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A178789
a(n) = 4^(n-1) + 2: Number of acute angles after n iterations of the Koch snowflake construction.
7
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
1,1
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
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.
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
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 *)
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
STATUS
approved