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Least k such that the sum of the distinct prime divisors of k equals m^n for some m > 1.
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%I #8 May 13 2013 01:49:55

%S 2,14,15,39,87,183,2071,1255,1527,3063,28546,12279,106327,49143,

%T 622231,1113823,1703767,1310695,9961111,3145719,29360002,12582903,

%U 218103418,50331639,2046816631,335544295,9932108986,23890747663,1610612727

%N Least k such that the sum of the distinct prime divisors of k equals m^n for some m > 1.

%C The sequence A063869 gives the least k such that sigma(k)=m^n for some m>1.

%C In this sequence, except n=2 -> m=3, the program gives m = 2 for n = 1 to 30.

%C All terms are squarefree. [_Charles R Greathouse IV_, Aug 19 2011]

%F a(n)=Min{x : A008472 (x)= m^n} for some m.

%e a(11) = 28546 because the sum of the distinct prime divisors {2, 7, 2039} is 2048 = 2^11.

%p with(numtheory):for n from 1 to 12 do:ii:=0:for k from 1 to 1000000 while(ii=0) do: ii:=0:x:=factorset(k):p1:=sum(x[i],i=1..nops(x)):jj:=0:for m from 2 to 10 while(jj=0) do :if p1=m^n then ii:=1:jj:=1: printf ( "%d %d \n", n, k):else fi:od:od:od:

%Y Cf. A008472 A063869.

%K nonn

%O 1,1

%A _Michel Lagneau_, Aug 19 2011