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A369366
Least number of inequivalent cells in a polyiamond with n cells.
6
1, 1, 2, 2, 3, 1, 3, 3, 3, 3, 6, 2, 4, 4, 4, 4, 9, 2, 5, 6
OFFSET
1,3
COMMENTS
Equivalently, a(n) is the least number of free rooted (or pointed) polyiamonds (A369365) corresponding to a given polyiamond with n cells.
From Peter Exley, Apr 07 2026: (Start)
a(19) = 5 and a(20) = 6 were computed by exhaustive enumeration of all free polyiamonds combined with Burnside orbit counting under the dihedral group D_6. Two independent verifiers with disjoint orbit-counting code confirmed all values for n = 1..20.
The maximum possible symmetry group order for a polyiamond depends on n mod 6. For n = 6k+5, no rotational symmetry is possible (n = 2 mod 3 blocks 3-fold rotation; n odd blocks 180-degree rotation), giving |G_max| = 2 and a(6k+5) >= ceiling(n/2) = 3(k+1). For n = 6k+2, 180-degree rotation is available but 3-fold rotation is not, giving |G_max| = 4 (the Klein 4-group) and a(6k+2) >= ceiling(n/4).
The extremal 19-iamonds (achieving a(19) = 5) have D_3 symmetry (|G| = 6), while the extremal 20-iamonds (achieving a(20) = 6) have Klein 4-group symmetry (|G| = 4). The number of minimizers (consistent with the companion sequence A369367): 2 for n = 19, 17 for n = 20. Our minimizer counts were verified against all 18 known A369367 terms.
a(19) = 5 and a(20) = 6 break the constant-block pattern seen for k = 1, 2: at those blocks a(6k+1) through a(6k+4) were all equal (3,3,3,3 and 4,4,4,4), but at k = 3 the first two positions diverge.
Conjecture: a(6k+5) = 3(k+1) for all k >= 0. Observed for k = 0, 1, 2 (n = 5, 11, 17). The next test case is a(23), predicted to be 12.
(End)
CROSSREFS
KEYWORD
nonn,more
AUTHOR
EXTENSIONS
a(18) from John Mason, Sep 20 2024
a(19)-a(20) from Peter Exley, Apr 07 2026
STATUS
approved