

A117302


Number of cases in which the first player gets killed in a Russian roulette game when 7 players use a gun with n chambers and the number of the bullets can be from 1 to n. In the game they do not rotate the cylinder after the game starts.


6



1, 2, 4, 8, 16, 32, 64, 129, 258, 516, 1032, 2064, 4128, 8256, 16513, 33026, 66052, 132104, 264208, 528416, 1056832, 2113665, 4227330, 8454660, 16909320, 33818640, 67637280, 135274560, 270549121, 541098242
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OFFSET

1,2


COMMENTS

We denote by U[7,n,m] the number of cases that the first player gets killed in a Russian roulette game when 7 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 (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 8th chamber and the rest of the bullets are in {9,..,n}. We have binomial(n8,m1) cases for this. We continue to calculate and the last is (t), where t = floor((nm)/7). (t) The first gets killed when one bullet is in (7t+1)st chamber and the remaining bullets are in {7t+2,...,n}. We have binomial(n7t1,m1) cases for this. Therefore U[7,n,m] = Sum_{z=0..t} binomial(n7z1,m1), where t = floor((nm)/7). Let A[7,n] be the number of cases in which the first player gets killed when 7 players use a gun with n chambers and the number of bullets can be from 1 to n. Then A[7,n] = Sum_{m=1..n} U[7,n,m].


REFERENCES

Miyadera, R. "General Theory of Russian Roulette." Mathematica source.
Miyadera, R. Mathematical Theory of Magic Fruits Archimedeslab.


LINKS

G. C. Greubel, Table of n, a(n) for n = 1..1000
R. Miyadera, General Theory of Russian Roulette, MathSource
R. Miyadera, Daisuke Minematsu, Satoshi Hashiba and Munetoshi Hashiba, Theory of Magic Fruits, Archimedeslab, Interesting patterns of fractions
Index entries for linear recurrences with constant coefficients, signature (2,0,0,0,0,0,1,2).


FORMULA

a(n) = (2^(n + 6)  2^((n1) mod 7))/(2^7  1).
a(n) = floor(2^(n+6)/127).  Mircea Merca, Dec 22 2010
From Joerg Arndt, Jan 08 2011: (Start)
G.f.: x/( (x1)*(2*x1)*(x^6 + x^5 + x^4 + x^3 + x^2 + x + 1) ).
a(n) = 2*a(n1) + a(n7)  2*a(n8). (End)


EXAMPLE

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


MAPLE

A117302 := proc(n) floor(2^(n+6)/127) ; end proc:


MATHEMATICA

U7[n_, m_]:= Block[{t}, t=Floor[(nm)/7]; Sum[Binomial[n17z, m1], {z, 0, t}]]; A7[n_]:= Sum[U7[n, m], {m, 1, n}]; Table[A7[n], {n, 1, 40}]
LinearRecurrence[{2, 0, 0, 0, 0, 0, 1, 2}, {1, 2, 4, 8, 16, 32, 64, 129}, 40] (* G. C. Greubel, May 07 2019 *)


PROG

(PARI) a(n)=2^(n+6)\127 \\ Charles R Greathouse IV, Oct 07 2015
(Magma) [Floor(2^(n+6)/127): n in [1..40]]; // G. C. Greubel, May 07 2019
(Sage) [floor(2^(n+6)/127) for n in (1..40)] # G. C. Greubel, May 07 2019


CROSSREFS

Sequence in context: A109912 A079845 A278995 * A265407 A023422 A084638
Adjacent sequences: A117299 A117300 A117301 * A117303 A117304 A117305


KEYWORD

nonn,easy


AUTHOR

Tomohide Hashiba, Akihiro Hyogu, Hiroshi Matsui, Ryohei Miyadera, Yuta Nakagawa, Apr 24 2006


EXTENSIONS

Edited by G. C. Greubel, May 07 2019


STATUS

approved



