Let c^i(n) be the number of Khalimskycontinuous functions f from [0,n1]_Z to [0,3]_Z such that f(n1)=i for i=0,1,2,3 and let a(n) be their sum. Then a(n) = a(n1)+2a(n2)+c^1(n3)+c^2(n3)
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(n1)+c^2(n1)), (c^0(n)+c^3(n))/(c^0(n1)+c^3(n1)) ans a(n)/a(n1) all tend to 1/2( sqrt(7+ sqrt(5)+ sqrt(38+14 sqrt(5)))) =~ 2.095293985.
Conjectures from Colin Barker, Jan 13 2018: (Start)
G.f.: x*(4 + 3*x  4*x^2  x^3) / (1  x  3*x^2 + x^3 + x^4).
a(n) = a(n1) + 3*a(n2)  a(n3)  a(n4) 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]
