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

Author

Mohammed Yaseen, Aug 08 2021

Keywords

• nonn
• hard
• more

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

Examples

			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.
		

Links

• Sean A. Irvine, Java program (github)

Crossrefs

Cf. A261491 (2D analog).

Cf. A000162, A003211, A193416.

Formula

a(n) < A193416(n) for n>2.

Extensions

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