

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 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 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 the (pt+1)st chamber and the remaining bullets are in {pt+2,...,n}. We have binomial(npt1,m1) cases for this. Therefore U[p,n,m] = Sum_{z=0..floor((nm)/p)} binomial(npz1,m1). 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(n1) + a(n4)  2*a(n5).
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(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



