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 A131935 a(n) is the number of Khalimsky-continuous functions with four-point codomain and an n-point range. 4
 4, 7, 15, 31, 65, 136, 285, 597, 1251, 2621, 5492, 11507, 24111, 50519, 105853 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 LINKS Table of n, a(n) for n=1..15. Shiva Samieinia, Digital straight line segments and curves. Licentiate Thesis. Stockholm University, Department of Mathematics, Report 2007:6. FORMULA 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. 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(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] CROSSREFS Cf. A131887. Sequence in context: A116969 A131090 A178615 * A119749 A201498 A145970 Adjacent sequences: A131932 A131933 A131934 * A131936 A131937 A131938 KEYWORD nonn,more AUTHOR Shiva Samieinia (shiva(AT)math.su.se), Oct 05 2007, Oct 09 2007 EXTENSIONS a(11)-a(15) from Neo Scott, Jan 12 2018 STATUS approved

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Last modified July 13 02:27 EDT 2024. Contains 374260 sequences. (Running on oeis4.)