan 17 2024 *)

%o (Sage)

%o def a(n):

%o return binomial(2*n,n)*(2*n+1)/2+n*binomial(2*n-2,n)+(n-1)*binomial(2*n-2,n+1)

%Y Cf. A344136 for the type A, A344191 for a similar sequence.

%K nonn

%O 1,1

%A _F. Chapoton_, May 12 2021