A269380 -id:A269380 - cp's OEIS frontend

A269172 Permutation of natural numbers: a(1) = 1, a(2n) = 2*a(n), a(2n+1) = A250469(a(A269380(2n+1))).

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 25, 20, 21, 22, 19, 24, 23, 26, 27, 28, 29, 30, 49, 32, 33, 34, 35, 36, 31, 50, 39, 40, 37, 42, 41, 44, 45, 38, 43, 48, 55, 46, 51, 52, 47, 54, 121, 56, 57, 58, 77, 60, 53, 98, 63, 64, 65, 66, 59, 68, 69, 70, 61, 72, 169, 62, 75, 100, 67, 78, 85, 80, 81

Offset: 1

Antti Karttunen, Mar 03 2016

nonn

Antti Karttunen, Table of n, a(n) for n = 1..2244
Index entries for sequences that are permutations of the natural numbers

Inverse: A269171.
Cf. A000035, A003309, A008578, A083221, A250469, A260738, A260739, A269380.
Related or similar permutations: A260741, A260742, A269356, A269358, A255422.
Cf. also A269394 (a(3n)/3) and A269396.
Differs from A255408 for the first time at n=38, where a(38) = 50, while A255408(38) = 38.

a(1) = 1, then after for even n, a(n) = 2*a(n/2), and for odd n, A250469(a(A269380(n))).
a(1) = 1, for n > 1, a(n) = A083221(A260738(n), a(A260739(n))).
As a composition of other permutations:
a(n) = A252755(A269386(n)).
a(n) = A252753(A269388(n)).
Other identities. For all n >= 1:
A000035(a(n)) = A000035(n). [This permutation preserves the parity of n.]
a(A003309(n)) = A008578(n). [Maps Ludic numbers to noncomposites.]

A269388 Permutation of natural numbers: a(1) = 0, after which, a(2n) = 1 + 2a(n), a(2n+1) = 2 a(A269380(n)).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 8, 7, 6, 9, 16, 11, 32, 17, 10, 15, 64, 13, 12, 19, 14, 33, 128, 23, 256, 65, 18, 35, 512, 21, 24, 31, 22, 129, 20, 27, 1024, 25, 34, 39, 2048, 29, 4096, 67, 30, 257, 8192, 47, 28, 513, 26, 131, 16384, 37, 48, 71, 38, 1025, 40, 43, 32768, 49, 66, 63, 36, 45, 65536, 259, 46, 41, 131072, 55, 96, 2049, 130, 51, 262144, 69, 44

Offset: 1

Antti Karttunen, Mar 01 2016

nonn

Note the indexing: Domain starts from 1, range from 0.

Antti Karttunen, Table of n, a(n) for n = 1..1024
Index entries for sequences that are permutations of the natural numbers

Inverse: A269387.
Cf. A269380.
Related permutations: A260742, A269386, A269172.
Cf. also A252754, A269378.
Differs from A156552, A252754 and A246677(n-1) for the first time at n=19, which here a(19)=12, instead of 128.

a(1) = 0, after which, a(2n) = 1 + 2*a(n), a(2n+1) = 2 * a(A269380(n)).
As a composition of other permutations:
a(n) = A252754(A269172(n)).
a(n) = A269378(A260742(n)).

A269386 Permutation of nonnegative integers: a(1) = 0, a(2) = 1, a(2n) = 2a(n), a(2n+1) = 1 + 2a(A269380(2n+1)).

Original entry on oeis.org

0, 1, 3, 2, 7, 6, 15, 4, 5, 14, 31, 12, 63, 30, 13, 8, 127, 10, 11, 28, 9, 62, 255, 24, 511, 126, 29, 60, 1023, 26, 23, 16, 25, 254, 27, 20, 2047, 22, 61, 56, 4095, 18, 8191, 124, 17, 510, 16383, 48, 19, 1022, 21, 252, 32767, 58, 47, 120, 57, 2046, 55, 52, 65535, 46, 125, 32, 59, 50, 131071, 508, 49, 54, 262143, 40, 95, 4094, 253, 44

Offset: 1

Antti Karttunen, Mar 01 2016

nonn

Note the indexing: Domain starts from 1, range from 0.

