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 A083593 Expansion of 1/((1-2*x)*(1-x^4)). 4
 1, 2, 4, 8, 17, 34, 68, 136, 273, 546, 1092, 2184, 4369, 8738, 17476, 34952, 69905, 139810, 279620, 559240, 1118481, 2236962, 4473924, 8947848, 17895697, 35791394, 71582788, 143165576, 286331153, 572662306, 1145324612, 2290649224 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS 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}. 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 LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..1000 Index entries for linear recurrences with constant coefficients, signature (2,0,0,1,-2). FORMULA a(n) = 2*a(n-1) + a(n-4) - 2*a(n-5). a(n) = -1/4 + (1/20 - i/10)*i^n + (1/12)*(-1)^n + (16/15)*2^n + (1/20 + i/10)*(-i)^n, for n >= 0, where i=sqrt(-1). - Paolo P. Lava, Jun 10 2008 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 a(n) = floor((2^(n+5) + 1)/30). - Tani Akinari, Jul 09 2013 a(n) = 2*a(n-1) + floor(((n-1) mod 4) /3), with a(0)=1. - Andres Cicuttin, Mar 29 2016 a(n) = 2*a(n-1) + 1 - ceiling((n mod 4)/4), with a(0)=1. - Andres Cicuttin, Mar 29 2016 15*a(n) = 2^(n+4) - A133145(n). - R. J. Mathar, Feb 27 2019 MATHEMATICA 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 *) CoefficientList[Series[1/((1-2x)(1-x^4)), {x, 0, 40}], x] (* Vincenzo Librandi, Apr 04 2012 *) a[n_] := FromDigits[Table[(Mod[j, 4]/4) // Round, {j, 1, n + 3}], 2] (* Andres Cicuttin, Mar 25 2016 *) a[n_] := a[n] = 2 a[n - 1] + 1 - Ceiling[Mod[n, 4]/4]; a = 1; Table[a[n], {n, 0, 31}] (* Andres Cicuttin, Mar 27 2016 *) LinearRecurrence[{2, 0, 0, 1, -2}, {1, 2, 4, 8, 17}, 40] (* Harvey P. Dale, Apr 03 2018 *) PROG (PARI) Vec(1/((1-2*x)*(1-x^4))+O(x^99)) \\ Charles R Greathouse IV, May 15 2013 (PARI) a(n)=(16<

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Last modified September 23 02:37 EDT 2019. Contains 327327 sequences. (Running on oeis4.)