A075777 Minimal surface area of a rectangular solid with volume n and integer sides.

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

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: A193416 A338703 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