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 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 (list; graph; refs; listen; history; text; internal format)
 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 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

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