

A033138


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


13



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
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OFFSET

1,2


COMMENTS

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 (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 (p+1)st chamber and the rest of the bullets are in {p+2,..,n}. We have binomial(np1,m1) cases for this. We continue to calculate and the last is (t), where t = floor((nm)/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(npt1,m1) cases for this.
Therefore U[p,n,m] = Sum_{z=0..t} binomial(npz1,m1), where t = floor((nm)/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(n1) where b=A078010.  Michael Somos, Nov 18 2020


LINKS

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 22322094.
Index entries for linear recurrences with constant coefficients, signature (2,0,1,2).


FORMULA

a(n) = 2*a(n1) + a(n3)  2*a(n4). John W. Layman
G.f.: 1/((1x^3)*(12*x)); a(n) = sum{k=0..floor(n/3), 2^(n3*k)}; a(n) = Sum_{k=0..n} 2^k*( cos(2*Pi*(nk)/3 + Pi/3)/3 + sqrt(3)*sin(2*Pi*(nk)/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(n3) + 2^(n1), 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


MAPLE

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


MATHEMATICA

U[p_, n_, m_, v_]:=Block[{t}, t=Floor[(1+pm+nv)/p]; Sum[Binomial[nvp*z, m1], {z, 0, t1}]]; 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 *)


PROG

(MAGMA) [Round((4*2^n2)/7): n in [1..40]]; // Vincenzo Librandi, Jun 25 2011
(PARI) a(n)=2^(n+2)\7 \\ Charles R Greathouse IV, Oct 07 2015


CROSSREFS

Cf. A000975, A078010, A083593, A195904, A117302, A023001, A111662, A077947.
Sequence in context: A253585 A119027 A145139 * A155803 A293352 A293327
Adjacent sequences: A033135 A033136 A033137 * A033139 A033140 A033141


KEYWORD

nonn,easy


AUTHOR

Clark Kimberling


EXTENSIONS

Edited by Jeremy Gardiner, Oct 08 2011
New name (using formula form Gary Detlefs) from Joerg Arndt, Mar 10 2015


STATUS

approved



