A260316 - OEIS

%I #14 May 20 2018 11:56:29

%S 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,

%T 27,28,10,30,31,11,33,34,12,36,37,13,39,40,14,42,43,15,45,46,16,48,49,

%U 17,51,52,18,54,55,19,57,58,20,60,61,21,63,64,22,66,67,23

%N n/3 if 3 divides n, else n-1.

%H Peter Kagey, <a href="/A260316/b260316.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,2,0,0,-1).

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

%F a(n) = 2*a(n-3) - a(n-6) for n>5. - _Colin Barker_, Jul 23 2015

%F 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

%e a(5) = 5 - 1 = 4 because 5 is not divisible by 3.

%e a(12) = 12/3 = 4 because 12 is divisible by 3.

%t Table[If[Mod[n, 3] == 0, n/3, n - 1], {n, 0, 69}] (* or *)

%t 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 *)

%t LinearRecurrence[{0,0,2,0,0,-1},{0,0,1,1,3,4},80] (* _Harvey P. Dale_, May 20 2018 *)

%o (Ruby) def a(n);(n%3==0)?n/3:n-1 end

%o (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

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

%K nonn,easy

%O 0,5

%A _Peter Kagey_, Jul 22 2015