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Least k >= 2 such that tau(n) divides tau(n^k), where tau(n) is the number of divisors of n, A000005.
1

%I #9 May 28 2017 09:13:48

%S 2,3,3,4,3,3,3,5,4,3,3,5,3,3,3,6,3,5,3,5,3,3,3,3,4,3,5,5,3,3,3,7,3,3,

%T 3,4,3,3,3,3,3,3,3,5,5,3,3,9,4,5,3,5,3,3,3,3,3,3,3,5,3,3,5,8,3,3,3,5,

%U 3,3,3,13,3,3,5,5,3,3,3,9,6,3,3,5,3,3,3,3,3,5,3,5,3,3,3,5,3,5,5,4,3,3,3,3

%N Least k >= 2 such that tau(n) divides tau(n^k), where tau(n) is the number of divisors of n, A000005.

%H Antti Karttunen, <a href="/A077603/b077603.txt">Table of n, a(n) for n = 1..10000</a>

%o (PARI) A077603(n) = { my(k=2,nd=numdiv(n)); while((numdiv(n^k)%nd),k=k+1); k; } \\ _Antti Karttunen_, May 28 2017

%Y Cf. A000005.

%K nonn

%O 1,1

%A _Benoit Cloitre_, Dec 01 2002