cp's OEIS Frontend

Examples

			Frem _M. F. Hasler_, Nov 06 2024: (Start)
For ludic numbers 1, 2, 3, 5, 7, 11, 13, 17, 23, 25, 29, 37, ..., a(n) = 1.
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.
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.
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.
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).
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).
(End)
To illustrate how this sequence allows one to compute the complete "ludic factorization" of a number, we consider n = 100.
For n = 100, its Ludic factor A272565(100) is 2, and we have seen that a(n) = 100/2 = 50.
For n = 50, its Ludic factor A272565(50) is 2 again, and again a(50) = 50/2 = 25.
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.
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.
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].