A119610 Number of cases in which the first player is killed in a Russian roulette game where 5 players use a gun with n chambers and the number of bullets can be from 1 to n. Players do not rotate the cylinder after the game starts.

1, 2, 4, 8, 16, 33, 66, 132, 264, 528, 1057, 2114, 4228, 8456, 16912, 33825, 67650, 135300, 270600, 541200, 1082401, 2164802, 4329604, 8659208, 17318416, 34636833, 69273666, 138547332, 277094664, 554189328, 1108378657, 2216757314

Offset: 1

Author

Ryohei Miyadera, Jun 04 2006

Keywords

• easy
• nonn

Comments

Denote by U(p,n,m) the number of the cases in which the first player is killed in a Russian roulette game where 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}.

Here we let p = 5 to produce the above sequence, but p can be an arbitrary positive integer. By letting p = 2, 3, 4, 6, 7 we can produce sequences A000975, A033138, A083593, A195904 and A117302, respectively.

The number of cases for each of the situations identified below by (0), (1), ..., (t), where t = floor((n-m)/p), can be calculated separately:

(0) The first player is killed when one bullet is in the first chamber and the remaining m-1 bullets are in chambers {2,3,...,n}. There are binomial(n-1,m-1) cases for this situation.

(1) The first player is killed when one bullet is in the (p+1)-th chamber and the rest of the bullets are in chambers {p+2,...,n}. There are binomial(n-p-1,m-1) cases for this situation.

...

(t) The first player is killed when one bullet is in the (p*t+1)-th chamber and the remaining bullets are in chambers {p*t+2,...,n}. There are binomial(n-p*t-1,m-1) cases for this situation.

Therefore U(p,n,m) = Sum_{z=0..t} binomial(n-p*z-1,m-1), where t = floor((n-m)/p). Let A(p,n) be the number of the cases in which the first player is killed when p players use a gun with n chambers and the number of bullets can be from 1 to n. Then A(p,n) = Sum_{m=1..n} U(p,n,m).

Examples

			If the number of chambers is 3, then the number of the bullets can be 1, 2, or 3. The first player is killed when one bullet is in the first chamber, and the remaining bullets are in the second and third chambers. The only cases are {{1, 0, 0}, {1, 1, 0}, {1, 0, 1}, {1, 1, 1}}, where we denote by 1 a chamber that contains a bullet. Therefore a(3) = 4.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 1..1000
• R. Miyadera, General Theory of Russian Roulette, Mathematica source.
• R. Miyadera, Interesting patterns of fractions, Archimedes' Laboratory.
• Index entries for linear recurrences with constant coefficients, signature (2,0,0,0,1,-2).

Crossrefs

Cf. A000975, A033138, A083593, A195904, A117302.

Partial sums of A349842.

Programs

Magma

I:=[1,2,4,8,16,33]; [n le 6 select I[n] else 2*Self(n-1)+Self(n-5)-2*Self(n-6): n in [1..40]]; // Vincenzo Librandi, Mar 18 2015

Maple

seq(floor(2^(n+4)/31), n = 1..32); # Mircea Merca, Dec 22 2010

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 = 5 to produce the above sequence, but this code can produce A000975, A033138, A083593, A195904, A117302 for p = 2, 3, 4, 6, 7. *)
Table[B[5,n,1],{n,1,20}] (* end of program *)
CoefficientList[ Series[ 1/(2x^6 - x^5 - 2x + 1), {x, 0, 32}], x] (* or *)
LinearRecurrence[{2, 0, 0, 0, 1, -2}, {1, 2, 4, 8, 16, 33}, 32] (* Robert G. Wilson v, Mar 12 2015 *)

PARI

for(n=1,50, print1(floor(2^(n+4)/31), ", ")) \\ G. C. Greubel, Oct 11 2017

Formula

a(n) = floor(2^(n+4)/31), which is obtained by letting p=5 in a_p(n) = (2^(n + p-1) - 2^((n-1) mod p))/(2^p - 1).

From Joerg Arndt, Jan 08 2011: (Start)

G.f.: x / ( (x-1)*(2*x-1)*(x^4+x^3+x^2+x+1) ).

a(n) = +2*a(n-1) +a(n-5) -2*a(n-6). (End)