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A362706
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Number of squares formed by first n vertices of the infinite-dimensional hypercube.
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1
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0, 0, 0, 1, 1, 2, 3, 6, 6, 7, 9, 13, 16, 21, 27, 36, 36, 37, 40, 45, 50, 57, 66, 78, 85, 94, 106, 121, 136, 154, 175, 200, 200, 201, 205, 211, 219, 229, 242, 258, 271, 286, 305, 327, 351, 378, 409, 444, 463, 484, 510, 539, 571, 606, 646, 690, 729, 771, 819
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OFFSET
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1,6
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COMMENTS
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We can take the coordinates of a vertex to represent a binary number, so we define the n-th point to have coordinates represented by the binary expansion of n-1.
Let d(m) = a(m+1) - a(m) be the shifted first differences of a(n), so that d(m) represents the additional squares introduced by the (m+1)-th vertex. Then d(0) = d(2^x) = 0; when m = 2^x + 2^y, x > y, d(m) = A115990(x - 1, x - y - 1); generally, d(m) = sum d(k) for all k formed by selecting two 1's from the binary expansion of m. Thus d(7) = d(3) + d(5) + d(6).
a(n) is a lower bound for an infinite-dimensional extension of A051602. Peter Munn notes that it is not an upper bound: for example, the vertices of a regular {k-1}-simplex duplicated at unit distance in any orthogonal direction gives T_k squares from 2k+2 points, which exceeds a(n) at 6, 10 and 12 points.
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LINKS
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FORMULA
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EXAMPLE
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The 6 points (0,0,0), (1,0,0), (0,1,0), (1,1,0), (0,0,1), (1,0,1) give the squares (0,0,0), (1,0,0), (0,1,0), (1,1,0) and (0,0,0), (1,0,0), (0,0,1), (1,0,1). So a(6) = 2.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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