# [seqfan] Corrections to A072368

martin_n_fuller martin_n_fuller at btinternet.com
Fri Jul 21 15:16:40 CEST 2023

```A072368 "Minimal total volume of n bricks with integer sides, all sides
being different. Lowest value of sum of products of triples p*q*r chosen
from [1,3n]."

I have found lower values for A072368(12) and A072368(14).  Can anyone
confirm that these are the only corrections?  My method is described
below along with some proofs.
There are further corrections and many missing sums in the "Illustration
of initial terms" linked to A072368.  Should I replace it with my file
(up to n=33) or should I add a comment on the existing file that some
terms are not the lowest value or are incomplete?

a(12) <= 36021 (was 36022)
1*35*36 + 2*33*34 + 3*31*32 + 4*28*29 + 5*22*30 + 6*21*26 + 7*19*25 +
8*17*24 + 9*16*23 + 10*12*27 + 11*15*20 + 13*14*18

a(14) <= 65269 (was 65270)
1*41*42 + 2*39*40 + 3*37*38 + 4*35*36 + 5*30*34 + 6*26*33 + 7*24*31 +
8*20*32 + 9*21*27 + 10*22*23 + 11*16*29 + 12*17*25 + 13*14*28 + 15*18*19
and 5 other sums

Also improvements for n=26, 31, 32, 33 in the "Illustration of initial
terms".  My calculations only go to n=33.
a(26) <= 729244 (was 729245)
a(31) <= 1456176 (was 1456177)
a(32) <= 1650123 (was 1650124)
a(33) <= 1862831 (was 1862833)
The worst case for missing sums is a(30).  "Illustration of initial
terms" gives 66 optimal sums but my program finds 1943.  NB My file size
with n=30 is 700 KB, file size up to n=29 is 120 KB.

Method: My program uses the branch and bound algorithm.  A branch is to
try each possible product that contains the smallest remaining number.
The bound for the remaining set is defined as follows.
Let S be the set of 3m remaining numbers, and define a(S) to be the
minimal sum of products of triples covering S.  Claim that the following
are lower bounds:
LB1: a(S) >= m * product(S)^(1/m).
LB2: Let p be the smallest element in S and let q<r be the largest 2
elements in S.  Let S2 = S\{p,q,r}.  If pqr < product(S)^(1/m), then
a(S) >= pqr + (m-1) * product(S2)^(1/(m-1))
LB3: If LB2 applies, then let t be the smallest element in S2 and let
u<v be the largest elements.  Let S3 = S2\{t,u,v}.  If tuv <
product(S2)^(1/(m-1)), and tu > pr, then a(S) >= pqr + tuv + (m-2) *
product(S3)^(1/(m-2))

Proof for LB1:
This is based on the AM-GM inequality. The geometric mean of the
products is product(S)^(1/m), which is a lower bound for the arithmetic
mean. This gives the stated bound for the sum.
OEIS currently states this for the initial case: a(n) >=
ceiling(n*(3n!)^(1/n)).  This is exact for n=1..3, but by n=19 the gap
is over 9000.

Proof for LB2:
Let x be the unknown product containing the smallest element of S.  Then
AM-GM inequality says that the a(S) >= x + (m-1) *
(product(S)/x)^(1/(m-1)).
Treat product(S) and m as fixed and differentiate wrt x.  Can show that
RHS is strictly decreasing for x < product(S)^(1/n), and strictly
increasing for x > product(S)^(1/n).
Therefore to minimise RHS, want to take x as close as possible to
product(S)^(1/n).
This bound is exact for the initial case for n<=6.  At n=19 the gap is
3207.

Proof for LB3:
Let a<b<c<d<e<f.  Claim that aef+bcd is the minimal
sum-of-products-of-triples iff bc>=af.
There are 10 sums-of-products-of-triples (there are 5-choose-2 pairs
that can be multiplied by a).  aef+bcd is one option; the other 9 can be
formed by swapping one letter between the summands.
Consider swapping a and b, leaving ef and cd untouched.  aef+bcd minimal
requires 0 <= (bef+acd) - (aef+bcd) = ef(b-a) + cd(a-b) = (ef-cd)(b-a).
This is true by the initial assumption a<b<c<d<e<f.
Same result swapping a and either c or d, because we get a similar
inequality, and cd>bd>bc and d>c>b from the initial assumption.
Now consider swapping e and d, leaving af and bc untouched.  aef+bcd
minimal requires 0 <= (adf+bce) - (aef+bcd) = bc(e-d) + af(d-e) =
(bc-af)(e-d).  e-d>0 by initial assumption. Therefore require bc>=af.
Initial assumption gives that cd>bd>bc and af>ae.  Therefore bc>=af
implies that aef+bcd is minimal compared to all remaining swaps.
This bound is exact for the initial case for n<=10 except n=7.  At n=19
the gap is 1008.

Unproved extension to LB3:  It would be nice to repeat LB3 as many times
as it applies, but I have no proof that this greedy algorithm is
globally optimal.  The proof for LB3 only shows that it is pairwise
optimal.
This greedy algorithm is exact for n=1..6 and 8..11, and is less than
the best known value by no more than 3 up to n=33.  My program uses this
as an initial upper bound for the branch and bound algorithm.

Martin Fuller
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