OFFSET
2,3
COMMENTS
a(n) appears to be the number of 021-avoiding ascent sequences (A022493) with exactly one repeated nonzero entry, where repeated means two consecutive equal entries. For example, a(4) = 4 counts 0011, 0110, 0112, 0122, and a(5) = 14 counts 00011, 00110, 00112, 00122, 01011, 01022, 01100, 0110 1, 01102, 01120, 01123, 0122 0, 01223, 01233. - David Callan, Nov 21 2021
LINKS
M. Bousquet-Mélou and A. Rechnitzer, The site-perimeter of bargraphs, Adv. in Appl. Math. 31 (2003), 86-112.
Emeric Deutsch and S. Elizalde, Statistics on bargraphs viewed as cornerless Motzkin paths, arXiv preprint arXiv:1609.00088 [math.CO], 2016.
FORMULA
G.f.: g = (1 - 2z - z^2 - Q)/(2Q), where Q = sqrt(1 - 4z + 2z^2 + z^4).
a(n) = Sum_{k>0} k*A273713(n,k).
From Benedict W. J. Irwin, May 29 2016: (Start)
Let y(0)=1, y(1)=2, y(2)=5, y(3)=14,
Let (n+2)*y(n) + (2*n+6)*y(n+2) - (4*n+14)*y(n+3) + (n+4)*y(n+4)=0,
a(n) = (y(n+2)-2*y(n+1)-y(n))/2.
(End)
D-finite with recurrence n*a(n) +6*(-n+1)*a(n-1) +9*(n-2)*a(n-2) -6*a(n-3) +(-n+8) * a(n-4) +2*(-n+4)*a(n-5) +(-n+6)*a(n-6)=0. - R. J. Mathar, Jun 06 2016
EXAMPLE
a(4) = 4 because the 5 (=A082582(4)) bargraphs of semiperimeter 4 correspond to the compositions [1,1,1], [1,2], [2,1], [2,2], [3] and the corresponding drawings show that they have 0, 0, 1, 1, 2 doublerises.
MAPLE
g := ((1-2*z-z^2-sqrt(1-4*z+2*z^2+z^4))*(1/2))/sqrt(1-4*z+2*z^2+z^4): gser := series(g, z = 0, 40): seq(coeff(gser, z, n), n = 2 .. 35);
MATHEMATICA
F[k_] := DifferenceRoot[Function[{y, n}, {(2 + n) y[n] + (6 + 2 n) y[2 + n] + (-14 - 4 n) y[3 + n] + (4 + n) y[4 + n] == 0, y[0] == 1, y[1] == 2, y[2] == 5, y[3] == 14}]][k]; Table[1/2 (-F[n] - 2 F[n + 1] + F[n + 2]), {n, 0, 20}] (* Benedict W. J. Irwin, May 29 2016 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Emeric Deutsch, May 28 2016
STATUS
approved