%I #30 Nov 07 2024 00:48:00
%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 Discard the least ludic factor of n: a(n) = A255127(A260738(c) + r - 1, A260739(c)), where r = A260738(n), c = A260739(n) are the row and the column index of n in the table A255127; a(n) = 1 if c = 1.
%C Original definition: 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.
%C The definition of "discard the least ludic factor" is based on the table A255127 of the ludic sieve, where row r lists the (r+1)-th ludic number k = A003309(r+1), determined at the r-th step of the sieve, followed by the numbers crossed out at this step, namely, every k-th of the numbers remaining so far after k. If the number n is in row r = A260738(n), column c = A260739(n) of that table, then its least ludic factor is A272565(n) = A003309(r+1), the 1st entry of the r-th row. To discard that factor means to consider the number which is r-1 rows below the number c in that table, whence a(n) = A255127(A260738(c)+r-1, A260739(c)) - unless n is a ludic number, in which case a(n) = 1. - _M. F. Hasler_, Nov 06 2024
%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))).
%F From _M. F. Hasler_, Nov 06 2024: (Start)
%F a(n) = 1 if n is a ludic number, i.e., in A003309. Otherwise:
%F a(n) = A255127(A260738(c) + r - 1, A260739(c)), with r = A260738(n), c = A260739(n).
%F In particular, a(2n) = n for all n. (End)
%e Frem _M. F. Hasler_, Nov 06 2024: (Start)
%e For ludic numbers 1, 2, 3, 5, 7, 11, 13, 17, 23, 25, 29, 37, ..., a(n) = 1.
%e For n = 4, an even number, we have r = A260738(4) = 1: It is listed in row 1 of the table A255127, which lists all numbers that were crossed out at the first step: namely, the ludic number k = 2 and every other larger number. Also, in this row 1, the number 4 is in column c = A260739(4) = 2. Therefore, we apply r-1 = 0 times the map A269379 to c = 2, whence a(4) = 2.
%e The number n = 6 is also even and therefore listed in row r = 1, now in column c = 3, whence a(6) = 3. Similarly, a(8) = 4 and a(2k) = k for all k >= 1.
%e The number n = 9 was crossed out at the 2nd step (so r = A260738(9) = 2), when k = 3 was added to the ludic numbers and every 3rd remaining number crossed out; 9 was the first of these (after k = 3) so it is in column c = A260739(9) = 2. Now we have to apply r-1 = 1 times the map A269379 to c. That map yields the number which is located just below the argument (here c = 2) in the table A255127. Since 2 is a ludic number, in the first column, we get the next larger ludic number, 3, whence a(9) = 3.
%e The number 15 was the (c = 3)rd number to be crossed out at the (r = 2)nd step. Hence a(15) = A269379^{r-1} (c) = A269379(3) = 5 (again, the next larger ludic number).
%e The number 19 was the (c = 2)nd number to be crossed out at the (r = 3)rd step (when k = 5, its least ludic factor, was added to the list of ludic numbers). Hence a(19) = A269379^2(2) = A269379(3) = 5 again (skipping twice to the next larger ludic number).
%e (End)
%e To illustrate how this sequence allows to compute the complete "ludic factorization" of a number, we consider n = 100.
%e For n = 100, its Ludic factor A272565(100) is 2, and we have seen that a(n) = 100/2 = 50.
%e For n = 50, its Ludic factor A272565(50) is 2 again, and again a(50) = 50/2 = 25.
%e Since n = 25 = A003309(1+9) is a ludic number, it equals its Ludic factor A272565(25) = 25. Because it appeared at the A260738(25) = 9th step, we apply A269379 eight times to the column index A260739(25) = 1, a fixed point, so a(25) = A269379^8(1) = 1.
%e Collecting the Ludic factors given by A272565 we get the multiset of factors: [2, 2, 25] = [A003309(1+1), A003309(1+1), A003309(1+9)]. By definition, A302026(100) = prime(1)*prime(1)*prime(9) = 2*2*23 = 92, the product of the corresponding primes.
%e If we start from n = 100, iterating the map n -> A302034(n) [instead of A302032] 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