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 A302032 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. 11
 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, 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, 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,4 COMMENTS 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. LINKS Antti Karttunen, Table of n, a(n) for n = 1..32768 FORMULA 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. a(n) = A302025(A032742(A302026(n))). EXAMPLE For n = 100, A272565(100) [its Ludic factor] is 2. Because A260738(100) = 1, a(100) is just A260739(100) = 100/2 = 50. For n = 50, A272565(50) [its Ludic factor] is 2. Because A260738(50) = 1, a(50) = A260739(50) = 50/2 = 25. 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. 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). 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]. PROG (PARI) \\ Assuming A269379 and its inverse A269380 have been precomputed, then the following is reasonably fast: 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)); CROSSREFS Cf. A003309, A255127, A269379, A269380, A272565, A260738, A260739, A269379, A269380, A302025, A302026, A302034, A302035. Cf. the following analogs A302031 (omega), A302037 (bigomega). Cf. also A032742, A302042. Sequence in context: A280498 A280496 A291327 * A291326 A291325 A116512 Adjacent sequences:  A302029 A302030 A302031 * A302033 A302034 A302035 KEYWORD nonn AUTHOR Antti Karttunen, Mar 31 2018 STATUS approved

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Last modified September 24 23:13 EDT 2020. Contains 337325 sequences. (Running on oeis4.)