1,2

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 (maxal(AT)cs.ucsd.edu), Nov 27 2006

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 (j.mccranie(AT)comcast.net), Oct 25 2001

a(n) = a(n-1)+a(n-2)-a(n-3)+1 = A002620(n+2)+A004526(n+2) - Henry Bottomley (se16(AT)btinternet.com), 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.: (2-x)/(1-2*x+2*x^3-x^4). - Vladeta Jovovic (vladeta(AT)Eunet.yu), 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

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.

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: A098393 A103567 A131723 this_sequence A122378 A111242 A133582

Adjacent sequences: A035103 A035104 A035105 this_sequence A035107 A035108 A035109

nonn,nice

njas, revised Oct 30, 2001