>>266
>p(k+3) = (1/3((p(k+2)+p(k+1)+p(k))
q(k+2)=p(k+3)-p(k+2)=-2/3(p(k+2)-p(k+1))-1/3(p(k+1)-p(k))=-2/3q(k+1)-1/3q(k)
3q(k+2)+2q(k+1)+q(k)=0
3t^2+2t+1=0
t=(-1±i√2)/3=t±
q(k)=A(t+)^k+B(t-)^k
p(k)-p(0)=A(1-(t+)^(k+1))/(1-t+)+B(1-(t-)^(k+1))/(1-t-)
p(∞)-1=A/(1-t+)+B/(1-t-)=A/(5/3+t-)+B/(5/3+t+)=(A(5/3+t+)+B(5/3+t-))/(5/35/3+(t++t-)+t+t-)=(5/3q(0)+q(1))/(25/9-2/3+1/3)=(5/3(p(1)-p(0))+(p(2)-p(1)))/(22/9)=(p(2)+2/3p(1)-5/3p(0))/(22/9)=(4/9+2/31/3-5/3)/(22/9)=(4+2-15)/22=-9/22
p(∞)=1-9/22=11/22=1/2