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

4, 7, 15, 31, 65, 136, 285, 597, 1251, 2621, 5492, 11507, 24111, 50519, 105853

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