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A239352
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van Heijst's upper bound on the number of squares inscribed by a real algebraic curve in R^2 of degree n, if the number is finite.
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3
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0, 0, 1, 12, 48, 130, 285, 546, 952, 1548, 2385, 3520, 5016, 6942, 9373, 12390, 16080, 20536, 25857, 32148, 39520, 48090, 57981, 69322, 82248, 96900, 113425, 131976, 152712, 175798, 201405, 229710, 260896, 295152, 332673, 373660, 418320, 466866, 519517
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OFFSET
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0,4
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COMMENTS
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In 1911 Toeplitz conjectured the Square Peg (or Inscribed Square) Problem: Every continuous simple closed curve in the plane contains 4 points that are the vertices of a square. The conjecture is still open. Many special cases have been proved; see Matschke's beautiful 2014 survey.
Recently van Heijst proved that any real algebraic curve in R^2 of degree d inscribes either at most (d^4 - 5d^2 + 4d)/4 or infinitely many squares. He conjectured that a generic complex algebraic plane curve inscribes exactly (d^4 - 5d^2 + 4d)/4 squares.
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REFERENCES
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Otto Toeplitz, Über einige Aufgaben der Analysis situs, Verhandlungen der Schweizerischen Naturforschenden Gesellschaft in Solothurn, 4 (1911), 197.
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LINKS
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FORMULA
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a(n) = (n^4 - 5*n^2 + 4*n)/4 = n*(n - 1)*(n^2 + n - 4)/4 = A000217(n-1)*A034856(n-1), which shows the formula is an integer.
G.f.: x^2 * (1 + 7*x - 2*x^2) / (1 - x)^5. - Michael Somos, Mar 21 2014
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EXAMPLE
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A point or a line has no inscribed squares, so a(0) = a(1) = 0.
A circle has infinitely many inscribed squares, and an ellipse that is not a circle has exactly one, agreeing with a(2) = 1.
G.f. = x^2 + 12*x^3 + 48*x^4 + 130*x^5 + 285*x^6 + 546*x^7 + 952*x^8 + ...
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MATHEMATICA
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Table[(n^4 - 5 n^2 + 4 n)/4, {n, 0, 38}]
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PROG
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(PARI) for(n=0, 50, print1((n^4 - 5*n^2 + 4*n)/4, ", ")) \\ G. C. Greubel, Aug 07 2018
(Magma) [(n^4 - 5*n^2 + 4*n)/4: n in [0..50]]; // G. C. Greubel, Aug 07 2018
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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