%I #74 Sep 08 2022 08:44:51

%S 1,2,4,9,18,36,73,146,292,585,1170,2340,4681,9362,18724,37449,74898,

%T 149796,299593,599186,1198372,2396745,4793490,9586980,19173961,

%U 38347922,76695844,153391689,306783378,613566756,1227133513,2454267026

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

%C Previous name was: "Base 2 digits are, in order, the first n terms of the periodic sequence with initial period 1,0,0".

%C 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.

%C 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.

%C 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

%C Partial sums of A077947. - _Mircea Merca_, Dec 28 2010

%C 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

%C A078010(n) = b(n+1) - 2*b(n) + b(n-1) where b=A078010. - _Michael Somos_, Nov 18 2020

%H Vincenzo Librandi, <a href="/A033138/b033138.txt">Table of n, a(n) for n = 1..1000</a>

%H INRIA Algorithms Project, <a href="http://ecs.inria.fr/services/structure?nbr=926">Encyclopedia of Combinatorial Structures 926</a>

%H S. Klavzar, <a href="http://www.imfm.si/preprinti/PDF/01150.pdf">Structure of Fibonacci cubes: a survey</a>, Institute of Mathematics, Physics and Mechanics Jadranska 19, 1000 Ljubljana, Slovenia; Preprint series Vol. 49 (2011), 1150 ISSN 2232-2094.

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,0,1,-2).

%F a(n) = 2*a(n-1) + a(n-3) - 2*a(n-4). -_John W. Layman_

%F 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

%F a(n) = floor(2^(n+2)/7). - _Gary Detlefs_, Sep 06 2010

%F 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

%F 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

%p seq(iquo(2^n,7),n=3..34); # _Zerinvary Lajos_, Apr 20 2008

%t 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 *)

%o (Magma) [Round((4*2^n-2)/7): n in [1..40]]; // _Vincenzo Librandi_, Jun 25 2011

%o (PARI) a(n)=2^(n+2)\7 \\ _Charles R Greathouse IV_, Oct 07 2015

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

%K nonn,easy

%O 1,2

%A _Clark Kimberling_

%E Edited by _Jeremy Gardiner_, Oct 08 2011

%E New name (using formula form _Gary Detlefs_) from _Joerg Arndt_, Mar 10 2015