A224835 - OEIS

%I #31 Nov 10 2017 09:56:38

%S 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 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 17,1,36,9,9,1,44,1,9,9

%N 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 Robert Israel, <a href="/A224835/b224835.txt">Table of n, a(n) for n = 1..10000</a>

%H 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 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 a(n) = (Sum_{d|n} d <= n^(1/3)) tau(d)^3.

%F If p is prime, a(p^k) = A000537(1 + floor(k/3)). - _Robert Israel_, Nov 30 2016

%e 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 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 f:= proc(n) add(numtheory:-tau(d)^3, d = select(t -> (t^3<=n), numtheory:-divisors(n))) end proc:

%p map(f, [$1..100]); # _Robert Israel_, Nov 30 2016

%t 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 (PARI) a(n) = sumdiv(n, d, (d^3<=n)*numdiv(d)^3) \\ _Michel Marcus_, Jul 21 2013

%Y Cf. A000537, A007425, A224834.

%K nonn

%O 1,8

%A _Michel Marcus_, Jul 21 2013