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A356016 Consider the exponents in the prime factorization of n, and replace each run of k consecutive e's by a unique k; the resulting list corresponds to the exponents in the prime factorization of a(n).

%I A356016 #9 Jul 25 2022 10:43:24
%S A356016 1,2,6,2,12,4,24,2,6,30,48,6,96,90,18,2,192,6,384,30,210,270,768,6,12,
%T A356016 810,6,90,1536,8,3072,2,1050,2430,36,4,6144,7290,5250,30,12288,60,
%U A356016 24576,270,30,21870,49152,6,24,30,26250,810,98304,6,420,90,131250
%N A356016 Consider the exponents in the prime factorization of n, and replace each run of k consecutive e's by a unique k; the resulting list corresponds to the exponents in the prime factorization of a(n).
%C A356016 We ignore the exponents (all 0's) for the prime numbers beyond the greatest prime factor of n.
%C A356016 There are only two fixed points: a(1) = 1 and a(2) = 2.
%C A356016 Iterating the sequence starting from any n > 1 will always eventually reach the fixed point 2.
%F A356016 a(n) = A319522(A356008(n)).
%F A356016 a(n^k) = a(n) for any k > 0.
%F A356016 a(n) = 2 iff n is a power of 2 > 1.
%F A356016 a(n) = 4 iff n is a power of 6 > 1.
%F A356016 a(n) = 2^k iff n is a power of A002110(k) > 1 (with k > 0).
%F A356016 a(prime(n)) = 3*2^(n-1) for any n > 1.
%e A356016 For n = 99:
%e A356016 - 99 = 11^1 * 7^0 * 5^0 * 3^2 * 2^0,
%e A356016 - the list of exponents is: 1 0 0 2 0,
%e A356016 - the run lengths are: 1 2 1 1,
%e A356016 - so a(99) = 7^1 * 5^2 * 3^1 * 2^1 = 1050.
%o A356016 (PARI) a(n) = { my (v=1, e=-1, k=0, r=0); forprime (p=2, oo, if (n==1, return (v*if (r, prime(k++)^r, 1)), if (e!=e=valuation(n,p), if (r, v*=prime(k++)^r; r=0)); r++; n/=p^e)) }
%Y A356016 Cf. A002110, A067255, A318928, A319522, A356008, A356014.
%K A356016 nonn
%O A356016 1,2
%A A356016 _Rémy Sigrist_, Jul 23 2022