|
%I #11 Nov 21 2016 06:21:17
%S 2,4,8,16,32,64,128,256,432,1024,864,4096,1728,2592,3456,65536,6912,
%T 262144,10368,14400,27648,4194304,21600,32400,110592,50400,43200,
%U 268435456,64800,1073741824,86400,230400,1769472,129600,151200,68719476736,7077888,921600
%N Least number k such that the number of its divisors is n times, with n>1, the number of its prime factors, counted without multiplicity.
%C The number of divisors of p^(n-1) is n times the number of prime factors of p^(n-1), where p is prime. It means that a solution exists for every n>1.
%H Giovanni Resta, <a href="/A275816/b275816.txt">Table of n, a(n) for n = 2..100</a>
%F Least solution k of the equation A000005(k) = n * A001221(k).
%e a(10) = 432 because the number of divisors of 432 is 20, the number of different prime factors of 432 is 2 (2, 3), and 20 = 10 * 2.
%p with(numtheory): P:=proc(q) local k,n; for n from 2 to q do for k from 1 to 2^(n-1) do
%p if tau(k)=n*nops(factorset(k)) then print(k); break; fi; od; od; end: P(10^9);
%t a[n_] := Block[{k = 2}, If[PrimeQ[n], 2^n/2, While[ DivisorSigma[0, k]/ PrimeNu[k] != n, k++]; k]]; a /@ Range[2, 25] (* _Giovanni Resta_, Nov 16 2016 *)
%Y Cf. A000005, A001221, A275819.
%K nonn,easy
%O 2,1
%A _Paolo P. Lava_, Nov 15 2016
%E a(29)-a(39) from _Giovanni Resta_, Nov 16 2016
|
