A196382 Number of sequences of n coin flips, that win on the last flip, if the sequence of flips ends with (1,1,0) or (1,0,1).
0, 0, 2, 3, 4, 7, 11, 16, 24, 36, 53, 78, 115, 169, 248, 364, 534, 783, 1148, 1683, 2467, 3616, 5300, 7768, 11385, 16686, 24455, 35841, 52528, 76984, 112826, 165355, 242340, 355167, 520523, 762864, 1118032, 1638556, 2401421, 3519454, 5158011
Offset: 1
Examples
For n=6 the a(6)=7 solutions are (0,0,0,1,1,0),(1,0,0,1,1,0),(0,0,1,1,1,0),(0,1,1,1,1,0),(1,1,1,1,1,0) for Abel and (0,0,0,1,0,1),(1,0,0,1,0,1) for Kain.
References
- A. Engel, Wahrscheinlichkeit und Statistik, Band 2, Klett, 1978, pages 25-26.
Links
- G. C. Greubel, Table of n, a(n) for n = 1..1000
- Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-1).
Programs
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Maple
a(1):=0: a(2):=0: a(3):=2: a(4):=3: a(5):=4: for n from 6 to 100 do a(n):=a(n-1)+a(n-2)-a(n-5): end do: seq(a(n),n=1..100);
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Mathematica
Rest[CoefficientList[Series[x^3*(2 - x)/((1 - x)*(1 - x - x^3)), {x,0,50}], x]] (* G. C. Greubel, May 02 2017 *)
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PARI
x='x+O('x^50); concat([0,0], Vec(x^3*(2 - x)/((1 - x)*(1 - x - x^3)))) \\ G. C. Greubel, May 02 2017
Formula
a(n) = +2*a(n-1) -a(n-2) +a(n-3) -a(n-4), n>=5.
G.f.: x^3*(2-x)/((1-x)*(1-x-x^3)).
a(n) = a(n-1) + a(n-3) + 1, n>3. - Greg Dresden, Feb 09 2020
Comments