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%I #24 Sep 18 2019 20:51:39
%S 2,60,1056,1260,1441440,551350800,42226984800,111924212400,
%T 11251629148359600,284440457440339200,582249616380374342400,
%U 621260340677859423340800,621260340677859423340800,921088919608373507667359523840000000
%N Least k>1 such that all powers k^e, 1 <= e <= n, are divisible by the number of their divisors, d(k^e).
%F 2 / d(2) = 2 / 2 = 1 but 2^2 / d(2^2) = 4 / 3;
%F 60 / d(60) = 60 / 12 = 5, 60^2 / d (60^2) = 3600 / 45 = 80 but 60^3 / d(60^3) = 216000 / 112 = 13500 / 7.
%p with(numtheory): P:= proc(q) local a,j,k,ok,n,p; a:=2;
%p for k from 1 to q do for n from a to q do ok:=1;
%p for j from 1 to k do if not type(n^j/tau(n^j),integer) then ok:=0; break; fi; od;
%p if ok=1 then a:=n; print(n); break; fi; od; od; end: P(10^9);
%t Table[SelectFirst[Range[2, 2*10^6], AllTrue[#^Range@ n, Divisible[#, DivisorSigma[0, #]] &] &], {n, 5}] (* _Michael De Vlieger_, May 12 2016, Version 10 *)
%Y Cf. A000005, A272981.
%K nonn
%O 1,1
%A _Paolo P. Lava_, May 12 2016
%E a(6)-a(14) from _Giovanni Resta_, May 12 2016
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