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 A064796 Largest integer m such that every permutation (p_1, ..., p_n) of (1, ..., n) satisfies p_i * p_{i+1} >= m for some i, 1 <= i <= n, where p_{n+1} = p_1. 5
 1, 2, 6, 8, 12, 15, 20, 24, 30, 35, 42, 48, 56, 63, 72, 80, 90, 99, 110, 120, 132, 143, 156, 168, 182, 195, 210, 224, 240, 255, 272, 288, 306, 323, 342, 360, 380, 399, 420, 440, 462, 483, 506, 528, 552, 575, 600, 624, 650, 675, 702, 728, 756, 783, 812, 840, 870 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Conjecture: a(n) = (n+1)(n+3)/4 for odd n, a(n) = (n)(n+4)/4 for even n. - Jud McCranie, Oct 25 2001 LINKS Vincenzo Librandi, Table of n, a(n) for n = 1..10000 Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1). FORMULA For odd n > 2, a(n) = (n+1)(n+3)/4. For even n > 2, a(n) = n(n+4)/4. - David Wasserman, Aug 19 2002 a(n) = 2*a(n-1)-2*a(n-3)+a(n-4) for n>6. G.f.: -x*(x^5-x^4-2*x^3+2*x^2+1) / ((x-1)^3*(x+1)). - Colin Barker, Aug 28 2013 For n>2: a(n)=1/8*(3-3*(-1)^n+8*n+2*n^2). - Tom Edgar, Aug 28 2013 EXAMPLE n=5: we must arrange the numbers 1..5 in a circle so that the max of the products of pairs of adjacent terms is minimized. The answer is 15243, with max product = 12, so a(5) = 12. MATHEMATICA Join[{1, 2}, LinearRecurrence[{2, 0, -2, 1}, {6, 8, 12, 15}, 60]] (* Harvey P. Dale, Sep 17 2013 *) PROG (PARI) a(n)=if(n<3, n, if(n%2, (n+1)*(n+3), (n+4)*n)/4) \\ Charles R Greathouse IV, Feb 19 2017 (MAGMA) [1, 2] cat [1/8*(3-3*(-1)^n+8*n+2*n^2): n in [3..60]]; // Vincenzo Librandi, Feb 24 2017 CROSSREFS Cf. A064764, A035106, A064797. Sequence in context: A190344 A287000 A282358 * A304483 A268177 A083769 Adjacent sequences:  A064793 A064794 A064795 * A064797 A064798 A064799 KEYWORD nonn,nice,easy AUTHOR N. J. A. Sloane, Oct 21 2001 EXTENSIONS More terms from Naohiro Nomoto and Vladeta Jovovic, Oct 22 2001 More terms from David Wasserman, Aug 19 2002 STATUS approved

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Last modified September 19 23:31 EDT 2019. Contains 327207 sequences. (Running on oeis4.)