A260316 n/3 if 3 divides n, else n-1.

0, 0, 1, 1, 3, 4, 2, 6, 7, 3, 9, 10, 4, 12, 13, 5, 15, 16, 6, 18, 19, 7, 21, 22, 8, 24, 25, 9, 27, 28, 10, 30, 31, 11, 33, 34, 12, 36, 37, 13, 39, 40, 14, 42, 43, 15, 45, 46, 16, 48, 49, 17, 51, 52, 18, 54, 55, 19, 57, 58, 20, 60, 61, 21, 63, 64, 22, 66, 67, 23

Offset

0,5

Links

Peter Kagey, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (0,0,2,0,0,-1).

Formula

a(3k) = k; a(3k + 1) = 3k; a(3k + 2) = 3k + 1.

a(n) = 2*a(n-3) - a(n-6) for n>5. - Colin Barker, Jul 23 2015

G.f.: x^2*(2*x^3+3*x^2+x+1) / ((x-1)^2*(x^2+x+1)^2). - Colin Barker, Jul 23 2015

Example

a(5) = 5 - 1 = 4 because 5 is not divisible by 3.
a(12) = 12/3 = 4 because 12 is divisible by 3.

Mathematica

Table[If[Mod[n, 3] == 0, n/3, n - 1], {n, 0, 69}] (* or *)
CoefficientList[Series[x^2*(2 x^3 + 3 x^2 + x + 1)/((x - 1)^2*(x^2 + x + 1)^2), {x, 0, 69}], x] (* Michael De Vlieger, Jul 23 2015 *)
LinearRecurrence[{0, 0, 2, 0, 0, -1}, {0, 0, 1, 1, 3, 4}, 80] (* Harvey P. Dale, May 20 2018 *)

Prog

(Ruby) def a(n); (n%3==0)?n/3:n-1 end
(PARI) concat([0, 0], Vec(x^2*(2*x^3+3*x^2+x+1) / ((x-1)^2*(x^2+x+1)^2) + O(x^100))) \\ Colin Barker, Jul 23 2015

Crossrefs

A029578 is an analogous case where the divisor is 2 instead of 3.

Sequence in context: A378127 A049277 A214917 * A258742 A372862 A143052

Adjacent sequences: A260313 A260314 A260315 * A260317 A260318 A260319

Keyword

nonn,easy

Author

Peter Kagey, Jul 22 2015

Status

approved