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%I #9 Dec 19 2017 02:33:04
%S 0,0,4,6,15,19,33,38,57,64,87,97,124,135,168,181,218,232,274,291,337,
%T 355,407,427,482,504,565,589,654,679,749,777,851,880,959,991,1074,
%U 1107,1196,1230,1323,1360,1458,1496,1599,1639,1746,1788,1900,1944,2060,2106
%N Sum of all the diagonals of the distinct rectangles that can be made with positive integer sides such that L + W = n, W < L. The sums are then rounded to the nearest integer.
%F a(n) = round(2 * Sum_{i=1..floor((n-1)/2)} sqrt(i^2 + (n-i)^2)).
%e a(4) = 6; There is only one 1 X 3 rectangle (there is no 2 X 2 rectangle since W < L) and sqrt(1^2 + 3^2) = sqrt(10). Since there are two diagonals in a rectangle, the total length is 2*sqrt(10). Then we have round(2*sqrt(10)) = round(6.32455532..) = 6.
%t Table[Round[2*Sum[Sqrt[i^2 + (n - i)^2], {i, Floor[(n-1)/2]}]], {n, 80}]
%Y Cf. A050187.
%K nonn,easy
%O 1,3
%A _Wesley Ivan Hurt_, Oct 30 2017
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