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

Author

Noah Priluck (npriluck(AT)gmail.com), May 02 2007

Keywords

• more
• nonn

Comments

A similar pattern of construction to A000330 (dividing a square), A002717 (dividing a triangle), dividing a hexagon and any other polygon in a similar fashion (sequences pending).

Use 1 midpoint (resp. 2 points) on each side placed to divide each side into 2 (resp. 3) equally-sized segments or so on, do the same construction for every side of the pentagon so that each side is equally divided in the same way. Connect all such points to each other with lines that are parallel to at least 1 side of the polygon.

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