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A087179 a(n) = (...(((x1^x2)^x3)^x4)^...) where x1,x2,... are the exponents in the prime factorization of n, a(1) = 0.

%I A087179 #17 Apr 25 2025 05:30:00
%S A087179 0,1,1,2,1,1,1,3,2,1,1,2,1,1,1,4,1,1,1,2,1,1,1,3,2,1,3,2,1,1,1,5,1,1,
%T A087179 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 A087179 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 A087179 a(n) = (...(((x1^x2)^x3)^x4)^...) where x1,x2,... are the exponents in the prime factorization of n, a(1) = 0.
%H A087179 Antti Karttunen, <a href="/A087179/b087179.txt">Table of n, a(n) for n = 1..16384</a>
%H A087179 MathMedics, <a href="http://www.sosmath.com/tables/factor/factor.html">The Prime Factorization of the First 1000 Integers</a>
%F A087179 If A001221(n) <= 1, a(n) = A067029(n) [i.e., when n is a prime power, p^k, a(n) = k], otherwise a(n) = a(A051119(n)) ^ A071178(n). - _Antti Karttunen_, Aug 27 2017
%e A087179 a(75) = a((3^1)*(5^2)) = 1^2 = 1.
%e A087179 a(108) = a((2^2)*(3^3)) = 2^3 = 8.
%e A087179 a(300) = a(2^2 * 3^1 * 5^2) = (2^1)^2 = 4. - _Antti Karttunen_, Aug 27 2017
%o A087179 (Scheme) (define (A087179 n) (if (<= (A001221 n) 1) (A067029 n) (expt (A087179 (A051119 n)) (A071178 n)))) ;; _Antti Karttunen_, Aug 27 2017
%Y A087179 Cf. A001221, A051119, A067029, A071178.
%Y A087179 After a(1) = 0 differs from A290109 for the next time at n=300.
%K A087179 nonn
%O A087179 1,4
%A A087179 _Sam Alexander_, Oct 19 2003
%E A087179 Term a(1) = 0 prepended by _Antti Karttunen_, Aug 27 2017