A253895 Total number of octagons in two variants of an octagon expansion after n iterations: either "side-to-side" or "vertex-to-vertex", respectively.
1, 3, 7, 14, 25, 41, 63, 90, 120, 154, 192, 233, 278, 328, 382, 439, 500, 566, 636, 709, 786, 868, 954, 1043, 1136, 1234, 1336, 1441, 1550, 1664, 1782, 1903, 2028, 2158, 2292, 2429, 2570, 2716, 2866, 3019, 3176, 3338, 3504, 3673, 3846, 4024, 4206, 4391, 4580, 4774, 4972
Offset: 1
Keywords
Links
- Kival Ngaokrajang, Illustration of initial terms, Rare type polygons
Crossrefs
Cf. A253896, A061777 (Triangle expansion, vertex-to-vertex, 3 vertices), A179178 (Triangle expansion, side-to-side, 2 sides), A253687 (Pentagon expansion, side-to-side, 2 consecutive sides and 1 isolated side), A253688 (Pentagon expansion, vertex-to-vertex, 2 consecutive vertices and 1 isolated vertex), A253547 (Hexagon expansion, vertex-to-vertex, 2 vertices separated by 1 vertex).
Programs
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PARI
{ a=1;d1=0;p=a;print1(a,", ");\\8s2a, total oct. for(n=2,100, if(n<=7,d1=n-1, if(n<9,d1=5, if(n<10,d1=3, if(n<11,d1=4, if(Mod(n,4)==0,d1=3, if(Mod(n,4)==1,d1=4, if(Mod(n,4)==2,d1=5,d1=4 ) ) ) ) ) ) ); a=a+d1;p=p+a; print1(p,", ") ) }
Formula
Conjectures from Colin Barker, Jan 17 2015: (Start)
a(n) = (-4-i*(-i)^n+i*i^n-18*n+8*n^2)/4 for n>8, where i=sqrt(-1).
G.f.: -x*(x^12-2*x^10-x^8+2*x^6+2*x^5+2*x^4+x^3+2*x^2+1) / ((x-1)^3*(x^2+1)).
(End)
Comments