%I #20 Apr 06 2018 10:15:39

%S 1,1,1,2,1,3,1,4,3,5,1,6,1,7,5,8,1,9,5,10,9,11,1,12,1,13,7,14,1,15,7,

%T 16,15,17,7,18,1,19,11,20,1,21,1,22,21,23,1,24,19,25,19,26,1,27,11,28,

%U 27,29,11,30,1,31,13,32,11,33,1,34,33,35,1,36,13,37,17,38,1,39,35,40,39,41,1,42,31,43,35,44,1,45,1,46,45,47,13,48,1,49,23,50

%N A032742 analog for a nonstandard factorization process based on the Ludic sieve (A255127); Discard a single instance of the Ludic factor A272565(n) from n.

%C Like [A020639(n), A032742(n)] or [A020639(n), A302042(n)], also ordered pair [A272565(n), a(n)] is unique for each n. Iterating n, a(n), a(a(n)), a(a(a(n))), ..., until 1 is reached, and taking the Ludic factor (A272565) of each term gives a multiset of Ludic numbers (A003309) in ascending order, unique for each natural number n >= 1. Permutation pair A302025/A302026 maps between this "Ludic factorization" and the ordinary prime factorization of n. See also comments in A302034.

%H Antti Karttunen, <a href="/A302032/b302032.txt">Table of n, a(n) for n = 1..32768</a>

%H <a href="/index/Si#sieve">Index entries for sequences generated by sieves</a>

%F For n > 1, a(n) = A269379^(r)(A260739(n)), where r = A260738(n)-1 and A269379^(r)(n) stands for applying r times the map x -> A269379(x), starting from x = n.

%F a(n) = A302025(A032742(A302026(n))).

%e For n = 100, A272565(100) [its Ludic factor] is 2. Because A260738(100) = 1, a(100) is just A260739(100) = 100/2 = 50.

%e For n = 50, A272565(50) [its Ludic factor] is 2. Because A260738(50) = 1, a(50) = A260739(50) = 50/2 = 25.

%e For n = 25, A272565(25) [its Ludic factor] is 25 = A003309(1+9). Because A260738(25) = 9, a(25) = A269379^8(A260739(25)) = A269379^8(1) = 1.

%e Collecting the Ludic factors given by A272565 we get a multiset of factors: [2, 2, 25] = [A003309(1+1), A003309(1+1), A003309(1+9)]. Note that prime(1)*prime(1)*prime(9) = 2*2*23 = 92 = A302026(100).

%e If we start from n = 100, iterating the map n -> A302034(n) [instead of n -> A302032(n)] and apply A272565 to each term obtained we get just a single instance of each Ludic factor: [2, 25]. Then by applying A302035 to the same terms we get the corresponding exponents (multiplicities) of those factors: [2, 1].

%o (PARI)

%o \\ Assuming A269379 and its inverse A269380 have been precomputed, then the following is reasonably fast:

%o A302032(n) = if(1==n,n,my(k=0); while((n%2), n = A269380(n); k++); n = n/2; while(k>0, n = A269379(n); k--); (n));

%Y Cf. A003309, A255127, A269379, A269380, A272565, A260738, A260739, A269379, A269380, A302025, A302026, A302034, A302035.

%Y Cf. the following analogs A302031 (omega), A302037 (bigomega).

%Y Cf. also A032742, A302042.

%K nonn

%O 1,4

%A _Antti Karttunen_, Mar 31 2018