A302035 -id:A302035 - cp's OEIS frontend

A260739 Column index to A255127: a(1) = 1; for n > 1, a(n) = the position at the stage where n is removed in the sieve which produces Ludic numbers.

1, 1, 1, 2, 1, 3, 1, 4, 2, 5, 1, 6, 1, 7, 3, 8, 1, 9, 2, 10, 4, 11, 1, 12, 1, 13, 5, 14, 1, 15, 2, 16, 6, 17, 3, 18, 1, 19, 7, 20, 1, 21, 1, 22, 8, 23, 1, 24, 4, 25, 9, 26, 1, 27, 2, 28, 10, 29, 3, 30, 1, 31, 11, 32, 5, 33, 1, 34, 12, 35, 1, 36, 2, 37, 13, 38, 1, 39, 6, 40, 14, 41, 1, 42, 4, 43, 15, 44, 1, 45, 1, 46, 16, 47, 7, 48, 1, 49, 17, 50, 2

Offset: 1

Antti Karttunen, Jul 30 2015

nonn

Ordinal transform of A272565 (Ludic factor), and also of A260738. - Antti Karttunen, Apr 03 2018

Antti Karttunen, Table of n, a(n) for n = 1..10105

Column index to array A255127.
Cf. A003309, A254100, A272565.
Cf. A260738 (corresponding row index).
Cf. A302035, A302036 (positions of terms that are powers of 2).
Cf. A078898, A246277, A260429, A260439 for column indices to other arrays similar to A255127.
Differs from A246277 (and also after the initial term from A078898) for the first time at n=19.

(define (A260739 n) (cond ((= 1 n) 1) ((even? n) (/ n 2)) (else (let searchrow ((row 2)) (let searchcol ((col 1)) (cond ((>= (A255127bi row col) n) (if (= (A255127bi row col) n) col (searchrow (+ 1 row)))) (else (searchcol (+ 1 col))))))))) ;; Code for A255127bi given in A255127.

Other identities. For all n >= 2:
a(A003309(n)) = 1. [In Ludic sieve each Ludic number (after 1) is the first among the numbers removed at stage k.]
a(A254100(n)) = 2.
A255127(A260738(n), a(n)) = n.
For n > 1, A001511(a(n)) = A302035(n). - Antti Karttunen, Apr 03 2018

Term a(1) changed from 0 to 1 to match with the definition of A078898 and the interpretation as an ordinal transform - Antti Karttunen, Apr 03 2018

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.

Original entry on oeis.org

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

Antti Karttunen, Mar 31 2018

nonn

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

			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].

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

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.

\\ 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))

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)

A302045 a(1) = 0, for n > 1, a(n) = A001511(A078898(n)); Number of instances of the smallest prime factor A020639(n) in nonstandard factorization of n that is based on the sieve of Eratosthenes (A083221).

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 4, 1, 1, 1, 2, 3, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 5, 2, 1, 1, 2, 1, 1, 1, 3, 1, 1, 1, 2, 4, 1, 1, 4, 2, 1, 1, 2, 1, 1, 3, 3, 2, 1, 1, 2, 1, 1, 1, 6, 1, 1, 1, 2, 3, 1, 1, 3, 1, 1, 1, 2, 1, 1, 1, 4, 2, 1, 1, 2, 2, 1, 1, 3, 1, 1, 3, 2, 5, 1, 1, 5, 1, 1, 1, 2, 1, 1, 1, 3, 2

Offset: 1

Antti Karttunen, Mar 31 2018

nonn

Iterating the map n -> A302044(n) until 1 is reached, and taking the smallest prime factor (A020639) of each term gives a sequence of distinct primes in ascending order, while applying this function (A302045) to those terms gives the corresponding "exponents" of those primes, that is, the count of consecutive occurrences of each prime when iterating the map n -> A302042(n), which gives the same primes with repetitions. Permutation pair A250245/A250246 maps between this non-standard prime factorization of n and the ordinary factorization of n. See also comments and examples in A302042.

Antti Karttunen, Table of n, a(n) for n = 1..65537
Index entries for sequences generated by sieves

Cf. A001511, A067029, A078898, A250245, A250246, A268674, A302042, A302044.

Cf. also A302035, A302050, A302051.

A020639(n) = { if(1==n,n,vecmin(factor(n)[, 1])); };
A078898(n) = { if(n<=1,n, my(spf=A020639(n),k=1,m=n/spf); while(m>1,if(A020639(m)>=spf,k++); m--); (k)); };
A001511(n) = 1+valuation(n,2);
A302045(n) = if(1==n,0,A001511(A078898(n)));

a(1) = 0, for n > 1, a(n) = A001511(A078898(n)).

For n > 1, a(n) = A250245(A067029(A250246(n))).

A302034 A028234 analog for a factorization process based on the Ludic sieve (A255127); Discard all instances of the (smallest) Ludic factor A272565(n) from n.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 1, 1, 5, 1, 3, 1, 7, 5, 1, 1, 9, 1, 5, 1, 11, 1, 3, 1, 13, 7, 7, 1, 15, 1, 1, 5, 17, 7, 9, 1, 19, 11, 5, 1, 21, 1, 11, 1, 23, 1, 3, 1, 25, 19, 13, 1, 27, 1, 7, 7, 29, 11, 15, 1, 31, 13, 1, 11, 33, 1, 17, 5, 35, 1, 9, 1, 37, 17, 19, 1, 39, 7, 5, 11, 41, 1, 21, 1, 43, 35, 11, 1, 45, 1, 23, 1, 47, 13, 3, 1, 49, 23, 25, 1, 51, 13, 13, 19

Offset: 1

Antti Karttunen, Apr 01 2018

nonn

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 sequence of distinct Ludic numbers (A003309) in ascending order, while applying A302035 to the same terms gives the corresponding "exponents" of these Ludic factors in this nonstandard "Ludic factorization of n", 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 and examples in A302032.

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

Cf. A003309, A255127, A260738, A260739, A269379, A269380, A302025, A302026, A302032, A302035.
Cf. A302036 (gives the positions of 1's).
Cf. also A028234, A302044.

\\ Assuming A269379 and its inverse A269380 have been precomputed, then the following is reasonably fast:
A302034(n) = if(1==n,n,my(k=0); while((n%2), n = A269380(n); k++); n = (n/2^valuation(n, 2)); while(k>0, n = A269379(n); k--); (n));

For n > 1, a(n) = A269379^(r)(A000265(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(A028234(A302026(n))).

cp's OEIS Frontend

A260739 Column index to A255127: a(1) = 1; for n > 1, a(n) = the position at the stage where n is removed in the sieve which produces Ludic numbers.

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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.

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A302045 a(1) = 0, for n > 1, a(n) = A001511(A078898(n)); Number of instances of the smallest prime factor A020639(n) in nonstandard factorization of n that is based on the sieve of Eratosthenes (A083221).

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A302034 A028234 analog for a factorization process based on the Ludic sieve (A255127); Discard all instances of the (smallest) Ludic factor A272565(n) from n.

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