Antti Karttunen, Table of n, a(n) for n = 1..1024
Index entries for sequences that are permutations of the natural numbers

Inverse: A269385.
Cf. A269380.
Related permutations: A260742, A269172, A269388.
Cf. also A252756, A269376.
Differs from A243071 and A252756 for the first time at n=19, which here a(19) = 11, instead of 255.

a(1) = 0, a(2) = 1, a(2n) = 2*a(n), a(2n+1) = 1 + 2*a(A269380(2n+1)).
As a composition of related permutations:
a(n) = A252756(A269172(n)).
a(n) = A269376(A260742(n)).

A269355 Permutation of natural numbers: a(n) = A269380(A250469(n)).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 23, 10, 11, 12, 13, 14, 15, 16, 9, 18, 17, 20, 31, 22, 25, 24, 21, 26, 27, 28, 19, 30, 29, 32, 49, 34, 71, 36, 37, 38, 39, 40, 41, 42, 43, 44, 107, 46, 47, 48, 119, 50, 51, 52, 35, 54, 89, 56, 101, 58, 53, 60, 61, 62, 63, 64, 115, 66, 67, 68, 173, 70, 55, 72, 33, 74, 75, 76, 131, 78, 77, 80, 167

Offset: 1

Antti Karttunen, Mar 03 2016

nonn

			For n=9 we first find what number is below 9 in square array A083221, which is 25, then we find what number is above 25 in square array A255127, which is 23, thus a(9) = 23.

Antti Karttunen, Table of n, a(n) for n = 1..10000
Index entries for sequences that are permutations of the natural numbers

Inverse: A269356.
Cf. A250469, A269380.
Cf. also arrays A083221 & A255127.
More recursed variant: A269357. Cf. also permutations A266645, A255407, A269171.

(define (A269355 n) (A269380 (A250469 n)))

a(n) = A269380(A250469(n)).
Other identities. For all n >= 1:
a(2n) = 2n. [Fixes the even numbers.]

A269357 Permutation of natural numbers: a(1) = 1, a(2n) = 2*a(n), a(2n+1) = A269380(A250469(2n+1)).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 23, 10, 11, 12, 13, 14, 15, 16, 9, 46, 17, 20, 31, 22, 25, 24, 21, 26, 27, 28, 19, 30, 29, 32, 49, 18, 71, 92, 37, 34, 39, 40, 41, 62, 43, 44, 107, 50, 47, 48, 119, 42, 51, 52, 35, 54, 89, 56, 101, 38, 53, 60, 61, 58, 63, 64, 115, 98, 67, 36, 173, 142, 55, 184, 33, 74, 75, 68, 131, 78, 77, 80, 167

Offset: 1

Antti Karttunen, Mar 03 2016

nonn

This is a variant of A269355, from which it differs for the first time at n=18.

Antti Karttunen, Table of n, a(n) for n = 1..16384
Index entries for sequences that are permutations of the natural numbers

Inverse: A269358.

Cf. A000035, A250469, A269380, A269355, A269172.

a(1) = 1, after which, for even n, a(n) = 2*a(n/2) and for odd n, a(n) = A269355(n) = A269380(A250469(n)).
Other identities. For all n >= 1:
A000035(a(n)) = A000035(n). [This permutation preserves the parity of n.]

A255127 Ludic array: square array A(row,col), where row n lists the numbers removed at stage n in the sieve which produces Ludic numbers. Array is read by antidiagonals A(1,1), A(1,2), A(2,1), A(1,3), A(2,2), A(3,1), ...

Original entry on oeis.org

2, 4, 3, 6, 9, 5, 8, 15, 19, 7, 10, 21, 35, 31, 11, 12, 27, 49, 59, 55, 13, 14, 33, 65, 85, 103, 73, 17, 16, 39, 79, 113, 151, 133, 101, 23, 18, 45, 95, 137, 203, 197, 187, 145, 25, 20, 51, 109, 163, 251, 263, 281, 271, 167, 29, 22, 57, 125, 191, 299, 325, 367, 403, 311, 205, 37, 24, 63, 139, 217, 343, 385, 461, 523, 457, 371, 253, 41

Offset: 2

Antti Karttunen, Feb 22 2015

The starting offset of the sequence giving the terms of square array is 2. However, we can tacitly assume that a(1) = 1 when the sequence is used as a permutation of natural numbers. However, term 1 itself is out of the array.

