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%I #16 Sep 30 2014 17:23:35
%S 2,3,2,5,8,7,2,3,32,11,8,13,128,243,2,17,8,19,32,2187,2048,23,8,5,
%T 8192,3,128,29,
%U 14134776518227074636666380005943348126619871175004951664972849610340958208,31,2,177147,131072,78125,8,37,524288,1594323,32,41
%N a(n)=p1^(p2^(p3^(p4^...)))... where p1<p2<p3<... are the distinct prime factors of n.
%H Reinhard Zumkeller, <a href="/A248012/b248012.txt">Table of n, a(n) for n = 2..65</a>, 65 = 2*3*11-1
%e To find a(14) we first find the distinct prime factors of 14 which are 2 and 7, which leads to a(14)=2^7=128.
%e To find a(8) we find 8's prime factors, 8=2*2*2, the distinct prime factor is 2 therefore a(8)=2.
%e 30 has 3 distinct prime factors {2,3,5}, so a(30)=2^(3^5)=14134776518227074636666380005943348126619871175004951664972849610340958208.
%o (Haskell)
%o a248012 = foldr1 (^) . a027748_row -- _Reinhard Zumkeller_, Sep 29 2014
%Y Cf. A027748.
%K nonn
%O 2,1
%A _Talha Ali_, Sep 29 2014
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