A261956 Start with a single equilateral triangle for n=0; for the odd n-th generation add a triangle at each expandable vertex of the triangles of the (n-1)-th generation (this is the "side to vertex" version); for the even n-th generation use the "side to side" version; a(n) is the number of triangles added in the n-th generation.
1, 3, 6, 9, 12, 18, 15, 21, 21, 36, 39, 54, 36, 54, 39, 57, 45, 72, 63, 90, 60, 90, 63, 93, 69, 108, 87, 126, 84, 126, 87, 129, 93, 144, 111, 162, 108, 162, 111, 165, 117, 180, 135, 198, 132, 198, 135, 201, 141, 216, 159
Offset: 0
Keywords
Links
- Kival Ngaokrajang, Illustration of initial terms
Programs
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PARI
{e=12; o=18; print1("1, 3, 6, 9, ", e, ", ", o, ", "); for(n=6, 100, if (Mod(n,2)==0, if (Mod(n,8)==6, e=e+3); if (Mod(n,8)==0, e=e+6); if (Mod(n,8)==2, e=e+18); if (Mod(n,8)==4, e=e-3); print1(e, ", "), if (Mod(n,8)==7, o=o+3); if (Mod(n,8)==1,o=o+15); if (Mod(n,8)==3, o=o+18); if (Mod(n,8)==5, o=o+0); print1(o, ", ")))}
Formula
Conjectures from Colin Barker, Sep 10 2015: (Start)
G.f.: -(9*x^13 +9*x^12 -12*x^11 -13*x^10 -12*x^9 -5*x^8 -3*x^7 -3*x^6 -9*x^5 -6*x^4 -6*x^3 -5*x^2 -3*x -1) / ((x-1)^2*(x+1)^2*(x^2+1)*(x^4+1)).
a(n) = a(n-2) + a(n-8) - a(n-10) for n > 13. (End)
Comments