cp's OEIS Frontend

A116969 If n mod 2 = 0 then 3*2^(n-1)+n-1 else 3*2^(n-1)+n.

%I A116969 #11 Jun 13 2015 00:52:04
%S A116969 4,7,15,27,53,101,199,391,777,1545,3083,6155,12301,24589,49167,98319,
%T A116969 196625,393233,786451,1572883,3145749,6291477,12582935,25165847,
%U A116969 50331673,100663321,201326619,402653211,805306397,1610612765,3221225503,6442450975,12884901921
%N A116969 If n mod 2 = 0 then 3*2^(n-1)+n-1 else 3*2^(n-1)+n.
%C A116969 Number of moves to solve Easy Pagoda puzzle.
%D A116969 Richard I. Hess, Compendium of Over 7000 Wire Puzzles, privately printed, 1991.
%D A116969 Richard I. Hess, Analysis of Ring Puzzles, booklet distributed at 13th International Puzzle Party, Amsterdam, Aug 20 1993.
%H A116969 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (3,-1,-3,2).
%F A116969 a(n) = 3*a(n-1)-a(n-2)-3*a(n-3)+2*a(n-4). G.f.: -x*(x^3-2*x^2-5*x+4) / ((x-1)^2*(x+1)*(2*x-1)). - _Colin Barker_, Jul 18 2013
%p A116969 f:=n-> if n mod 2 = 0 then 3*2^(n-1)+n-1 else 3*2^(n-1)+n; fi;
%t A116969 f[n_]:=If[EvenQ[n],3*2^(n-1)+n-1,3*2^(n-1)+n]; f/@Range[40] (* _Harvey P. Dale_, Sep 21 2012 *)
%K A116969 nonn,easy
%O A116969 1,1
%A A116969 _N. J. A. Sloane_, Apr 01 2006
%E A116969 More terms from _Colin Barker_, Jul 18 2013