OFFSET
1,1
LINKS
Indranil Ghosh, Table of n, a(n) for n = 1..4182
Joseph Myers, BMO 2008--2009 Round 1 Problem 1---Generalisation
Index entries for linear recurrences with constant coefficients, signature (0,3).
FORMULA
a(2n) = 4*3^(n-1), a(2n+1) = 8*3^(n-1) for n > 0, a(1)=3.
From Colin Barker, May 10 2012: (Start)
a(n) = 3*a(n-2) for n > 3.
G.f.: x*(3+4*x-x^2)/(1-3*x^2). (End)
MATHEMATICA
a[1]=3; a[2]=4; a[3]=8; a[n_]:=3 a[n-2]; Table[a[n], {n, 1, 34}] (* or *) LinearRecurrence[{0, 3}, {3, 4, 8}, 34] (* Indranil Ghosh, Feb 20 2017 *)
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Joseph Myers, Dec 24 2008
STATUS
approved