A162484 a(1) = 2, a(2) = 8; a(n) = 2 a(n - 1) + a(n - 2) - 4*(n mod 2).
2, 8, 14, 36, 82, 200, 478, 1156, 2786, 6728, 16238, 39204, 94642, 228488, 551614, 1331716, 3215042, 7761800, 18738638, 45239076, 109216786, 263672648, 636562078, 1536796804, 3710155682, 8957108168, 21624372014, 52205852196, 126036076402, 304278005000
Offset: 1
Examples
a(3) = 2 a(2) + a(1) - 4*(3 mod 2) = 2*8 + 2 - 4 = 14.
Links
- Colin Barker, Table of n, a(n) for n = 1..1000
- Sarah-Marie Belcastro, Domino Tilings of 2 X n Grids (or Perfect Matchings of Grid Graphs) on Surfaces, J. Integer Seq. 26 (2023), Article 23.5.6.
- Index entries for linear recurrences with constant coefficients, signature (2,2,-2,-1).
Crossrefs
Cf. A000129.
Programs
-
Mathematica
Fold[Append[#1, 2 #1[[#2 - 1]] + #1[[#2 - 2]] - 4 Mod[#2, 2]] &, {2, 8}, Range[3, 30]] (* or *) Rest@ CoefficientList[Series[-2 x (-1 - 2 x + 3 x^2 + 2 x^3)/((x - 1) (1 + x) (x^2 + 2 x - 1)), {x, 0, 30}], x] (* Michael De Vlieger, Dec 16 2017 *) LinearRecurrence[{2,2,-2,-1},{2,8,14,36},30] (* Harvey P. Dale, Aug 24 2018 *)
Formula
for n > 2, (1/2) ((1 + sqrt(2))^n (2 - (-2 + sqrt(2)) (-1 + sqrt(2))^(2 floor(n/2))) + (1 - sqrt(2))^n (2 + (1 + sqrt(2))^(2 floor(n/2)) (2 + sqrt(2)))) (from Mathematica's solution to the recurrence).
Pell(n) + Pell(n-2) + 2*((n-1) mod 2).
From R. J. Mathar, Jul 26 2009: (Start)
a(n)= 2*a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) = 2*A100828(n-1).
G.f.: -2*x*(-1-2*x+3*x^2+2*x^3)/((x-1)*(1+x)*(x^2+2*x-1)).
(End)
a(n) = 1 + (-1)^n + (1-sqrt(2))^n + (1+sqrt(2))^n. - Colin Barker, Dec 16 2017
Comments