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_{n>=1} Abel(n)/2^n = 7/8.
Win probability for Kain: Kain(3)/8 = 1/8.
Mean length of the game: Sum_{n>=1} n*a(n)/2^n = 7.
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
Tian-Xiao He, Peter J.-S. Shiue, Zihan Nie, Minghao Chen, Recursive sequences and Girard-Waring identities with applications in sequence transformation, Electronic Research Archive (2020) Vol. 28, No. 2, 1049-1062.
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
E.g.f.: 2 - 2*x + 2*exp(x/2)*(3*sqrt(5)*sinh(sqrt(5)*x/2) - 5*cosh(sqrt(5)*x/2))/5. - Stefano Spezia, Feb 19 2023
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
KEYWORD
nonn,easy
AUTHOR
Paul Weisenhorn, Oct 30 2011
STATUS
approved