%I
%S 1,2,4,8,17,34,68,136,273,546,1092,2184,4369,8738,17476,34952,69905,
%T 139810,279620,559240,1118481,2236962,4473924,8947848,17895697,
%U 35791394,71582788,143165576,286331153,572662306,1145324612,2290649224
%N Expansion of 1/((12*x)*(1x^4)).
%C Here we let p =4 to produce the above sequence, but p can be an arbitrary natural number. By letting p = 2, 3, 6, 7 we produce A000975, A033138, A195904 and A117302. We denote by U[p,n,m] the number of the cases that the first player gets killed in a Russian roulette game when p players use a gun with nchambers and mbullets. They never rotate the cylinder after the game starts. The chambers can be represented by the list {1,2,...,n}.
%C We are going to calculate the following (0), (1),...(t) separately. (0) The first player gets killed when one bullet is in the first chamber and the remaining (m1) bullets are in {2,3,...,n}. We have binomial[n1,m1]cases for this. (1) The first gets killed when one bullet is in the (p+1)th chamber and the rest of the bullets are in {p+2,..,n}. We have binomial[np1,m1]cases for this. We continue to calculate and the last is (t), where t = Floor[(nm)/ p]. (t) The first gets killed when one bullet is in (pt+1)th chamber and the remaining bullets are in {pt+2,...,n}. We have binomial[npt 1,m1]cases for this. Therefore U[p,n,m] = Sum[binomial[npz1,m1], for z = 0 to t, where t = Floor[(nm)/p]. Let A[p,n] be the number of the cases that the first player gets killed when pplayer use a gun with nchambers and the number of the bullets can be from 1 to n. Then A[p,n] = Sum[U[p,n,m], m = 1 to n].  _Ryohei Miyadera_, Tomohide Hashiba, Yuta Nakagawa, Hiroshi Matsui, Jun 04 2006
%H Vincenzo Librandi, <a href="/A083593/b083593.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (2,0,0,1,2).
%F a(n) = 2*a(n1) +a(n4) 2*a(n5).
%F a(n) = 1/4+[1/20(1/10)*I]*I^n+(1/12)*(1)^n+(16/15)*2^n+[1/20+(1/10)*I]*(I)^n, with n>=0 and I=sqrt(1).  _Paolo P. Lava_, Jun 10 2008
%F If n is a multiple of 4, then a(n) = 2*a(n1) + 1, otherwise a(n) = 2*a(n1).  _Gerald McGarvey_, Oct 14 2008
%F a(n) = floor((2^(n+5)+1)/30).  _Tani Akinari_, Jul 09 2013
%F a(n) = 2*a(n1) + floor(((n1) mod 4) /3), with a(0)=1.  _Andres Cicuttin_, Mar 29 2016
%F a(n) = 2*a(n1) + 1  ceiling[(n mod 4)/4], with a(0)=1.  _Andres Cicuttin_, Mar 29 2016
%t U[p_,n_,m_,v_]:=Block[{t},t=Floor[(1+pm+nv)/p];Sum[Binomial[nvp*z,m1],{z,0,t1}]]; A[p_,n_,v_]:=Sum[U[p,n,k,v],{k,1,n}]; (* Here we let p = 4 to produce the above sequence, but this code can produce A000975, A033138, A195904, A117302 for p=2,3,6,7.*) Table[A[4,n,1], {n,1,20}] (* _Ryohei Miyadera_, Tomohide Hashiba, Yuta Nakagawa, Hiroshi Matsui, Jun 04 2006 *)
%t CoefficientList[Series[1/((12x)(1x^4)),{x,0,40}],x] (* _Vincenzo Librandi_, Apr 04 2012 *)
%t a[n_] := FromDigits[Table[(Mod[j, 4]/4) // Round, {j, 1, n + 3}], 2] (* _Andres Cicuttin_, Mar 25 2016 *)
%t a[n_] := a[n] = 2 a[n  1] + 1  Ceiling[Mod[n, 4]/4]; a[0] = 1;
%t Table[a[n], {n, 0, 31}] (* _Andres Cicuttin_, Mar 27 2016 *)
%t LinearRecurrence[{2,0,0,1,2},{1,2,4,8,17},40] (* _Harvey P. Dale_, Apr 03 2018 *)
%o (PARI) Vec(1/((12*x)*(1x^4))+O(x^99)) \\ _Charles R Greathouse IV_, May 15 2013
%o (PARI) a(n)=(16<<n)\15 \\ _Charles R Greathouse IV_, Mar 27 2016
%Y Cf. A033138, A000975, A033138, A195904, A117302.
%K easy,nonn
%O 0,2
%A _Paul Barry_, May 02 2003
