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A033138
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Base 2 digits are, in order, the first n terms of the periodic sequence with initial period 1,0,0.
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7
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| 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 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]. - Ryohei Miyadera, Tomohide Hashiba, Yuta Nakagawa, Hiroshi Matsui (miyadera1272000(AT)yahoo.co.jp), Jun 04 2006
Partial sums of A077947. [From Mircea Merca, Dec 28 2010]
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REFERENCES
| 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; http://www.imfm.si/preprinti/PDF/01150.pdf.
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LINKS
| Vincenzo Librandi, Table of n, a(n) for n = 1..1000
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 926
Index entries for sequences related to linear recurrences with constant coefficients, signature (2,0,1,-2).
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FORMULA
| a(n) = 2*a(n-1) + a(n-3) - 2*a(n-4) -John W. Layman (layman(AT)math.vt.edu)
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 (pbarry(AT)wit.ie), Apr 16 2005
a(n) = floor(2^(n+2)/7) [From Gary Detlefs (gdetlefs(AT)aol.com), Sep 06 2010]
a(n)=floor((4*2^n-1)/7)=ceil((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. [From Mircea Merca, Dec 28 2010]
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MAPLE
| seq(iquo(2^n, 7), n=3..34); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Apr 20 2008
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MATHEMATICA
| 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 (miyadera1272000(AT)yahoo.co.jp), Jun 04 2006
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PROG
| (MAGMA) [Round((4*2^n-2)/7): n in [1..40]]; // Vincenzo Librandi, Jun 25 2011
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CROSSREFS
| Cf. A000975, A083593, A195904, A117302, A023001, A111662, A077947.
Sequence in context: A111662 A119027 A145139 * A155803 A056185 A152537
Adjacent sequences: A033135 A033136 A033137 * A033139 A033140 A033141
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KEYWORD
| nonn,base,easy
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AUTHOR
| Clark Kimberling (ck6(AT)evansville.edu)
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EXTENSIONS
| Edited by Jeremy Gardiner (jeremy.gardiner(AT)btinternet.com), Oct 08 2011
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