The choice of offset = 2 for the terms starting in rows >= 1 is motivated by the desire to have a permutation of the integers n -> a(n) with a(n) = A(A002260(n-1), A004736(n-1)) for n > 1 and a(1) := 1. However, since this sequence is declared as a "table", offset = 2 would mean that the first *row* (not element) has index 2. I think the sequence should have offset = 1 and the permutation of the integers would be n -> a(n-1) with a(0) := 1 (if a(1) = A(1,1) = 2). Or, the sequence could have offset 0, with an additional row 0 of length 1 with the only element a(0) = A(0,1) = 1, the permutation still being n -> a(n-1) if a(n=0, 1, 2, ...) = (1, 2, 4, ...). This would be in line with considering 1 as the first ludic number, and A(n, 1) = A003309(n+1) for n >= 0. - M. F. Hasler, Nov 12 2024

			The top left corner of the array:
   2,   4,   6,   8,  10,  12,   14,   16,   18,   20,   22,   24,   26
   3,   9,  15,  21,  27,  33,   39,   45,   51,   57,   63,   69,   75
   5,  19,  35,  49,  65,  79,   95,  109,  125,  139,  155,  169,  185
   7,  31,  59,  85, 113, 137,  163,  191,  217,  241,  269,  295,  323
  11,  55, 103, 151, 203, 251,  299,  343,  391,  443,  491,  539,  587
  13,  73, 133, 197, 263, 325,  385,  449,  511,  571,  641,  701,  761
  17, 101, 187, 281, 367, 461,  547,  629,  721,  809,  901,  989, 1079
  23, 145, 271, 403, 523, 655,  781,  911, 1037, 1157, 1289, 1417, 1543
  25, 167, 311, 457, 599, 745,  883, 1033, 1181, 1321, 1469, 1615, 1753
  29, 205, 371, 551, 719, 895, 1073, 1243, 1421, 1591, 1771, 1945, 2117
...

Antti Karttunen, Table of n, a(n) for n = 2..10441; the first 144 antidiagonals of array, flattened

Transpose: A255129.
Inverse: A255128. (When considered as a permutation of natural numbers with a(1) = 1).
Cf. A260738 (index of the row where n occurs), A260739 (of the column).
Main diagonal: A255410.
Column 1: A003309 (without the initial 1). Column 2: A254100.
Row 1: A005843, Row 2: A016945, Row 3: A255413, Row 4: A255414, Row 5: A255415, Row 6: A255416, Row 7: A255417, Row 8: A255418, Row 9: A255419.
A192607 gives all the numbers right of the leftmost column, and A192506 gives the composites among them.
Cf. A272565, A271419, A271420 and permutations A269379, A269380, A269384.
Cf. also related or derived arrays A260717, A257257, A257258 (first differences of rows), A276610 (of columns), A276580.
Analogous arrays for other sieves: A083221, A255551, A255543.
Cf. A376237 (ludic factorials), A377469 (ludic analog of A005867).

rows = 12; cols = 12; t = Range[2, 3000]; r = {1}; n = 1; While[n <= rows, k = First[t]; AppendTo[r, k]; t0 = t; t = Drop[t, {1, -1, k}]; ro[n++] = Complement[t0, t][[1 ;; cols]]]; A = Array[ro, rows]; Table[ A[[n - k + 1, k]], {n, 1, rows}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Mar 14 2016, after Ray Chandler *)

