A269962 Start with a square; at each stage add a square at each expandable vertex so that the ratio of the side of the squares at stage n+1 and at stage n is the golden ratio phi=0.618...; a(n) is the number of squares at n-th stage.

1, 5, 17, 45, 105, 237, 537, 1229, 2825, 6493, 14905, 34189, 78409, 179837, 412505, 946221, 2170473, 4978653, 11420025, 26195213, 60086537, 137826493, 316146457, 725176813, 1663410601, 3815531165, 8752065209, 20075486925, 46049151561, 105627543165

Offset: 1

Author

Paolo Franchi, Mar 08 2016

Keywords

• nonn
• easy

Comments

The ratio phi=0.618... is chosen so that from the fourth stage on some squares overlap perfectly. The figure displays some kind of fractal behavior. See illustration.

Links

• Colin Barker, Table of n, a(n) for n = 1..1000
• Paolo Franchi, Illustration of initial terms
• Index entries for linear recurrences with constant coefficients, signature (4,-5,2,2,-2).

Crossrefs

Cf. A247618.

Auxiliary sequences: A269963, A269964, A269965.

Programs

Mathematica

RecurrenceTable[{a[n + 1] ==
   4 a[n] - 5 a[n - 1] + 2 a[n - 2] + 2 a[n - 3] - 2 a[n - 4],
  a[1] == 1, a[2] == 5, a[3] == 17, a[4] == 45, a[5] == 105}, a, {n,
  1, 30}]
RecurrenceTable[{a[n + 1] ==
   2 a[n] + a[n - 1] - 2 a[n - 2] + 2 a[n - 3] + 2 a[n - 4] + 4,
  a[1] == 1, a[2] == 5, a[3] == 17, a[4] == 45, a[5] == 105}, a, {n,
  1, 30}]

PARI

Vec(x*(1+x)*(1+2*x^2-2*x^3)/((1-x)*(1-3*x+2*x^2-2*x^4)) + O(x^50)) \\ Colin Barker, Mar 09 2016

Formula

a(1)=1, for n>=1, a(n) = 2*A269963(n) + 2*A269963(n-1) - 1.

Linear non-homogeneous recurrence relation:

a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) + 2*a(n-4) + 2*a(n-5) + 4.

Linear homogeneous recurrence relation:

a(n) = 4*a(n-1) - 5*a(n-2) + 2*a(n-3) + 2*a(n-4) - 2*a(n-5).

G.f.: x*(1+x)*(1+2*x^2-2*x^3) / ((1-x)*(1-3*x+2*x^2-2*x^4)). - Colin Barker, Mar 09 2016