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A279847 a(n) = Sum_{k=1..n} k^2*(floor(n/k) - 1).

%I A279847 #19 Jan 11 2025 03:19:43
%S A279847 0,1,2,7,8,22,23,44,54,84,85,151,152,206,241,326,327,458,459,605,664,
%T A279847 790,791,1065,1091,1265,1356,1622,1623,2023,2024,2365,2496,2790,2865,
%U A279847 3480,3481,3847,4026,4636,4637,5373,5374,6000,6341,6875,6876,7982,8032,8787,9086,9952,9953,11137,11284
%N A279847 a(n) = Sum_{k=1..n} k^2*(floor(n/k) - 1).
%C A279847 Sum of all squares of proper divisors of all positive integers <= n.
%C A279847 Total volume of all rectangular prisms with dimensions (x, x, z) and integers x and y, such that x + y = n, 0 < x <= y, and z = floor(y/x). - _Wesley Ivan Hurt_, Dec 21 2020
%H A279847 <a href="/index/Su#sums_of_divisors">Index entries for sequences related to sums of divisors</a>
%F A279847 G.f.: -x*(1 + x)/(1 - x)^4 + (1/(1 - x))*Sum_{k>=1} k^2*x^k/(1 - x^k).
%F A279847 a(n) = A064602(n) - A000330(n).
%F A279847 a(n) = Sum_{k=1..n} A067558(k).
%F A279847 a(n) = Sum_{k=1..n} (A001157(k) - A000290(k)).
%F A279847 a(p^k) = a(p^k-1) + (p^(2*k) - 1)/(p^2 - 1), when p is prime.
%F A279847 a(n) ~ ((zeta(3) - 1)/3)*n^3.
%F A279847 a(n) = Sum_{k=1..floor(n/2)} k^2 * floor((n-k)/k). - _Wesley Ivan Hurt_, Dec 21 2020
%e A279847 For n = 7 the proper divisors of the first seven positive integers are {0}, {1}, {1}, {1, 2}, {1}, {1, 2, 3}, {1} so a(7) = 0^2 + 1^2 + 1^2 + 1^2 + 2^2 + 1 ^2 + 1^2 + 2^2 + 3^2 + 1^2 = 23.
%t A279847 Table[Sum[k^2 (Floor[n/k] - 1), {k, 1, n}], {n, 55}]
%t A279847 Table[Sum[DivisorSigma[2, k] - k^2, {k, 1, n}], {n, 55}]
%o A279847 (PARI) a(n) = sum(k=1, n, k^2*(floor(n/k)-1)) \\ _Felix Fröhlich_, Dec 20 2016
%o A279847 (Python)
%o A279847 from math import isqrt
%o A279847 def A279847(n): return (-n*(n+1)*(2*n+1)-(s:=isqrt(n))**2*(s+1)*(2*s+1) + sum((q:=n//k)*(6*k**2+q*(2*q+3)+1) for k in range(1,s+1)))//6 # _Chai Wah Wu_, Oct 21 2023
%Y A279847 Cf. A000290, A000330, A001157, A064602, A067558, A153485.
%K A279847 nonn,easy
%O A279847 1,3
%A A279847 _Ilya Gutkovskiy_, Dec 20 2016