a255127 = lambda n: A255127(A002260(k-1), A004736(k-1))
def A255127(n, k):
    A = A255127; R = A.rows
    while len(R) <= n or len(R[n]) < min(k, A.P[n]): A255127_extend(2*n)
    return R[n][(k-1) % A.P[n]] + (k-1)//A.P[n] * A.S[n]
A=A255127; A.rows=[[1],[2],[3]]; A.P=[1]*3; A.S=[0,2,6]; A.limit=30
def A255127_extend(rMax=9, A=A255127):
    A.limit *= 2; L = [x+5-x%2 for x in range(0, A.limit, 3)]
    for r in range(3, rMax):
        if len(A.P) == r:
            A.P += [ A.P[-1] * (A.rows[-1][0] - 1) ]  # A377469
            A.rows += [[]]; A.S += [ A.S[-1] * L[0] ] # ludic factorials
        if len(R := A.rows[r]) < A.P[r]: # append more terms to this row
            R += L[ L[0]*len(R) : A.S[r] : L[0] ]
        L = [x for i, x in enumerate(L) if i%L[0]] # M. F. Hasler, Nov 17 2024

(define (A255127 n) (if (<= n 1) n (A255127bi (A002260 (- n 1)) (A004736 (- n 1)))))
(define (A255127bi row col) ((rowfun_n_for_A255127 row) col))
;; definec-macro memoizes its results:
(definec (rowfun_n_for_A255127 n) (if (= 1 n) (lambda (n) (+ n n)) (let* ((rowfun_for_remaining (rowfun_n_for_remaining_numbers (- n 1))) (eka (rowfun_for_remaining 0))) (COMPOSE rowfun_for_remaining (lambda (n) (* eka (- n 1)))))))
(definec (rowfun_n_for_remaining_numbers n) (if (= 1 n) (lambda (n) (+ n n 3)) (let* ((rowfun_for_prevrow (rowfun_n_for_remaining_numbers (- n 1))) (off (rowfun_for_prevrow 0))) (COMPOSE rowfun_for_prevrow (lambda (n) (+ 1 n (floor->exact (/ n (- off 1)))))))))

From M. F. Hasler, Nov 12 2024: (Start)
A(r, c) = A(r, c-P(r)) + S(r) = A(r, ((c-1) mod P(r)) + 1) + floor((c-1)/P(r))*S(r) with periods P = (1, 1, 2, 8, 48, 480, 5760, ...) = A377469, and shifts S = (2, 6, 30, 210, 2310, 30030, 510510) = A376237(2, 3, ...). For example:
A(1, c) = A(1, c-1) + 2 = 2 + (c-1)*2 = 2*c,
A(2, c) = A(2, c-1) + 6 = 3 + (c-1)*6 = 6*c - 3,
A(3, c) = A(3, c-2) + 30 = {5 if c is odd else 19} + floor((c-1)/2)*30 = 15*c - 11 + (c mod 2),
A(4, c) = A(4, c-8) + 210 = A(4, ((c-1) mod 8)+1) + floor((c-1)/8)*210, etc. (End)

A269379 a(1) = 1; for n > 1, a(n) = A255127(A260738(n)+1, A260739(n)).

Original entry on oeis.org

1, 3, 5, 9, 7, 15, 11, 21, 19, 27, 13, 33, 17, 39, 35, 45, 23, 51, 31, 57, 49, 63, 25, 69, 29, 75, 65, 81, 37, 87, 55, 93, 79, 99, 59, 105, 41, 111, 95, 117, 43, 123, 47, 129, 109, 135, 53, 141, 85, 147, 125, 153, 61, 159, 73, 165, 139, 171, 103, 177, 67, 183, 155, 189, 113, 195, 71, 201, 169, 207, 77, 213, 101, 219, 185, 225, 83

Offset: 1

Antti Karttunen, Mar 01 2016

nonn

a(n) = the number located immediately below n in A255127 (square array generated by Ludic sieve) in the same column where n itself is, or in other words, the number removed in the next filtering stage at the same step as when n was removed in the A260738(n)-th stage.

Permutation of odd numbers.

Antti Karttunen, Table of n, a(n) for n = 1..32999

Cf. A255127, A260738, A260739.
Cf. A269171, A269356, A269358, A269382, A269385, A269387 (sequences that use this function).
Cf. A269380 (left inverse).
Cf. also A250469, A269369.

