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 A198834 Number of sequences of n coin flips that win on the last flip, if the sequence of flips ends with (0,1,1) or (1,1,1). 1
 0, 0, 2, 2, 4, 6, 10, 16, 26, 42, 68, 110, 178, 288, 466, 754, 1220, 1974, 3194, 5168, 8362, 13530, 21892, 35422, 57314, 92736, 150050, 242786, 392836, 635622, 1028458, 1664080, 2692538, 4356618, 7049156, 11405774, 18454930, 29860704, 48315634, 78176338 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS If the sequence ends with (011) Abel wins; if it ends with (111) Kain wins. Kain(n)=0 for n <>3; Kain(3)=1. Abel(n)=A128588(n-2) for n>2. a(n)=A006355(n-1) for n>2. Win probability for Abel=sum(Abel(n)/2^n)=7/8. Win probability for Kain=Kain(3)/8=1/8. Mean length of the game=sum(n*a(n)/2^n)=7. Appears to be essentially the same as A163733, A118658, A055389. - R. J. Mathar, Oct 31 2011 REFERENCES A. Engel, Wahrscheinlichkeit und Statistik, Band 2, Klett, 1978, pages 25-26. LINKS Vincenzo Librandi, Table of n, a(n) for n = 1..1000 Index entries for linear recurrences with constant coefficients, signature (1,1). FORMULA a(n)=a(n-1)+a(n-2) for n>3. G.f.: 2*x^3/(1-x-x^2). a(n) = 2*A000045(n-2). - R. J. Mathar, Jan 11 2017 EXAMPLE For n=6 the a(6)=6 solutions are (0,0,0,0,1,1), (1,0,0,0,1,1); (0,1,0,0,1,1), (1,1,0,0,1,1), (0,0,1,0,1,1), (1,0,1,0,1,1) all for Abel. MAPLE a(1):=0: a(2):=0: a(3):=2: ml:=0.75: pot:=8: for n from 4 to 100 do   pot:=2*pot:   a(n):=a(n-1)+a(n-2):   ml:=ml+n*a(n)/pot: end do: printf("%12.8f", ml); seq(a(n), n=1..100); MATHEMATICA Join[{0, 0}, Table[2*Fibonacci[n], {n, 70}]] (* Vladimir Joseph Stephan Orlovsky, Feb 10 2012 *) Join[{0}, LinearRecurrence[{1, 1}, {0, 2}, 50]] (* Vincenzo Librandi, Feb 19 2012 *) CROSSREFS Cf. A006355, A128588. Sequence in context: A006355 A055389 A163733 * A270925 A084202 A300865 Adjacent sequences:  A198831 A198832 A198833 * A198835 A198836 A198837 KEYWORD nonn,easy AUTHOR Paul Weisenhorn, Oct 30 2011 STATUS approved

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Last modified June 25 03:56 EDT 2019. Contains 324338 sequences. (Running on oeis4.)