%I #31 Aug 19 2018 07:37:21
%S 1,2,3,4,5,6,7,8,8,10,8,12,12,14,8,16,16,18,12,20,18,22,16,24,18,26,
%T 27,27,27,27,24,27,32,27,27,36,36,27,36,40,36,42,36,27,45,45,36,48,48,
%U 48,48,45,27,54,48,48,50,58,48,60,60,36,60,64,48,64,64,60,64,63,64
%N Choose the lexically first tuple of six nonincreasing positive integers (a, b, c, d, e, f) such that a*b*c + d*e*f = n. Then a(n) = a*b*c.
%C Previous name: Numbers n such that n is the sum of the volumes of two rectangular cuboids, abc + def where a >= b >= c >= d >= e >= f >= 1. a(n) = abc. (Additional constraints below)
%C In the case of multiple solutions:
%C a is made as small as possible  then
%C b is made as small as possible  then
%C c is made as small as possible  then
%C ...
%C f is made as small as possible.
%C a(n) = abc
%C <Calculations done by hand  can someone please confirm before posting>
%H Charlie Neder, <a href="/A274016/b274016.txt">Table of n, a(n) for n = 2..10000</a>
%H David A. Corneth, <a href="/A274016/a274016.gp.txt">PARI program</a>
%e a(33) = 27 because 3*3*3 + 3*2*1 = 33.
%e a(33) != 32 because although 4*4*2 + 1*1*1 = 33 in the case of multiple solutions, you must choose a minimal value for a.
%o (PARI) See Corneth link \\ _David A. Corneth_, Aug 14 2018
%o (Python)
%o #by _Charlie Neder_, using an algorithm from _David A. Corneth_, Aug 14 2018
%o limit = 10000
%o res = [0 for i in range(limit1)]
%o a = 1
%o while not all(i > 0 for i in res):
%o ..for b in range(1,a+1):
%o ....for c in range(1,b+1):
%o ......for d in range(1,c+1):
%o ........for e in range(1,d+1):
%o ..........for f in range(1,e+1):
%o ............if a*b*c + d*e*f in range(2,limit+1):
%o ..............if not res[a*b*c + d*e*f  2]:
%o ................res[a*b*c + d*e*f  2] = a*b*c
%o ..a += 1
%o for i in range(limit1):
%o ..print(i+2, res[i])
%K nonn
%O 2,2
%A _Gordon Hamilton_, Jun 06 2016
%E New title, corrected a(32) and more terms added by _Charlie Neder_, Aug 13 2018
