A195904 -id:A195904 - cp's OEIS frontend

A033138 a(n) = floor(2^(n+2)/7).

1, 2, 4, 9, 18, 36, 73, 146, 292, 585, 1170, 2340, 4681, 9362, 18724, 37449, 74898, 149796, 299593, 599186, 1198372, 2396745, 4793490, 9586980, 19173961, 38347922, 76695844, 153391689, 306783378, 613566756, 1227133513, 2454267026

Offset: 1

Clark Kimberling

Previous name was: "Base 2 digits are, in order, the first n terms of the periodic sequence with initial period 1,0,0".
Here we let p = 3 to produce the above sequence, but p can be an arbitrary natural number. By letting p = 2, 4, 6, 7 we produce A000975, A083593, A195904 and A117302. We denote by U[p,n,m] the number of cases in which the first player in a game of Russian roulette gets killed 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)st 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 the (pt+1)st 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_{z=0..t} binomial(n-pz-1,m-1), where t = floor((n-m)/p). 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 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
Partial sums of A077947. - Mircea Merca, Dec 28 2010
a(n+1) is the number of partitions of n into two kinds of part 1 and one kind of part 2. - Joerg Arndt, Mar 10 2015
A078010(n) = b(n+1) - 2*b(n) + b(n-1) where b=A078010. - Michael Somos, Nov 18 2020

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 926
S. Klavzar, Structure of Fibonacci cubes: a survey, Institute of Mathematics, Physics and Mechanics Jadranska 19, 1000 Ljubljana, Slovenia; Preprint series Vol. 49 (2011), 1150 ISSN 2232-2094.
Index entries for linear recurrences with constant coefficients, signature (2,0,1,-2).

Cf. A000975, A078010, A083593, A195904, A117302, A023001, A111662, A077947.

[Round((4*2^n-2)/7): n in [1..40]]; // Vincenzo Librandi, Jun 25 2011

seq(iquo(2^n,7),n=3..34); # Zerinvary Lajos, Apr 20 2008

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 = 3 to produce the above sequence, but this code can produce A000975, A083593, A195904, A117302 for p = 2,4,6,7. *) Table[A[3,n,1], {n,1,20}] (* Ryohei Miyadera, Tomohide Hashiba, Yuta Nakagawa, Hiroshi Matsui, Jun 04 2006 *)

a(n)=2^(n+2)\7 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = 2*a(n-1) + a(n-3) - 2*a(n-4). -John W. Layman
G.f.: 1/((1-x^3)*(1-2*x)); a(n) = sum{k=0..floor(n/3), 2^(n-3*k)}; a(n) = Sum_{k=0..n} 2^k*( cos(2*Pi*(n-k)/3 + Pi/3)/3 + sqrt(3)*sin(2*Pi*(n-k)/3 + Pi/3)/3 + 1/3 ). - Paul Barry, Apr 16 2005
a(n) = floor(2^(n+2)/7). - Gary Detlefs, Sep 06 2010
a(n) = floor((4*2^n - 1)/7) = ceiling((4*2^n - 4)/7) = round((4*2^n - 2)/7) = round((8*2^n - 5)/14); a(n) = a(n-3) + 2^(n-1), n>3. - Mircea Merca, Dec 28 2010
a(n) = 4/7*2^n - 5/21*cos(2/3*Pi*n) + 1/21*3^(1/2)*sin(2/3*Pi*n)-1/3. - Leonid Bedratyuk, May 13 2012

Edited by Jeremy Gardiner, Oct 08 2011

New name (using formula form Gary Detlefs) from Joerg Arndt, Mar 10 2015

A083593 Expansion of 1/((1-2x)(1-x^4)).

Original entry on oeis.org

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

Offset: 0

Paul Barry, May 02 2003

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 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 the (pt+1)-st 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_{z=0..floor((n-m)/p)} binomial(n-pz-1,m-1). 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
A001045(n+5) without last digit. - Paul Curtz, Apr 21 2021
a(n) is the number of partitions of n into parts 1 and 4 where there are two colors of part 1 and the order of the colors of parts 1 matters. If the order of colors doesn't matter we get A001972. - Joerg Arndt, Jan 18 2024

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).

Cf. A000975, A033138, A195904, A117302.

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 = 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/((1-2x)(1-x^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 *)

Vec(1/((1-2*x)*(1-x^4))+O(x^99)) \\ Charles R Greathouse IV, May 15 2013

a(n)=(16<Charles R Greathouse IV, Mar 27 2016

def A083593(n): return ((32<Chai Wah Wu, Apr 25 2025

a(n) = 2*a(n-1) + a(n-4) - 2*a(n-5).
If n is a multiple of 4, then a(n) = 2*a(n-1) + 1, otherwise a(n) = 2*a(n-1). - Gerald McGarvey, Oct 14 2008
a(n) = floor((2^(n+5) + 1)/30). - Tani Akinari, Jul 09 2013
From Andres Cicuttin, Mar 29 2016: (Start)
a(n) = 2*a(n-1) + floor(((n-1) mod 4)/3), with a(0)=1.
a(n) = 2*a(n-1) + 1 - ceiling((n mod 4)/4), with a(0)=1. (End)
15*a(n) = 2^(n+4) - A133145(n). - R. J. Mathar, Feb 27 2019
E.g.f.: (3*cos(x) - 5*cosh(x) + 32*cosh(2*x) + 6*sin(x) - 10*sinh(x) + 32*sinh(2*x))/30. - Stefano Spezia, Apr 25 2025

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.

Original entry on oeis.org

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

Ryohei Miyadera, Jun 04 2006

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).

			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.

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).

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

Partial sums of A349842.

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

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

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 *)

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

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)

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A033138 a(n) = floor(2^(n+2)/7).

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A083593 Expansion of 1/((1-2*x)*(1-x^4)).

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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.

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A101333 A081254-A072762.

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A083593 Expansion of 1/((1-2x)(1-x^4)).