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Numbers k such that sopfr(k)=tau(k).
6

%I #23 Jun 22 2019 09:21:24

%S 2,60,72,168,270,450,560,990,1056,1188,1470,2496,3200,3402,3520,3672,

%T 4788,5700,8320,9856,9936,10164,10395,11000,11160,12180,12740,13056,

%U 13720,14616,16524,17400,17940,18522,19380,20000,20475,21252,21528

%N Numbers k such that sopfr(k)=tau(k).

%C It appears Sum_{n=1..m} a(n) =~ c*sqrt(m)*(log(m) - log(log(m))^2), at c ~ 0.038, asymptotically, after examining m up to 10,000,000, with data intervals of 10,000. - _Richard R. Forberg_, Feb 10 2015

%H Amiram Eldar, <a href="/A078511/b078511.txt">Table of n, a(n) for n = 1..10000</a>

%F A001414(a(n)) = A000005(a(n)).

%t s = {}; Do[If[DivisorSigma[0, n] == Total[Times @@@ FactorInteger[n]], AppendTo[s, n]], {n, 2, 22000}]; s (* _Amiram Eldar_, Jun 22 2019 *)

%o (PARI) isok(n) = {my(f = factor(n)); sum(i=1, #f~, f[i,1]*f[i,2]) == numdiv(n);} \\ _Michel Marcus_, Feb 11 2015

%Y Cf. A000005, A001414.

%K nonn

%O 1,1

%A _Benoit Cloitre_, Jan 11 2003