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%I #76 Jan 18 2024 02:39:33
%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/((1-2*x)*(1-x^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 cases in which the first player gets killed in a Russian roulette game when p players use a gun with n chambers and m bullets. 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 m-1 bullets are in {2,3,...,n}. We have binomial(n-1,m-1) 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(n-p-1,m-1) cases for this. We continue to calculate and the last is (t), where t = floor((n-m)/p). (t) The first gets killed when one bullet is in the (pt+1)-st chamber and the remaining bullets are in {pt+2,...,n}. We have binomial(n-pt-1,m-1) cases for this. Therefore U[p,n,m] = Sum_{z=0..floor((n-m)/p)} binomial(n-pz-1,m-1). Let A[p,n] be the number of the cases in which the first player gets killed when p players use a gun with n chambers and the number of the bullets can be from 1 to n. Then A[p,n] = Sum_{m=1..n} U[p,n,m]. - _Ryohei Miyadera_, Tomohide Hashiba, Yuta Nakagawa, Hiroshi Matsui, Jun 04 2006
%C A001045(n+5) without last digit. - _Paul Curtz_, Apr 21 2021
%C a(n) is the number of partitions of n into parts 1 and 4 where there are two colors of part 1 and the order of the colors of parts 1 matters. If the order of colors doesn't matter we get A001972. - _Joerg Arndt_, Jan 18 2024
%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(n-1) + a(n-4) - 2*a(n-5).
%F If n is a multiple of 4, then a(n) = 2*a(n-1) + 1, otherwise a(n) = 2*a(n-1). - _Gerald McGarvey_, Oct 14 2008
%F a(n) = floor((2^(n+5) + 1)/30). - _Tani Akinari_, Jul 09 2013
%F a(n) = 2*a(n-1) + floor(((n-1) mod 4) /3), with a(0)=1. - _Andres Cicuttin_, Mar 29 2016
%F a(n) = 2*a(n-1) + 1 - ceiling((n mod 4)/4), with a(0)=1. - _Andres Cicuttin_, Mar 29 2016
%F 15*a(n) = 2^(n+4) - A133145(n). - _R. J. Mathar_, Feb 27 2019
%t U[p_,n_,m_,v_]:=Block[{t},t=Floor[(1+p-m+n-v)/p];Sum[Binomial[n-v-p*z,m-1],{z,0,t-1}]]; 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/((1-2x)(1-x^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/((1-2*x)*(1-x^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. A000975, A033138, A195904, A117302.
%K easy,nonn
%O 0,2
%A _Paul Barry_, May 02 2003
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