A201631 a(n) is the number of Fibonacci meanders of length m*n and central angle 360/m degrees where m = 2.

1, 3, 6, 13, 30, 70, 167, 405, 992, 2450, 6090, 15214, 38165, 96069, 242530, 613811, 1556856, 3956316, 10070871, 25674210, 65541142, 167517654, 428635032, 1097874434, 2814611701, 7221917871, 18544968768, 47655572191, 122544150258, 315313433594, 811792614547

Offset: 1

Author

Peter Luschny, Jan 15 2012

Keywords

• nonn

Comments

Empirically the partial sums of A051291. - Sean A. Irvine, Jul 13 2022

The above conjecture was proved by Baril et al., which also give a formal definition of the Fibonacci meanders and describe a bijection with a certain class of peakless grand Motzkin paths of length n. - Peter Luschny, Mar 16 2023

Examples

			a(3) = 6 = card({100001, 100100, 110000, 111001, 111100, 111111}).
		

Links

• Jean-Luc Baril, Sergey Kirgizov, Rémi Maréchal, and Vincent Vajnovszki, Enumeration of Dyck paths with air pockets, arXiv:2202.06893 [cs.DM], 2022-2023.
• Peter Luschny, Fibonacci meanders.

Crossrefs

Cf. A110236, A110198, A202411, A203611, A051291, A358734.

Cf. A361574 (case m=3).

Programs

Maple

A201631 := n -> add(A202411(k),k=0..2*n-1): seq(A201631(i),i=1..9);
# Alternative, using the g.f. of Baril et al.:
S := (x^2 - x + 1 - R)/((x - 1)*(x^2 - x - 1 + R)*R):
R := (((x - 3)*x + 1)*(x^2 + x + 1))^(1/2): ser := series(S, x, 33):
seq(coeff(ser, x, n), n = 1..31); # Peter Luschny, Mar 16 2023
# Using a recurrence:
a := proc(n) option remember; if n < 5 then return [0, 1, 3, 6, 13][n + 1] fi;
(n*(2*n - 1)*(2*n - 3)*(n - 5)*a(n - 5) - (n - 4)*(2*n - 1)^2*(3*n - 5)*a(n - 4) + (2*n - 5)*(n - 3)*(2*n^2 - 3*n + 2)*a(n - 3) - (2*n - 3)*(n - 2)*(2*n^2 - 3*n + 5)*a(n - 2) + (3*n - 4)*(2*n - 1)*(2*n - 5)*(n - 1)*a(n - 1))/(n*(2*n - 3)*(2*n - 5)*(n - 1)) end: seq(a(n), n = 1..31); # Peter Luschny, Mar 16 2023

Mathematica

a[n_] := Sum[A202411[k], {k, 0, 2 n - 1}];
Array[a, 31] (* Jean-François Alcover, Jun 29 2019 *)

Formula

a(n) = Sum_{k=0..2n-1} A202411(k).

a(n) = [x^n] (x^2 - x + 1 - R)/((x - 1)*(x^2 - x - 1 + R) * R), where R = (((x - 3)*x + 1)*(x^2 + x + 1))^(1/2). (This is Theorem 21 in Baril et al.) - Peter Luschny, Mar 16 2023