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A318743
a(n) = Sum_{k=1..n} floor(n/k)^4.
7
1, 17, 83, 274, 644, 1396, 2502, 4388, 6919, 10743, 15385, 22407, 30233, 41209, 53853, 70650, 88636, 113308, 138654, 172332, 207984, 252416, 298002, 358654, 417873, 492065, 569061, 663427, 756053, 875541, 989063, 1130915, 1272967, 1441383, 1607147, 1817080
OFFSET
1,2
LINKS
FORMULA
a(n) = -A006218(n) + 4*A024916(n) - 6*A064602(n) + 4*A064603(n).
a(n) ~ zeta(4) * n^4.
a(n) ~ Pi^4 * n^4 / 90.
G.f.: (1/(1 - x)) * Sum_{k>=1} (2*k - 1) * (2*k^2 - 2*k + 1) * x^k/(1 - x^k). - Ilya Gutkovskiy, Jul 16 2019
MATHEMATICA
Table[Sum[Floor[n/k]^4, {k, 1, n}], {n, 1, 40}]
Accumulate[Table[-DivisorSigma[0, k] + 4*DivisorSigma[1, k] - 6*DivisorSigma[2, k] + 4*DivisorSigma[3, k], {k, 1, 40}]]
PROG
(PARI) a(n) = sum(k=1, n, (n\k)^4); \\ Michel Marcus, Sep 03 2018
(Magma) [&+[Floor(n/k)^4:k in [1..n]]:n in [1..36]]; // Marius A. Burtea, Jul 16 2019
(Python)
from math import isqrt
def A318743(n): return -(s:=isqrt(n))**5+sum((q:=n//k)*(k**4-(k-1)**4+q**3) for k in range(1, s+1)) # Chai Wah Wu, Oct 26 2023
KEYWORD
nonn
AUTHOR
Vaclav Kotesovec, Sep 02 2018
STATUS
approved