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A159344
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Number of Hamiltonian cycles in the n-hypercube up to automorphism.
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4
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OFFSET
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1,4
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COMMENTS
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Here we count equivalence classes under the full automorphism group of the n-cube. - N. J. A. Sloane, Sep 06 2012
a(4) is due to Gilbert and a(5) is due to Dejter & Delgado.
a(n) is, in Abbott's terminology, h*(n); see (2) and (3) which yield a(n) >= sqrt(294)^(2^n-4)/(n! * 2^n) [Note that we have written sqrt(294) for 7 sqrt(6)].
Unfortunately, the lower bound seems incompatible with the known values of a(n), even for a(3) and a(4) which were known to Abbott.
Looking at Abbot's paper, at least one error is the claim "it is easy to verify that (12) implies (3)."
(12) is h(m+3) >= 6^2^m h(m), which implies h(m) >= 6^2^(m-3) for m >= 4, or h(m) >= 2/5 * (6^2^(m-3)) for m >= 1, but certainly doesn't imply (3) h(m) >= (7 sqrt(6))^(2^n-4). (End)
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LINKS
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FORMULA
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a(n) < n^(2^n).
a(n) >= sqrt(294)^(2^n-4)/(n! * 2^n) and a(n) >= A066037(n)/A000165(n) due to Abbott 1966. [Warning: see Comments above!]
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EXAMPLE
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There are six Hamiltonian cycles in the cube, but they are isomorphic so a(3) = 1.
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CROSSREFS
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KEYWORD
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nonn,hard,more,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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