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A356014 Consider the exponents in the prime factorization of n, and replace each run of k consecutive e's by a unique e; the resulting list corresponds to the exponents in the prime factorization of a(n).

%I A356014 #9 Jul 25 2022 10:43:29
%S A356014 1,2,3,4,3,2,3,8,9,10,3,12,3,10,3,16,3,18,3,20,21,10,3,24,9,10,27,20,
%T A356014 3,2,3,32,21,10,3,4,3,10,21,40,3,10,3,20,45,10,3,48,9,50,21,20,3,54,
%U A356014 21,40,21,10,3,12,3,10,63,64,21,10,3,20,21,10,3,72,3
%N A356014 Consider the exponents in the prime factorization of n, and replace each run of k consecutive e's by a unique e; the resulting list corresponds to the exponents in the prime factorization of a(n).
%C A356014 We ignore the exponents (all 0's) for the prime numbers beyond the greatest prime factor of n.
%C A356014 This sequence operates on prime exponents as A090079 and A337864 operate on binary and decimal digits, respectively.
%F A356014 a(a(n)) = a(n).
%F A356014 a(n^k) = a(n)^k for any k >= 0.
%F A356014 a(n) = A319521(A356008(n)).
%F A356014 A007814(a(n)) = A007814(n).
%F A356014 a(n) = 3 iff n belongs to A294674 \ {1}.
%F A356014 a(n) = 4 iff n belongs to A061742 \ {1}.
%F A356014 a(n) = 8 iff n belongs to A115964.
%e A356014 For n = 99:
%e A356014 - 99 = 11^1 * 7^0 * 5^0 * 3^2 * 2^0,
%e A356014 - the list of exponents is: 1 0 0 2 0,
%e A356014 - compressing consecutive values, we obtain: 1 0 2 0,
%e A356014 - so a(99) = 7^1 * 5^0 * 3^2 * 2^0 = 63.
%o A356014 (PARI) a(n) = { my (v=1, e=-1, k=0); forprime (p=2, oo, if (n==1, return (v), if (e!=e=valuation(n,p), v*=prime(k++)^e); n/=p^e)) }
%Y A356014 Cf. A007814, A061742, A067255, A090079, A115964, A294674, A319521, A337864, A356008, A356021 (fixed points).
%K A356014 nonn
%O A356014 1,2
%A A356014 _Rémy Sigrist_, Jul 23 2022