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%I #13 Apr 06 2018 10:15:49
%S 1,1,1,1,1,3,1,1,1,5,1,3,1,7,5,1,1,9,1,5,1,11,1,3,1,13,7,7,1,15,1,1,5,
%T 17,7,9,1,19,11,5,1,21,1,11,1,23,1,3,1,25,19,13,1,27,1,7,7,29,11,15,1,
%U 31,13,1,11,33,1,17,5,35,1,9,1,37,17,19,1,39,7,5,11,41,1,21,1,43,35,11,1,45,1,23,1,47,13,3,1,49,23,25,1,51,13,13,19
%N A028234 analog for a factorization process based on the Ludic sieve (A255127); Discard all instances of the (smallest) Ludic factor A272565(n) from n.
%C 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 sequence of distinct Ludic numbers (A003309) in ascending order, while applying A302035 to the same terms gives the corresponding "exponents" of these Ludic factors in this nonstandard "Ludic factorization of n", 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 and examples in A302032.
%H Antti Karttunen, <a href="/A302034/b302034.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)(A000265(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(A028234(A302026(n))).
%o (PARI)
%o \\ Assuming A269379 and its inverse A269380 have been precomputed, then the following is reasonably fast:
%o A302034(n) = if(1==n,n,my(k=0); while((n%2), n = A269380(n); k++); n = (n/2^valuation(n, 2)); while(k>0, n = A269379(n); k--); (n));
%Y Cf. A003309, A255127, A260738, A260739, A269379, A269380, A302025, A302026, A302032, A302035.
%Y Cf. A302036 (gives the positions of 1's).
%Y Cf. also A028234, A302044.
%K nonn
%O 1,6
%A _Antti Karttunen_, Apr 01 2018
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