A356014 - OEIS

%I #9 Jul 25 2022 10:43:29

%S 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 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 21,40,21,10,3,12,3,10,63,64,21,10,3,20,21,10,3,72,3

%N 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 We ignore the exponents (all 0's) for the prime numbers beyond the greatest prime factor of n.

%C This sequence operates on prime exponents as A090079 and A337864 operate on binary and decimal digits, respectively.

%F a(a(n)) = a(n).

%F a(n^k) = a(n)^k for any k >= 0.

%F a(n) = A319521(A356008(n)).

%F A007814(a(n)) = A007814(n).

%F a(n) = 3 iff n belongs to A294674 \ {1}.

%F a(n) = 4 iff n belongs to A061742 \ {1}.

%F a(n) = 8 iff n belongs to A115964.

%e For n = 99:

%e - 99 = 11^1 * 7^0 * 5^0 * 3^2 * 2^0,

%e - the list of exponents is: 1 0 0 2 0,

%e - compressing consecutive values, we obtain: 1 0 2 0,

%e - so a(99) = 7^1 * 5^0 * 3^2 * 2^0 = 63.

%o (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 Cf. A007814, A061742, A067255, A090079, A115964, A294674, A319521, A337864, A356008, A356021 (fixed points).

%K nonn

%O 1,2

%A _Rémy Sigrist_, Jul 23 2022