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A290109 a(1) = 1; for n > 1, a(n) = x1^(x2^(x3^(x4^...))) where x1, x2, ... are the exponents of the primes present (listed from the smallest prime to the largest) in the prime factorization of n.

%I A290109 #6 Aug 28 2017 09:22:34
%S A290109 1,1,1,2,1,1,1,3,2,1,1,2,1,1,1,4,1,1,1,2,1,1,1,3,2,1,3,2,1,1,1,5,1,1,
%T A290109 1,4,1,1,1,3,1,1,1,2,2,1,1,4,2,1,1,2,1,1,1,3,1,1,1,2,1,1,2,6,1,1,1,2,
%U A290109 1,1,1,9,1,1,1,2,1,1,1,4,4,1,1,2,1,1,1,3,1,1,1,2,1,1,1,5,1,1,2,4,1,1,1,3,1,1,1,8,1,1,1,4,1,1,1,2,2,1,1,3
%N A290109 a(1) = 1; for n > 1, a(n) = x1^(x2^(x3^(x4^...))) where x1, x2, ... are the exponents of the primes present (listed from the smallest prime to the largest) in the prime factorization of n.
%H A290109 Antti Karttunen, <a href="/A290109/b290109.txt">Table of n, a(n) for n = 1..16384</a>
%F A290109 a(1) = 1; for n > 1, a(n) = A067029(n) ^ a(A028234(n)).
%e A290109 For n = 300 = 2^2 * 3^1 * 5^2 we have a(300) = 2^(1^2) = 2.
%e A290109 For n = 600 = 2^3 * 3^1 * 5^2 we have a(600) = 3^(1^2) = 3.
%o A290109 (Scheme) (define (A290109 n) (if (= 1 n) 1 (expt (A067029 n) (A290109 (A028234 n))))) ;; _Antti Karttunen_, Aug 27 2017
%Y A290109 Cf. A028234, A067029.
%Y A290109 After a(1) = 1 differs from A087179 for the next time at n=300.
%K A290109 nonn
%O A290109 1,4
%A A290109 _Antti Karttunen_, Aug 27 2017