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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.)