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%I #20 Mar 24 2024 03:23:12
%S 0,0,0,1,1,1,2,4,6,11,20,35,62,111,197,350,623,1108,1970,3504,6232,
%T 11083,19711,35056,62346,110881,197200,350716,623741,1109311,1972887,
%U 3508739,6240221,11098106,19737755,35103195,62430317
%N Number of sequences of n coin flips that win on the last flip, if the sequence of flips ends with (1,1,1,1).
%C If the sequence ends with (0,0,1,1) Abel wins.
%C Abel(n)=A199925(n); Winsum(n)=A199594(n).
%C Win probability of Kain = a(n)/2^n = 1/4.
%C Win probability of Abel = Abel(n)/2^n = 3/4.
%C Mean length of the game= n*Winsum(n)/2^n = 12.
%D A. Engel, Wahrscheinlichkeit und Statistik, Band 2, Klett, 1978, pages 25-26.
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,1,0,-1).
%F a(n) = a(n-1)+a(n-2)+a(n-3)-a(n-5) for n>7.
%F a(n) = 2*a(n-1)-a(n-4)-a(n-5)+a(n-6) for n>8.
%F G.f.: x^4*(1-x^2-x^3)/(1-x-x^2-x^3+x^5).
%e For n=7 the a(7)=2 solutions are (0,1,0,1,1,1,1) and (1,1,0,1,1,1,1).
%p a(1):=0: a(2):=0: a(3):=0: a(4):=1: a(5):=1:
%p a(6):=1: a(7):=2: pot:=2^3: pk:=0:
%p for n from 4 to 7 do
%p pot:=2*pot:
%p pk:=pk+a(n)/pot:
%p end do:
%p for n from 8 to 100 do
%p pot:=2*pot:
%p a(n):=a(n-1)+a(n-2)+a(n-3)-a(n-5):
%p pk:=pk+a(n)/pot:
%p end do:
%p printf("10.5f",pk):
%p seq(a(n),n=1..100);
%Y Cf. A199594, A199925.
%K nonn,easy
%O 1,7
%A _Paul Weisenhorn_, Nov 12 2011
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