A257164 Period 5 sequence: repeat [0, 2, 4, 1, 3].

0, 2, 4, 1, 3, 0, 2, 4, 1, 3, 0, 2, 4, 1, 3, 0, 2, 4, 1, 3, 0, 2, 4, 1, 3, 0, 2, 4, 1, 3, 0, 2, 4, 1, 3, 0, 2, 4, 1, 3, 0, 2, 4, 1, 3, 0, 2, 4, 1, 3, 0, 2, 4, 1, 3, 0, 2, 4, 1, 3, 0, 2, 4, 1, 3, 0, 2

Offset: 0

Author

Wesley Ivan Hurt, Apr 16 2015

Keywords

• nonn
• easy

Comments

Label the vertices of a regular pentagon from 0..4 going clockwise. Then, starting at vertex "0", a(n) gives the order in which the vertices must be connected to draw a clockwise inscribed, 5-pointed star that remains unbroken during construction.

Examples

			0 -> 2 -> 4 -> 1 -> 3 -> ..repeat
		

Links

• Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,1).

Crossrefs

Cf. A005843.

Bisection of A010874.

Programs

Magma

[(2*n mod 5) : n in [0..100]];

Maple

A257164:=n->(2*n mod 5): seq(A257164(n), n=0..100);

Mathematica

Mod[2 Range[0, 100], 5] (* or *)
CoefficientList[Series[x (2 + 4 x + x^2 + 3 x^3)/(1 - x^5), {x, 0, 100}], x]
LinearRecurrence[{0, 0, 0, 0, 1}, {0, 2, 4, 1, 3}, 105] (* or *)
NestList[# /. {0 -> 2, 1 -> 3, 2 -> 4, 3 -> 0, 4 -> 1} &, {0}, 104] // Flatten (* Robert G. Wilson v, Apr 30 2015 *)

PARI

a(n)=2*n%5 \\ Charles R Greathouse IV, Apr 21 2015

Formula

a(n) = (2n mod 5) = A010874(A005843(n)).

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

Recurrence: a(n) = a(n-5).

a(n) = a(a(a(a(a(n))))).

a(-n) = A010874(3n) = a(a(a(n))).

Bisections: a(2n) = A010874(-n) = a(a(n)); a(2n+1) = A010874(2-n).

Trisections: a(3n) = A010874(n) = a(a(a(a(n)))); a(3n+1) = A010874(n+2); a(3n+2) = A010874(n-1).