

A193416


Minimum surface area of polycubes with volume n.


0



6, 10, 14, 16, 20, 22, 24, 24, 28, 30, 32, 32, 36, 38, 40, 40, 42, 42, 46, 48, 50, 50, 52, 52, 54, 54, 54, 58, 60, 62, 62, 64, 64, 66, 66, 66, 70, 72, 74, 74, 76, 76, 78, 78, 78, 80, 80, 80, 84, 86, 88, 88, 90, 90, 92, 92, 92, 94, 94, 94, 96, 96, 96, 96, 100
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OFFSET

1,1


COMMENTS

First differences are 0, 2, or 4.  Charles R Greathouse IV, Aug 25 2011
From Juan Barajas Martin, Aug 27 2011  Sep 03 2011, Sep 26 2001: (Start)
Initial cubic shape polycube edge = e; volume = e^3; minimum surface area = 6e^2.
Target cubic shape polycube edge = e+1; volume = (e+1)^3; minimum surface area = 6(e+1)^2.
The target polycube is achieved by the successive addition of 3 orthogonal layers of unit cubes to the faces of the initial polycube. The number of unit cubes added in each successive layer take values e^2, e*(e+1) and (e+1)^2, with every layer fully completed before starting the following layer.
This is so because the first unit cube added on every layer will add a surface area of 4 to the previous polycube, which is greater than keeping the new unit cube in the current layer, where we ensure that area will increase by 2 at maximum.
Every layer is constructed as an Ulam spiral, starting by the center and until the completion of the layer.
Except for the first cube placed on every layer which will increase the surface area of the previous polycube by 4, we can count 2 surface area increments for the unit cubes placed at the positions given by the function of the Quartersquares + 1 sequence (i.e., A033638 = A002620 + 1, which set the locations of right angle turns in Ulam square spiral).
All other positions in the Ulam spiral will increase no area to the previous polycube.
An infinite sequence of polycubes with minimal surface area is approached between pairs of successive polycubes with cubic shape and edge values (e) and (e+1), having respectively polycube volumes e^3 and (e+1)^3 and minimum surface areas 6e^2 and 6(e+1)^2.
(End)


LINKS

Table of n, a(n) for n=1..65.
Eric Weisstein's World of Mathematics, Polycube
Eric Weisstein's World of Mathematics, Ulam Spiral


FORMULA

a(n^3) = 6n^2, a(n) ~ 6n^(2/3).  Charles R Greathouse IV, Aug 25 2011
From Juan Barajas Martin, Aug 28 2011: (Start)
The following formula is derived from the Mathematica program below:
smin[n]={6 Floor[(1+n)^(1/3)]^2+If[n==1,4,If[kz[n]>az[n]^2+az[n] (az[n]+1),12,If[kz[n]>az[n]^2,8,4]]]+If[n>1,2 (c1[n]+c2[n]+c3[n]),2]}
az[n_]:Floor[Power[n1, (3)^1]]
kz[n_]:=naz(n)^3
c1c3: number of unit cubes increasing the surface area by 2 in every layer (see Comments above). (End)


EXAMPLE

The unique polycube of volume 1 is a cube with surface area 6, so a(1) = 6. There are eight polycubes of volume 4, of which seven have surface area 18 and one has surface area 16, so a(4) = 16.  Charles R Greathouse IV, Aug 25 2011


MATHEMATICA

vals=100;
az[n_]:=Floor[(n1)^(1/3)];
kz[n_]:=naz[n]^3;
av[n_]:=6*az[n]^2;
bv[n_]:=If[n==1, 4, If[kz[n]>az[n]^2+(az[n]+1)*az[n], 12, If[kz[n]>az[n]^2, 8, 4]]];
pz[n_]:=If[kz[n]<az[n]^2, kz[n], az[n]^2];
qz[n_]:=If[kz[n]<az[n]^2+(az[n]+1)*az[n], kz[n]pz[n], az[n]*(az[n]+1)];
rz[n_]:=If[kz[n]<(az[n]+1)^2+az[n]*(az[n]+1)+az[n]^2, kz[n]pz[n]qz[n], (az[n]+1)^2];
f[j_]:=Ceiling[j/2]*Floor[j/2]+1;
Do[c1[g]=0; p=0; c1[1]=0; Do[If[f[j]<=pz[g], p++], {j, 2, pz[g]}]; c1[g]=p; , {g, 2, vals}];
Do[c2[g]=0; p=0; c2[1]=0; Do[If[f[j]<=qz[g], p++], {j, 2, qz[g]}]; c2[g]=p; , {g, 2, vals}];
Do[c3[g]=0; p=0; c3[1]=0; Do[If[f[j]<=rz[g], p++], {j, 2, rz[g]}]; c3[g]=p; , {g, 2, vals}];
cv[n_]:=If[n>1, 2*(c1[n]+c2[n]+c3[n]), 2];
smin[n_]:=av[n]+bv[n]+cv[n];
Table[smin[n], {n, vals}] (* Juan Barajas Martin, Sep 01 2011 *)


CROSSREFS

Cf. A033638 (A002620 + 1) identify the unit cubes which will increase the minimum area by 2 (locations of right angle turns in Ulam square spiral).
Sequence in context: A183072 A225704 A228301 * A075777 A207870 A248632
Adjacent sequences: A193413 A193414 A193415 * A193417 A193418 A193419


KEYWORD

nonn


AUTHOR

Juan Barajas Martin, Jul 25 2011


STATUS

approved



