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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 (list; graph; refs; listen; history; text; internal format)
 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

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Last modified September 10 17:29 EDT 2024. Contains 375792 sequences. (Running on oeis4.)