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 A233699 Ideal rectangle side length for packing squares with side 1/n. 1
 7, 7, 3, 9, 2, 0, 8, 8, 0, 2, 1, 7, 8, 7, 1, 7, 2, 3, 7, 6, 6, 8, 9, 8, 1, 9, 9, 9, 7, 5, 2, 3, 0, 2, 2, 7, 0, 6, 2, 7, 3, 9, 8, 8, 1, 4, 4, 8, 1, 5, 8, 1, 2, 5, 2, 8, 2, 6, 6, 9, 8, 7, 5, 2, 4, 4, 0, 0, 8, 9, 6, 4, 4, 8, 3, 8, 4, 1, 0, 4, 8, 6 (list; constant; graph; refs; listen; history; text; internal format)
 OFFSET 0,1 COMMENTS With one side s_1 = 1/2+1/3 = 5/6, and with area A = s_1*s_2 = sum(n=2,infinity, 1/n^2) = Pi^2/6 - 1 = A013661 - 1, the second side, s_2, can be solved. The current packing record holder is Marc Paulhus, who developed a packing algorithm (see Link). LINKS G. C. Greubel, Table of n, a(n) for n = 0..10000 M. M. Paulhus, An Algorithm for Packing Squares, Journal of Combinatorial Theory,1998, A,82(2), pages 147-157. Pegg Jr, Ed., Wolfram Demonstrations Project, Packing Squares with Side 1/n Wikipedia, Packing Squares with Side 1/n FORMULA Equals (Pi^2-6)/5 = A164102/10 - 6/5. EXAMPLE 0.77392088021787172376689819997523022706273988144815812528266987524400896448... MATHEMATICA RealDigits[(Pi^2-6)/5, 10, 120][[1]] (* Harvey P. Dale, Aug 21 2017 *) PROG (PARI) (Pi^2-6)/5; (MAGMA) C := ComplexField(); (Pi(C)^2-6)/5 // G. C. Greubel, Jan 26 2018 CROSSREFS Sequence in context: A212299 A193751 A290565 * A220052 A153204 A199508 Adjacent sequences:  A233696 A233697 A233698 * A233700 A233701 A233702 KEYWORD nonn,cons AUTHOR John W. Nicholson, Dec 15 2013 STATUS approved

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Last modified April 20 04:17 EDT 2018. Contains 302772 sequences. (Running on oeis4.)