%I #18 Oct 11 2020 10:30:58
%S 27,125,135,162,169,189,250,289,297,343,351,361,375,459,500,513,529,
%T 621,675,686,729,783,810,837,841,845,875,945,961,972,999,1000,1024,
%U 1029,1107,1125,1134,1161,1183,1250,1269,1323,1331,1369,1372,1375,1431,1445,1485
%N Numbers n such that d(n*k) = d(k^2) where k < n, is soluble (d = A000005).
%C The values of the least k such that d(n*k) = d(k^2) are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 12, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, ...
%C Perfect power terms in this sequence are 27, 125, 169, 289, 343, 361, 529, 729, 841, 961, 1000, 1024, 1331, 1369, 1681, 1849, 2187, 2197, 2209, ...
%C From _Robert Israel_, Aug 08 2016: (Start)
%C No terms are squarefree.
%C Contains p^2 where p is a prime >= 13 (with k = 144).
%C Contains p^3 where p is an odd prime (with k = 4 p).
%C Contains p^4 where p is a prime >= 11 (with k = 3600).
%C If n is in the sequence, with d(n*k) = d(k^2), and GCD(n*k,m) = 1, then n*m is in the sequence. (End)
%H Robert Israel, <a href="/A275430/b275430.txt">Table of n, a(n) for n = 1..1000</a>
%e 27 is a term because d(27*k) = d(k^2) with k = 12.
%p f:= proc(n) local k,r,S;
%p S:= select(t -> t[2]::odd, isqrfree(n)[2]);
%p r:= mul(t[1],t=S);
%p for k from r to n-1 by r do
%p if numtheory:-tau(n*k)=numtheory:-tau(k^2) then return true fi
%p od;
%p false
%p end proc:
%p select(f, [$1..2000]); # _Robert Israel_, Aug 08 2016
%t f[n_] := Module[{k = 1}, While[DivisorSigma[0, k n] != DivisorSigma[0, k^2], k++]; k];
%t Reap[For[n = 1, n <= 1500, n++, If[f[n] < n, Print[n]; Sow[n]]]][[2, 1]] (* _Jean-François Alcover_, Oct 11 2020, after PARI *)
%o (PARI) a(n) = {my(k = 1); while (numdiv(k*n) != numdiv(k^2), k++); k; }
%o lista(nn) = for(n=1, nn, if(a(n) < n, print1(n, ", ")));
%Y Cf. A000005.
%K nonn
%O 1,1
%A _Altug Alkan_, Jul 27 2016