A064602 Partial sums of A001157: Sum_{j=1..n} sigma_2(j).
1, 6, 16, 37, 63, 113, 163, 248, 339, 469, 591, 801, 971, 1221, 1481, 1822, 2112, 2567, 2929, 3475, 3975, 4585, 5115, 5965, 6616, 7466, 8286, 9336, 10178, 11478, 12440, 13805, 15025, 16475, 17775, 19686, 21056, 22866, 24566, 26776, 28458, 30958
Offset: 1
Keywords
Links
- Seiichi Manyama, Table of n, a(n) for n = 1..10000 (terms 1..1000 from T. D. Noe)
- Project Euler, Problem 401: Sum of squares of divisors, 2012.
Programs
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Mathematica
Accumulate@ Array[DivisorSigma[2, #] &, 42] (* Michael De Vlieger, Jan 02 2017 *)
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PARI
a(n) = sum(j=1, n, sigma(j, 2)); \\ Michel Marcus, Dec 15 2013
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PARI
f(n) = n*(n+1)*(2*n+1)/6; \\ A000330 a(n) = my(s=sqrtint(n)); sum(k=1, s, f(n\k) + k^2*(n\k)) - s*f(s); \\ Daniel Suteu, Nov 26 2020
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Python
from math import isqrt def f(n): return n*(n+1)*(2*n+1)//6 def a(n): s = isqrt(n) return sum(f(n//k) + k*k*(n//k) for k in range(1, s+1)) - s*f(s) print([a(k) for k in range(1, 43)]) # Michael S. Branicky, Oct 01 2022 after Daniel Suteu
Formula
a(n) = Sum_{i=1..n} i^2 * floor(n/i). - Enrique Pérez Herrero, Sep 15 2012
G.f.: (1/(1 - x))*Sum_{k>=1} k^2*x^k/(1 - x^k). - Ilya Gutkovskiy, Jan 02 2017
a(n) ~ zeta(3) * n^3 / 3. - Vaclav Kotesovec, Sep 02 2018
a(n) = Sum_{k=1..s} (A000330(floor(n/k)) + k^2*floor(n/k)) - s*A000330(s), where s = floor(sqrt(n)). - Daniel Suteu, Nov 26 2020
Comments