A178789 a(n) = 4^(n-1) + 2: Number of acute angles after n iterations of the Koch snowflake construction.

3, 6, 18, 66, 258, 1026, 4098, 16386, 65538, 262146, 1048578, 4194306, 16777218, 67108866, 268435458, 1073741826, 4294967298, 17179869186, 68719476738, 274877906946, 1099511627778, 4398046511106, 17592186044418, 70368744177666

Offset: 1

Author

Vladimir Joseph Stephan Orlovsky, Jun 14 2010

Keywords

• nonn
• easy

Comments

Starting from an equilateral triangle, at each step each straight segment is replaced by a "/\" shape of four segments of equal length, with the acute angle in the middle pointing to the exterior. The sequence counts the angles which are (i.e., already were) at both extremities, plus the one newly created acute angle in the middle of each former segment. At step n, there are 3*4^(n-1) straight segments, therefore a(n+1) = a(n) + 3*4^(n-1). - M. F. Hasler, Dec 17 2013

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
• Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
• Mario Raso, Integer Sequences in Cryptography: A New Generalized Family and its Application, Ph. D. Thesis, Sapienza University of Rome (Italy 2025). See p. 113.
• Larry Riddle, Koch Curve.
• Index entries for linear recurrences with constant coefficients, signature (5,-4).

Crossrefs

Cf. A002001, A000302, A010502, A047849, A164346.

Programs

Magma

[2^(2*(n-1)) + 2: n in [1..30]]; // Vincenzo Librandi, Feb 02 2013

Maple

A178789:=n->2+4^(n-1); seq(A178789(n), n=1..30); # Wesley Ivan Hurt, Dec 17 2013

Mathematica

a=b=3;lst={a};Do[a=a+b;b*=4;AppendTo[lst,a],{n,40}];lst
Flatten[Table[2^(2*(n-1)) + 2, {n, 1, 50}]](* or *)   CoefficientList[Series[(3 - 9*x)/(1 - 5*x + 4*x^2),{x, 0, 100}], x] (* Vincenzo Librandi, Feb 02 2013 *)

PARI

A178789=n->4^(n-1)+2  \\ - M. F. Hasler, Dec 17 2013

Formula

G.f.: 3*x*(1 - 3*x)/(1 - 5*x + 4*x^2).

a(n) = 3 * A047849(n-1).

a(n) = 2^(2*(n-1)) + 2. - Vincenzo Librandi, Feb 02 2013

a(n+1) = a(n) + 3*4^(n-1) = a(n) + A002001(n) for n > 0. - M. F. Hasler, Dec 17 2013

a(n) = 2 + A000302(n-1). - Omar E. Pol, Dec 18 2013