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Total area of all incongruent integer-sided rectangles of area <= n.
7

%I #26 Jul 11 2023 19:31:05

%S 0,1,3,6,14,19,31,38,54,72,92,103,139,152,180,210,258,275,329,348,408,

%T 450,494,517,613,663,715,769,853,882,1002,1033,1129,1195,1263,1333,

%U 1513,1550,1626,1704,1864,1905,2073,2116,2248,2383,2475,2522,2762,2860

%N Total area of all incongruent integer-sided rectangles of area <= n.

%H Nick MacKinnon, <a href="https://www.jstor.org/stable/2695719">Problem 10883</a>, Amer. Math. Monthly, 108 (2001) 565; <a href="https://www.jstor.org/stable/3647894">solution</a> by John C. Cock, 110 (2003) 343-344.

%F a(n) = Sum_{k=1..n} k*ceiling(d(k)/2), where d(k)=A000005(k) is the number of divisors of k.

%F a(n) = Sum_{r=1..floor(sqrt(n))} r*(r+floor(n/r))*(floor(n/r)+1-r)/2.

%F a(n) = ( A143127(n) + A000330(floor(sqrt(n))) ) / 2. - _Max Alekseyev_, Jan 31 2012

%F a(n) ~ n^2 * log(n) / 4

%F G.f.: x*f'(x)/(1 - x), where f(x) = Sum_{k>=1} x^k^2/(1 - x^k). - _Ilya Gutkovskiy_, Apr 12 2017

%e a(5)=19, the rectangles being 1 X 1, 1 X 2, 1 X 3, 1 X 4, 1 X 5 and 2 X 2.

%t a[n_] := Sum[r(r+Floor[n/r])(Floor[n/r]+1-r), {r, 1, Floor[Sqrt[n]]}]/2

%o (Python)

%o from math import isqrt

%o def A083356(n): return (k:=isqrt(n))*(k+1)*(2+4*k-3*k*(k+1))//24+sum(i*(m:=n//i)*(1+m)>>1 for i in range(1,k+1)) # _Chai Wah Wu_, Jul 11 2023

%Y Cf. A083357, A143127

%Y Partial sums of A060872.

%K nonn,easy

%O 0,3

%A _Dean Hickerson_, Apr 26 2003