%I #47 Apr 27 2019 20:35:04

%S 1,5,7,9,21,22,44,45,66,70,78,112,150,156,160,264,270,280,432,600,

%T 1080,1680

%N Numbers n such that the number of divisors of n equals the integer part of the geometric mean of the divisors of n.

%C Numbers k such that A000005(k) = floor(A007955(k)^(1/A000005(k))).

%C Numbers k such that A000005(k) = A000196(k).

%C Numbers k such that the number of divisors of k equals the number of squares <= k.

%C It is assumed that the sequence is finite.

%C Numbers k such that A000196(k)/A000005(k) = r; r is a rational number. This sequence has r = 1. Does an r exist for which the sequence is infinite? - _Ctibor O. Zizka_, Jan 01 2017

%C The sequence is complete. This follows easily from the upper bound on the number of divisors of k proved by Nicolas & Robin. - _Giovanni Resta_, Jul 30 2018

%H Ilya Gutkovskiy, <a href="/A276734/a276734.pdf">Illustration of dynamics of floor(sqrt(n)) - sigma_0(n)</a>

%H L. Nicolas and G. Robin, <a href="http://dx.doi.org/10.4153/CMB-1983-078-5">Majorations explicites pour le nombre de diviseurs de N</a>, Canadian Mathematical Bulletin 26 (1983), pp. 485-492.

%e a(10) = 70, because 70 has 8 divisors {1, 2, 5, 7, 10, 14, 35, 70} and floor((1*2*5*7*10*14*35*70)^(1/8)) = floor(sqrt(70)) = 8; equivalently, we have 8 squares {1, 4, 9, 16, 25, 36, 49, 64} <= 70.

%t Select[Range[10000], DivisorSigma[0, #1] == Floor[Sqrt[#1]] & ]

%Y Cf. A000005, A000196, A007955, A280235.

%K nonn,fini,full

%O 1,2

%A _Ilya Gutkovskiy_, Oct 03 2016