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A131935 a(n) is the number of Khalimsky-continuous functions with four-point codomain and an n-point range.

%I A131935 #10 Jan 14 2018 00:12:47
%S A131935 4,7,15,31,65,136,285,597,1251,2621,5492,11507,24111,50519,105853
%N A131935 a(n) is the number of Khalimsky-continuous functions with four-point codomain and an n-point range.
%H A131935 Shiva Samieinia, <a href="http://www.math.su.se/reports/2007/6/">Digital straight line segments and curves</a>. Licentiate Thesis. Stockholm University, Department of Mathematics, Report 2007:6.
%F A131935 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)
%F A131935 The sequence is determined by the above recurrence together with the following recurrences:
%F A131935 c^0(2k + 1) = c^0(2k) + c^1(2k),
%F A131935 c^1(2k + 1) = c^1(2k),
%F A131935 c^2(2k + 1) = c^1(2k) + c^2(2k) + c^3(2k),
%F A131935 c^3(2k + 1) = c^3(2k) and
%F A131935 c^0(2k) = c^0(2k - 1),
%F A131935 c^1(2k) = c^0(2k - 1) + c^1(2k - 1) + c^2(2k - 1),
%F A131935 c^2(2k) = c^2(2k - 1),
%F A131935 c^3(2k) = c^2(2k - 1) + c^3(2k - 1).
%F A131935 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.
%F A131935 Conjectures from _Colin Barker_, Jan 13 2018: (Start)
%F A131935 G.f.: x*(4 + 3*x - 4*x^2 - x^3) / (1 - x - 3*x^2 + x^3 + x^4).
%F A131935 a(n) = a(n-1) + 3*a(n-2) - a(n-3) - a(n-4) for n>4.
%F A131935 (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]
%Y A131935 Cf. A131887.
%K A131935 nonn,more
%O A131935 1,1
%A A131935 Shiva Samieinia (shiva(AT)math.su.se), Oct 05 2007, Oct 09 2007
%E A131935 a(11)-a(15) from _Neo Scott_, Jan 12 2018