A128127 The number of regular pentagons found by constructing n equally-spaced points on each side of the pentagon and drawing lines parallel to the pentagon side.
1, 3, 9, 21, 37, 59
Offset: 0
Examples
With 0 point, there is only 1 pentagon, the original one. With 1 point (a midpoint on each side), 3 regular pentagons are found. With two points, 9 regular pentagons are found in total.
Links
- Michel Marcus, Figure with 2 points on each side
- Noah Priluck, On Counting Regular Polygons Formed by Special Families of Parallel Lines, Geombinatorics Quarterly, Vol XVII (4), 2008, pp. 166-171. (note there is no document to download).
- Noah Priluck, On Counting Regular Polygons Formed by Special Families of Parallel Lines
Crossrefs
Cf. A128153 (same construction but with pentagon vertices also connected).
Programs
-
PARI
ldraw(w, vx, vy, np, with, ia, ib, jb, ja) = {if (with, kdeb = 0; kend = np, kdeb = 1; kend = np-1;); for (k=kdeb, kend, plotmove(w, vx[ia]+k*(vx[ib]-vx[ia])/np, vy[ia]+k*(vy[ib]-vy[ia])/np); plotlines(w, vx[ja]+k*(vx[jb]-vx[ja])/np, vy[ja]+k*(vy[jb]-vy[ja])/np););} modnv(i, nv) = {i = i % nv; if (i == 0, i = nv); return (i);} poly(nv, np, with) = {w = 2; s = plothsizes(); plotinit(w, s[1]-1, s[2]-1); plotscale(w, 0, 1000, 0, 1000); xc = 500; yc = 500; vx = vector(nv, i, xc + 500*sin(i*2*Pi/nv)); vy = vector(nv, i, yc + 500*cos(i*2*Pi/nv)); plotlines(w ,vx, vy, 1); plotmove(w, vx[nv], vy[nv]); plotlines(w, vx[1], vy[1]); np++; for (ia=1, nv, ia = modnv(ia, nv); ib = modnv(ia+1, nv); for (ja=1, nv, ja = modnv(ja, nv); if (ja != ia, jb = modnv(ja+1, nv); ldraw(w, vx, vy, np, with, ia, ib, ja, jb);););); plotdraw([w, 0, 0]); return(0);} \\ use poly(5, n, 0) to get figure with n points \\ Michel Marcus, Jul 09 2013
Formula
a(n) = (10*n^2 - 4*n + 5 -(-1)^n)/4 (conjectural).
Extensions
Edited by Michel Marcus, Jul 09 2013
a(4) and a(5) from Michel Marcus, Jul 21 2013
Comments