1,2

Here we let p = 5 to produce the above sequence, but p can be an arbitrary natural number. By letting p = 2, 3, 4, 6, 7 we can produce A000975, A033138, A083593, A101333 and A117302 in the On-Line Encyclopedia of Integer Sequences. 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 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 (pt+1)-th 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[binomial[n-pz-1,m-1], for z = 0 to t, where t = Floor[(n-m)/p]. Let A[p,n] be the number of the cases that the first player gets killed when p-player use a gun with n-chambers 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].

Miyadera, R. "General Theory of Russian Roulette." Mathematica source.

Miyadera, R. Mathematical Theory of Magic Fruits Archimedes-lab.

R.Miyadera, General Theory of Russian Roulette, Mathematica source.

R.Miyadera, nteresting patterns offractions,Archimedes-lab

Let p = 5 and a(n) = (2^(n + p-1) - 2^(Mod[n - 1, p]))/(2^p - 1), where Mod[,p] is the remaining part of the number when divided by p. In the followings we present the formula based on the theory we used to define our sequence as a Mathematica code. The above formula is a lot easier to use, but the Mathematica code has an important mathematical meaning in it.

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 is {{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.

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, A101333, A117302 for p = 2, 3, 4, 6, 7.*) Table[B[5, n, 1], {n, 1, 20}]

Cf. A000975, A033138, A083593, A101333, A117302.

Adjacent sequences: A119607 A119608 A119609 this_sequence A119611 A119612 A119613

Sequence in context: A138814 A137181 A036373 this_sequence A121485 A098588 A126683

easy,nonn,uned

Ryohei Miyadera (miyadera1272000(AT)yahoo.co.jp), Jun 04 2006