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
FORMULA
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
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, Mar 31 2018
STATUS
approved