

A083593


Expansion of 1/((12*x)*(1x^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
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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 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}.
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


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(n1) +a(n4) 2*a(n5).
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
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
a(n) = floor((2^(n+5)+1)/30).  Tani Akinari, Jul 09 2013
a(n) = 2*a(n1) + floor(((n1) mod 4) /3), with a(0)=1.  Andres Cicuttin, Mar 29 2016
a(n) = 2*a(n1) + 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+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 *)
CoefficientList[Series[1/((12x)(1x^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[0] = 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/((12*x)*(1x^4))+O(x^99)) \\ Charles R Greathouse IV, May 15 2013
(PARI) a(n)=(16<<n)\15 \\ Charles R Greathouse IV, Mar 27 2016


CROSSREFS

Cf. A033138, A000975, A033138, A195904, A117302.
Sequence in context: A056184 A098718 A018299 * A267045 A266446 A018093
Adjacent sequences: A083590 A083591 A083592 * A083594 A083595 A083596


KEYWORD

easy,nonn


AUTHOR

Paul Barry, May 02 2003


STATUS

approved



