A224835 Sum of the cubes of the number of divisors function for those divisors of n that are less than or equal to the cube root of n.
1, 1, 1, 1, 1, 1, 1, 9, 1, 9, 1, 9, 1, 9, 1, 9, 1, 9, 1, 9, 1, 9, 1, 9, 1, 9, 9, 9, 1, 17, 1, 9, 9, 9, 1, 17, 1, 9, 9, 9, 1, 17, 1, 9, 9, 9, 1, 17, 1, 9, 9, 9, 1, 17, 1, 9, 9, 9, 1, 17, 1, 9, 9, 36, 1, 17, 1, 36, 9, 9, 1, 44, 1, 9, 9
Offset: 1
Keywords
Examples
a(7) = 1 because the divisors of 7 are 1 and 7; only 1 is less than the cube root of 7, and tau(1^3) = 1, so the sum is 1. a(8) = 9 because the divisors of 8 are 1, 2, 4, 8; the cube root of 8 is 2, so only 1 and 2 are divisors less than or equal to the cube root, these divisors cubed are 1 and 8, which add up to 9.
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
- Sary Drappeau, Propriétés multiplicatives des entiers friables translatés, arXiv:1307.4250 [math.NT] (see page 9).
- B. Landreau, A New Proof of a Theorem of Van Der Corput, Bull. London Math. Soc. (1989) 21 (4): 366-368. doi: 10.1112/blms/21.4.366, see Lemma (3) page 1.
Programs
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Maple
f:= proc(n) add(numtheory:-tau(d)^3, d = select(t -> (t^3<=n), numtheory:-divisors(n))) end proc: map(f, [$1..100]); # Robert Israel, Nov 30 2016
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Mathematica
Table[selDivs = Select[Range[Floor[n^(1/3)]], IntegerQ[n/#]&]; Sum[DivisorSigma[0, selDivs[[m]]]^3, {m, Length[selDivs]}], {n, 100}] (* Alonso del Arte, Jul 21 2013 *) -
PARI
a(n) = sumdiv(n, d, (d^3<=n)*numdiv(d)^3) \\ Michel Marcus, Jul 21 2013
Formula
a(n) = (Sum_{d|n} d <= n^(1/3)) tau(d)^3.
If p is prime, a(p^k) = A000537(1 + floor(k/3)). - Robert Israel, Nov 30 2016