

A075777


Minimal surface area of a rectangular solid with volume n and integer sides.


4



6, 10, 14, 16, 22, 22, 30, 24, 30, 34, 46, 32, 54, 46, 46, 40, 70, 42, 78, 48, 62, 70, 94, 52, 70, 82, 54, 64, 118, 62, 126, 64, 94, 106, 94, 66, 150, 118, 110, 76, 166, 82, 174, 96, 78, 142, 190, 80, 126, 90, 142, 112, 214, 90, 142, 100, 158, 178, 238, 94, 246, 190
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OFFSET

1,1


COMMENTS

To find minimum surface area (incorrect, see below!), let s1_0 = [n^(1/3)]. Find largest integer s1 such that s1 <= s1_0 and s1  n. Then let s2_0 = [sqrt(n / s1)]. Find largest integer s2 such that s2 <= s2_0 and s2  (n / s1). Then s3 = n / (s1 * s2). And minimum surface area a(n) = 2 * (s1 * s2 + s1 * s3 + s2 * s3).
The above algorithm is not correct. Consider n=68: algorithm returns (s1,s2,s3) = (4,1,17) for minimal surface area 178. However, (2,2,17) has a surface area of 144. Consider n=246: algorithm gives (6,1,41) but (3,2,41) has a lower surface area. It appears that the algorithm, by picking s1 to be the highest divisor of n, does not get the chance to choose a pair (s1,s2) that is equal or nearly equal. I wrote a Python script for a new algorithm (posted below). However, it is much slower since it loops through every divisor s1 of n and divisor s2 of a given n/s1 while finding and keeping a minimum surface area. (Ceiling is used to avoid a floating point error on the roots.)  Scott B. Farrar, Sep 29 2015


LINKS

Robert Israel, Table of n, a(n) for n = 1..10000
Dan Meyer, The Math Problem that 1000 Math Teachers Couldn't Solve


EXAMPLE

a(12) = 32 because side lengths of 2, 2 and 3 will give volume 12 and surface area 32, which is the minimum surface area.


MAPLE

N:= 1000: # to get a(1) .. a(N)
A:= Vector(N, 1)*6*N;
for p1 from 1 to N do
for p2 from p1 to floor(N/p1) do
for p3 from p2 to floor(N/p1/p2) do
n:= p1*p2*p3;
a:= 2*(p1*p2 + p2*p3 + p1*p3);
if a < A[n] then A[n]:= a fi
od od od:
seq(A[i], i=1..N); # Robert Israel, Sep 30 2015


PROG

(PARI) a(n) = {s1_0 = floor(n^(1/3)); s1 = s1_0; while (n % s1 != 0, s1); s2_0 = floor(sqrt(n/s1)); nds1 = n/s1; s2 = s2_0; while (nds1 % s2 != 0, s2); s3 = n/(s1*s2); return (2 * (s1 * s2 + s1 * s3 + s2 * s3)); } \\ Michel Marcus, Apr 14 2013; script based on 1st algorithm now known to give ok terms up to n=67 only
(PARI) a(n) = {mins = 1; fordiv(n, x, q = n/x; fordiv(q, y, z = q/y; s = 2*(x*y + y*z +x*z); if (mins ==1, mins =s, mins = min(mins, s)); ); ); mins; } \\ Michel Marcus, Sep 30 2015
(Python)
import math
def cubestepdown(n):
s1_0 = int(math.ceil(n ** (1 / 3.0)))
minSA = 1
s1 = s1_0
while s1>=1:
while n % s1 > 0:
s1 = s1  1
s1quot = int(n/s1)
s2_0 = int(math.ceil(math.sqrt(n/s1)))
s2 = s2_0
while s2>=1:
while s1quot % s2 > 0:
s2 = s2  1
s3 = int(n / (s1 * s2))
SA = 2*(s1*s2 + s1*s3 + s2*s3)
if minSA==1:
minSA = SA
else:
if SA<minSA:
minSA = SA
s2 = s2  1
s1 = s1  1
return minSA
# Scott B. Farrar, Sep 29 2015


CROSSREFS

Cf. A135711.
Sequence in context: A228301 A193416 A315160 * A315161 A315162 A207870
Adjacent sequences: A075774 A075775 A075776 * A075778 A075779 A075780


KEYWORD

easy,nonn


AUTHOR

Robert A. Stump (bee_ess107(AT)msn.com), Oct 09 2002


STATUS

approved



