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%I #15 Sep 19 2019 11:06:22
%S 1,2,0,1,0,1,2,1,0,1,2,0,1,2,1,0,1,2,0,1,0,1,2,0,1,2,1,0,1,2,0,1,0,1,
%T 2,1,0,1,2,0,1,0,1,2,0,1,2,1,0,1,2,0,1,0,1,2,1,0,1,2,0,1,2,1,0,1,2,0,
%U 1,0,1,2,1,0,1,2,0,1,0,1,2,0,1,2,1,0
%N Start with 0 and repeatedly substitute 0->10, 1->12, 2->0.
%C The fixed point of the morphism 0->10, 1->12, 2->0. Let u be the sequence of positions of 0, and likewise, v for 1 and w for 2. Let U, V, W be the limits of u(n)/n, v(n)/n, w(n)/n, respectively. It appears that 1/U + 1/V + 1/W = 1, where
%C U = 3.079595623491438786010417...,
%C V = 2.324717957244746025960908...,
%C W = U + 1 = 4.079595623491438786010417....
%C From _Michel Dekking_, Sep 15 2019: (Start)
%C The incidence matrix of the morphism sigma: 0->10, 1->12, 2->0 has characteristic polynomial chi(u) = u^3-2u^2+u-1. The real root of chi is lambda := Q/6 + 2/3*1/Q + 2/3, where
%C Q = ( 100 + 12*sqrt(69) )^1/3.
%C An eigenvector of lambda is (1, lambda^2-lambda, lambda-1).
%C The Perron-Frobenius Theorem gives that the asymptotic frequencies f0, f1 and f2 of the letters 0, 1, and 2 are
%C f0 = 1/lambda^2,
%C f1 = (lambda^2 - lambda +1)/lambda^3,
%C f2 = (lambda - 1)/lambda^2.
%C Algebraic expressions for the constants U,V and W are then given by
%C U = 1/f0, V = 1/f1, W = 1/f2.
%C In particular, this shows that W = U + 1.
%C (End)
%C Conjecture: if n >=2, then u(n) - u(n-1) is in {2,3,4}, v(n) - v(n-1) is in {2,3}, and w(n) - w(n-1) is in {3,4,5}.
%C See A287105, A287106, and A287107 for proofs of these conjectures, with explicit expressions for u, v, and w. - _Michel Dekking_, Sep 15 2019
%H Clark Kimberling, <a href="/A287104/b287104.txt">Table of n, a(n) for n = 1..10000</a>
%H <a href="/index/Fi#FIXEDPOINTS">Index entries for sequences that are fixed points of mappings</a>
%t s = Nest[Flatten[# /. {0 -> {1, 0}, 1 -> {1, 2}, 2 -> 0}] &, {0}, 10] (* A287104 *)
%t Flatten[Position[s, 0]] (* A287105 *)
%t Flatten[Position[s, 1]] (* A287106 *)
%t Flatten[Position[s, 2]] (* A287107 *)
%Y Cf. A287105, A287106, A287107.
%K nonn,easy
%O 1,2
%A _Clark Kimberling_, May 21 2017
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