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%I #16 Jan 25 2015 22:13:47
%S 0,0,0,0,0,1,0,0,0,2,0,1,0,2,0,0,0,1,0,1,0,3,0,1,0,3,0,1,0,1,0,0,0,4,
%T 0,1,0,4,0,1,0,1,0,2,0,4,0,1,0,2,0,2,0,1,0,1,0,4,0,1,0,4,0,0,0,1,0,2,
%U 0,1,0,1,0,5,0,2,0,1,0,1,0,5,0,1,0,5,0,1,0,1,0,2,0,5,0,1,0,2,0,1
%N a(n) is the least natural number k such that n^k is abundant or perfect, or a(n) is 0 if all n^k are deficient numbers.
%C Let p_1, p_2, ..., p_m be the distinct primes dividing n. If (p_1/(p_1 - 1))*(p_2/(p_2 - 1))*...*(p_m/(p_m - 1)) > 2, then sufficiently high powers of n are abundant. Otherwise all powers of n are deficient, and we set a(n)=0.
%C The sequence is unbounded. In particular, for each N, we have a(A063765(N)) = N.
%H Jeppe Stig Nielsen, <a href="/A254001/b254001.txt">Table of n, a(n) for n = 1..10000</a>
%e a(38)=4 because 38^4 is abundant (A023196) while 38^3, 38^2, and 38 are all deficient (A005100).
%o (PARI) a(n) = {
%o primeVect = factor(n)[,1];
%o if(prod(i=1,#primeVect,1-1/primeVect[i])>=1/2,return(0));
%o for(k=1,10^99,t=n^k;if(sigma(t)>=2*t,return(k))); }
%K nonn,easy
%O 1,10
%A _Jeppe Stig Nielsen_, Jan 22 2015
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