%I #26 Sep 15 2017 09:39:30
%S 1,6,70,210,2622,9282,277134,1159710,8064030,56185590,186605430,
%T 2748628830,5053814978,72641163166
%N a(n) is the smallest k>1 such that d(n,k)^2 = d(n^2,k^2), where d(n,k) is the n-th divisor of a number k, for n>1; and a(1) = 1.
%C Or a(n) is the smallest number k such that, if d is the n-th divisor of k, then d^2 is the (n^2)-th divisor of k^2.
%C For n <= 14, a(n) is squarefree, and omega(a(n)) < 9. Is a(n) squarefree for all n? - _David A. Corneth_, May 12 2017
%e a(2) = 6 because the divisors of 6 and 36 are {1, 2, 3, 6} and {1, 2, 3, 4, 6, 9, 12, 18, 36} respectively, and the 2nd divisor of 6 is 2, and the 4th divisor of 36 is 2^2. Hence, d(2,6)^2 = d(4,36) = 4.
%t Do[k=1;While[!(Length[Divisors[k]]>=n&&Length[Divisors[k^2]]>=n^2&&Part[Divisors[k],n]^2==Part[Divisors[k^2],n^2]),k++];Print[n," ",k],{n,1,10}]
%o (PARI) a(n) = {if (n==1, return (1)); my(k=2); while (iferr(divisors(k)[n]^2 != divisors(k^2)[n^2], E, 1), k++); k;} \\ _Michel Marcus_, Sep 12 2017
%Y Cf. A027750.
%K nonn,more
%O 1,2
%A _Michel Lagneau_, May 11 2017
%E a(10)-a(12) from _Giovanni Resta_, May 12 2017
%E a(13)-a(14) from _Giovanni Resta_, May 16 2017
%E Name edited by _Michel Marcus_, Sep 15 2017