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A035106
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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-1.
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15
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1, 2, 3, 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
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OFFSET
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1,2
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COMMENTS
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Equivalently, smallest integer m such that there exists a permutation (p_1, ..., p_n) of (1, ..., n) satisfying p_i * p_{i+1} <= m for every i, 1 <= i <= n-1.
Nonsquare positive integers m such that [sqrt(m)] divides m. Numbers of the form k*(k+1) or k*(k+2). - Max Alekseyev, Nov 27 2006
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LINKS
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Table of n, a(n) for n=1..57.
Index to sequences with linear recurrences with constant coefficients, signature (2,0,-2,1).
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FORMULA
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Theorem: a(n)=n*(n+2)/4 if n is even and (n-1)*(n+3)/4 if n is odd, n>1. - Jud McCranie, Oct 25 2001
a(n) = a(n-1)+a(n-2)-a(n-3)+1 = A002620(n+2)+A004526(n+2) - Henry Bottomley, Mar 08 2000
a(n+2) = (2*n^2+12*n+3*(-1)^n+13)/8, i.e. a(n+2) = (n+2)*(n+4)/4 if n is even and (n+1)*(n+5)/4 if n is odd. G.f.: x + x^2*(-2+x) / ( (1+x)*(x-1)^3 ). - Vladeta Jovovic, Oct 23 2001
{a[n] == a[n - 2] + (n - 1), a[1] == 0, a[2] == 0}; (-4 - 3*(-1)^n - (-1)^(2*n) + 2*n - 2*(-1)^(2*n)*n + 2*n^2)/8 - Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 14 2004
a(n)= floor((n+1)^4/(4*(n+1)^2+1))...n>1 [Gary Detlefs, Feb 11 2010]
a(n) = n + ceil((1/4)*(n-1)^2) - 1, n >= 2. - Clark Kimberling, Jan 07 2011; corrected by Arkadiusz Wesolowski, Sep 25 2012
a(1)=1, a(2)=2, a(3)=3, a(4)=6, a(5)=8, a(n)=2*a(n-1)-2*a(n-3)+a (n-4) [Harvey P. Dale, May 03 2012]
G.f.: (x^4-2*x^3+x^2-1)/((x-1)^3*(x+1)) [From Harvey P. Dale, May 03 2012]
a(n) = floor(n/2)*(1 + ceil(n/2)), a(1) = 1. - Arkadiusz Wesolowski, Sep 25 2012
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EXAMPLE
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n=5: we must arrange the numbers 1..5 so that the max of the products of pairs of adjacent terms is minimized. The answer is 51324, with max product = 8, so a(5) = 8.
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MAPLE
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seq(floor((n+1)^4/(4*(n+1)^2+1)), n= 2..25); [From Gary Detlefs, Feb 11 2010]
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MATHEMATICA
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Join[{1}, LinearRecurrence[{2, 0, -2, 1}, {2, 3, 6, 8}, 60]] (* or *) Join[{1}, Table[ If[EvenQ[n], (n(n+2))/4, ((n-1)(n+3))/4], {n, 2, 60}]] (* Harvey P. Dale, May 03 2012 *)
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CROSSREFS
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Cf. A064764, A064796, A064797, A064817, A004652, A035104, A035107.
First differences give (essentially) A028242.
Bisections: A002378 (pronic numbers) and A005563.
Cf. A002378, A006446.
Sequence in context: A103567 A131723 A198442 * A122378 A181687 A194881
Adjacent sequences: A035103 A035104 A035105 * A035107 A035108 A035109
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KEYWORD
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nonn,nice,changed
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AUTHOR
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N. J. A. Sloane, revised Oct 30, 2001
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STATUS
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approved
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