A302032 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.

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

Offset: 1

Author

Antti Karttunen, Mar 31 2018

Keywords

• nonn

Comments

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.

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.

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

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].
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..32768
• Index entries for sequences generated by sieves

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.

Programs

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))

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))).

From M. F. Hasler, Nov 06 2024: (Start)

a(n) = 1 if n is a ludic number, i.e., in A003309. Otherwise:

a(n) = A255127(A260738(c) + r - 1, A260739(c)), with r = A260738(n), c = A260739(n).

In particular, a(2n) = n for all n. (End)