A238976 a(n) = ((3^(n-1)-1)^2)/4.

0, 1, 16, 169, 1600, 14641, 132496, 1194649, 10758400, 96845281, 871666576, 7845176329, 70607118400, 635465659921, 5719195722256, 51472775849209, 463255025689600, 4169295360346561, 37523658630539536, 337712928837117289, 3039416363020840000, 27354747277647913201, 246192725530212278416

Offset

1,3

Comments

If the Cantor square fractal is modified as shown in the illustration (see Links), then 4*a(n) is the total number of holes in the modified Cantor square fractal after n iterations. The total number of sides (outside) is 4*A171498(n-1). The total length of the sides (outside) converges to 20 when the initial total side length is 12 (starting with 5 unit squares).

For the Cantor square fractal, the total number of sides (outside) is 4*A168616(n+2). The total number of holes is 4*A060867(n-1) for n > 1. The total length of the sides (outside) converges to 12 with the same initial condition (i.e., 5 unit square); its maximum is 17.333... and is reached at n = 2, 3. The Cantor square fractal and modified one are not true fractals.

See illustrations in links.

Links

Table of n, a(n) for n=1..23.

Kival Ngaokrajang, Illustrations of initial terms

Eric Weisstein's World of Mathematics, Cantor Square Fractal

Index entries for linear recurrences with constant coefficients, signature (13,-39,27)

Formula

a(n) = (A024023(n-1))^2/4.

G.f.: x*(3*x + 1)/((1-x)*(1-3*x)*(1-9*x)). - Ralf Stephan, Mar 14 2014

Prog

(Small Basic)
For n = 1 To 30
  a = Math.Power(Math.Power(3, n-1)-1, 2)
  TextWindow.Write(a/4+", ")
EndFor
(PARI) a(n) = ((3^(n-1)-1)^2)/4; \\ Joerg Arndt, Mar 08 2014

Crossrefs

Cf. A024023, A060867, A168616, A171498.

Sequence in context: A283574 A067711 A068175 * A242078 A264092 A050628

Adjacent sequences: A238973 A238974 A238975 * A238977 A238978 A238979

Keyword

nonn,easy

Author

Kival Ngaokrajang, Mar 07 2014

Status

approved