A261953 Start with a single equilateral triangle for n=0; for the odd n-th generation add a triangle at each expandable side of the triangles of the (n-1)-th generation (this is the "side to side" version); for the even n-th generation use the "vertex to vertex" version; a(n) is the number of triangles added in the n-th generation.
1, 3, 9, 12, 18, 21, 27, 30, 36, 39, 45, 48, 54, 57, 63, 66, 72, 75, 81, 84, 90, 93, 99, 102, 108, 111, 117, 120, 126, 129, 135, 138, 144, 147, 153, 156, 162, 165, 171, 174, 180, 183, 189, 192, 198, 201, 207, 210, 216, 219, 225, 228, 234, 237, 243, 246, 252
Offset: 0
Examples
If the graph G in the comment by Manfred Boergens has 5 nearest neighbors there are at most 21 vertices in G (conjectured; it is proved that there are G with 5 nearest neighbors and 21 vertices but it is not yet proved that 21 is the maximum). - _Manfred Boergens_, Sep 21 2021
Links
- Kival Ngaokrajang, Illustration of initial terms
- Manfred Boergens, Next-neighbours
- Index entries for linear recurrences with constant coefficients, signature (1,1,-1).
Programs
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Mathematica
Join[{1}, Table[If[OddQ[n], (9 n - 3)/2, 9 n/2], {n, 1, 100}]] (* Manfred Boergens, Sep 21 2021 *)
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PARI
{a=3; print1("1, ", a, ", "); for(n=2, 100, if (Mod(n,2)==0, a=a+6, a=a+3); print1(a, ", "))}
Formula
a(0)=1, a(1)=3; for n >= 2, a(n) = a(n-1) + 6, if mod(n,2) = 0, otherwise a(n) = a(n-1) + 3.
From Colin Barker, Sep 10 2015: (Start)
a(n) = (3*(-1+(-1)^n+6*n))/4.
a(n) = a(n-1)+a(n-2)-a(n-3) for n>3.
G.f.: (x^3+5*x^2+2*x+1) / ((x-1)^2*(x+1)). (End)
a(n) = 3 * A032766(n) for n>=1. - Michel Marcus, Sep 13 2015
a(0)=1; for n >= 1, a(n) = 9n/2 for even n, a(n) = (9n-3)/2 for odd n. - Manfred Boergens, Sep 21 2021
Comments