OFFSET
1,1
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.
Larry Riddle, Koch Curve.
Index entries for linear recurrences with constant coefficients, signature (5,-4).
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
MAPLE
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 *)
PROG
(Magma) [2^(2*(n-1)) + 2: n in [1..30]]; // Vincenzo Librandi, Feb 02 2013
(PARI) A178789=n->4^(n-1)+2 \\ - M. F. Hasler, Dec 17 2013
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vladimir Joseph Stephan Orlovsky, Jun 14 2010
STATUS
approved