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Let c^i(n) be the number of Khalimsky-continuous functions f from [0,n-1]_Z to [0,3]_Z such that f(n-1)=i for i=0,1,2,3 and let a(n) be their sum. Then a(n) = a(n-1)+2a(n-2)+c^1(n-3)+c^2(n-3)
The sequence is determined by the above recurrence together with the following recurrences:
c^0(2k + 1) = c^0(2k) + c^1(2k),
c^1(2k + 1) = c^1(2k),
c^2(2k + 1) = c^1(2k) + c^2(2k) + c^3(2k),
c^3(2k + 1) = c^3(2k) and
c^0(2k) = c^0(2k - 1),
c^1(2k) = c^0(2k - 1) + c^1(2k - 1) + c^2(2k - 1),
c^2(2k) = c^2(2k - 1),
c^3(2k) = c^2(2k - 1) + c^3(2k - 1).
For the asymptotic behavior, (c^1(n)+c^2(n))/(c^1(n-1)+c^2(n-1)), (c^0(n)+c^3(n))/(c^0(n-1)+c^3(n-1)) ans a(n)/a(n-1) all tend to 1/2( sqrt(7+ sqrt(5)+ sqrt(38+14 sqrt(5)))) =~ 2.095293985.
G.f.: x*(4 + 3*x - 4*x^2 - x^3) / (1 - x - 3*x^2 + x^3 + x^4).
a(n) = a(n-1) + 3*a(n-2) - a(n-3) - a(n-4) for n>4.
(End) [Since we have an explicit set of recurrences that produce a(n), it should be straightforward to prove these conjectures. - N. J. A. Sloane, Jan 14 2018]
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