cp's OEIS Frontend

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.

%I A224835 #31 Nov 10 2017 09:56:38
%S A224835 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,
%T A224835 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,
%U A224835 17,1,36,9,9,1,44,1,9,9
%N 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.
%H A224835 Robert Israel, <a href="/A224835/b224835.txt">Table of n, a(n) for n = 1..10000</a>
%H A224835 Sary Drappeau, <a href="http://arxiv.org/abs/1302.4318">Propriétés multiplicatives des entiers friables translatés</a>, arXiv:1307.4250 [math.NT] (see page 9).
%H A224835 B. Landreau, <a href="http://blms.oxfordjournals.org/content/21/4/366.extract">A New Proof of a Theorem of Van Der Corput</a>, Bull. London Math. Soc. (1989) 21 (4): 366-368. doi: 10.1112/blms/21.4.366, see Lemma (3) page 1.
%F A224835 a(n) = (Sum_{d|n} d <= n^(1/3)) tau(d)^3.
%F A224835 If p is prime, a(p^k) = A000537(1 + floor(k/3)). - _Robert Israel_, Nov 30 2016
%e A224835 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.
%e A224835 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.
%p A224835 f:= proc(n) add(numtheory:-tau(d)^3, d = select(t -> (t^3<=n), numtheory:-divisors(n))) end proc:
%p A224835 map(f, [$1..100]); # _Robert Israel_, Nov 30 2016
%t A224835 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 *)
%o A224835 (PARI) a(n) = sumdiv(n, d, (d^3<=n)*numdiv(d)^3) \\ _Michel Marcus_, Jul 21 2013
%Y A224835 Cf. A000537, A007425, A224834.
%K A224835 nonn
%O A224835 1,8
%A A224835 _Michel Marcus_, Jul 21 2013