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A346958
a(n) is the minimal number of cubes required to make a void of volume n.
1
6, 10, 13, 15, 17, 18, 18, 21, 23, 25, 26, 26
OFFSET
1,1
COMMENTS
Following is an illustration of the first few voids in the form of polycubes (where an o represents a continuation upwards and an x represents a continuation downwards) each of which can be made by concealing it with a(n) cubes.
.---. .---.
| | | |
.---. .---.---. .---.---. .---.---.
| | | | | | | | | | o |
.---. .---.---. .---.---. .---.---.
n=1 n=2 n=3 n=4
.---. .---. .---.
| | | | | |
.---.---. .---.---.---. .---.---.---.
| | o | | | o | | | | ox| |
.---.---. .---.---.---. .---.---.---.
| | | | | |
.---. .---. .---.
n=5 n=6 n=7
Equivalently, the minimum perimeter size of any polycube of size n. - Sean A. Irvine, Aug 23 2021
Conjecture: When n is in A001845 the void is an octahedral crystal ball of volume n = A001845(m), which is concealed by a(n) = A005899(m+1) cubes. So a(A001845(m)) = A005899(m+1), m>=0. For example, a(1)=6 and a(7)=18. - Mohammed Yaseen, Sep 15 2022
LINKS
Sean A. Irvine, Java program (github)
FORMULA
a(n) < A193416(n) for n>2.
EXAMPLE
A cube-shaped void can be made by concealing it with 6 cubes, which is the minimal number to do so. So a(1)=6.
A dicube-shaped void can be made by concealing it with 10 cubes, which is the minimal number to do so. So a(2)=10.
CROSSREFS
Cf. A261491 (2D analog).
Sequence in context: A340874 A213428 A006187 * A288222 A004234 A121149
KEYWORD
nonn,hard,more
AUTHOR
Mohammed Yaseen, Aug 08 2021
EXTENSIONS
a(8)-a(12) from Sean A. Irvine, Aug 23 2021
STATUS
approved