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%I #18 Dec 25 2017 11:24:14
%S 3,2,3,1,3,1,3,2,1,0,2,0,1,1,3,0,2,0,1,2,0,0,3,1,0,2,1,0,1,0,3,0,0,1,
%T 2,0,0,0,2,0,2,0,0,1,0,0,2,1,1,0,0,0,2,0,1,0,0,0,1,0,0,1,3,0,0,0,0,0,
%U 1,0,2,0,0,1,0,0,0,0,1,2,0,0,2,0,0,0,1,0,1,1,1,0,0
%N Largest number m such that n^m divides tau(n), where tau(n)=A000594(n) is Ramanujan's tau function.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/TauFunction.html">Tau Function.</a>
%e tau(2) = -24 and 2^3 divides 24, so a(2) = 3.
%e tau(3) = 252 and 3^2 divides 252, so a(3) = 2.
%e tau(4) = -1472 and 4^3 divides 1472, so a(4) = 3.
%t f[n_] := Block[{m = 0}, While[Mod[RamanujanTau@n, n^m] == 0, m++]; m - 1]; Array[f, 93, 2] (* _Robert G. Wilson v_, Dec 23 2017 *)
%Y Cf. A063938 (a(n)>=1), A296991 (a(n)>=2), A296993 (a(n)>=3).
%Y Cf. A191599 (a(n)=0), A297000 (a(n)=1), A297001 (a(n)=2).
%K nonn
%O 2,1
%A _Seiichi Manyama_, Dec 22 2017
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