A346958 a(n) is the minimal number of cubes required to make a void of volume n.

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

Table of n, a(n) for n=1..12.

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).

Cf. A000162, A003211, A193416.

Sequence in context: A340874 A213428 A006187 * A288222 A004234 A121149

Adjacent sequences: A346955 A346956 A346957 * A346959 A346960 A346961

Keyword

nonn,hard,more

Author

Mohammed Yaseen, Aug 08 2021

Extensions

a(8)-a(12) from Sean A. Irvine, Aug 23 2021

Status

approved