(define (A269379 n) (if (= 1 n) n (A255127bi (+ (A260738 n) 1) (A260739 n)))) ;; Code for A255127bi given in A255127.

a(1) = 1; for n > 1, a(n) = A255127(A260738(n)+1, A260739(n)).
Other identities. For all n >= 1:
A269380(a(n)) = n.

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)

A302026 Permutation of natural numbers mapping "Ludic factorization" to ordinary factorization: a(1) = 1, a(2n) = 2*a(n), a(A269379(n)) = A003961(a(n)).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 25, 20, 27, 22, 19, 24, 23, 26, 21, 28, 29, 30, 49, 32, 45, 34, 35, 36, 31, 50, 33, 40, 37, 54, 41, 44, 81, 38, 43, 48, 125, 46, 75, 52, 47, 42, 121, 56, 63, 58, 77, 60, 53, 98, 39, 64, 55, 90, 59, 68, 135, 70, 61, 72, 169, 62, 51, 100, 67, 66, 175, 80, 99, 74, 71, 108, 343, 82, 105, 88

Offset: 1

Antti Karttunen, Apr 03 2018

nonn

See comments and examples in A302032 to see how Ludic factorization proceeds.

Cf. A302025 (inverse permutation).
Cf. A005940, A250246, A269172, A269388 (similar or related permutations).
Cf. A003961, A064989, A269379, A269380, A302031, A302032, A302034, A302037.

\\ With A269380 precomputed:
A003961(n) = my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); \\ From A003961
A302026(n) = if(1==n,n,if(!(n%2),2*A302026(n/2),A003961(A302026(A269380(n)))));
(Scheme, with memoization-macro definec)
(definec (A302026 n) (cond ((= 1 n) n) ((even? n) (* 2 (A302026 (/ n 2)))) (else (A003961 (A302026 (A269380 n))))))

a(1) = 1, a(2n) = 2*a(n), a(2n+1) = A003961(a(A269380(2n+1))).
a(n) = A250246(A269172(n)).
a(n) = A005940(1+A269388(n)).
Other identities. For all n >= 1:
A001221(a(n)) = A302031(n).
A001222(a(n)) = A302037(n).

A269370 a(1) = 1, after which, for odd n: a(n) = A260439(n)-th number k for which A260438(k) = A260438(n)-1, and for even n: a(n) = a(n/2).

Original entry on oeis.org

1, 1, 2, 1, 4, 2, 3, 1, 7, 4, 6, 2, 9, 3, 13, 1, 8, 7, 5, 4, 15, 6, 10, 2, 21, 9, 19, 3, 12, 13, 25, 1, 31, 8, 14, 7, 33, 5, 11, 4, 16, 15, 37, 6, 27, 10, 18, 2, 43, 21, 49, 9, 20, 19, 45, 3, 39, 12, 22, 13, 17, 25, 51, 1, 24, 31, 63, 8, 67, 14, 26, 7, 69, 33, 73, 5, 28, 11, 75, 4, 23, 16, 30, 15, 55, 37, 79, 6, 32, 27, 61, 10, 87, 18, 34, 2

Offset: 1

Antti Karttunen, Mar 01 2016

nonn

For odd numbers n > 1, a(n) tells which term is on the immediately preceding row of A255551 (square array generated by Lucky sieve), in the same column where n itself is.

Antti Karttunen, Table of n, a(n) for n = 1..11017

Cf. A255551, A260438, A260439, A269369.

Cf. also A268674, A269380.

(define (A269370 n) (cond ((= 1 n) n) ((even? n) (A269370 (/ n 2))) (else (A255551bi (- (A260438 n) 1) (A260439 n))))) ;; Code for A255551bi given in A255551.

a(1) = 1; after which for even n, a(n) = a(n/2), and for odd n, a(n) = A255551(A260438(n)-1, A260439(n)).
Other identities. For all n >= 1:
a(A269369(n)) = n.

cp's OEIS Frontend

A269172 Permutation of natural numbers: a(1) = 1, a(2n) = 2*a(n), a(2n+1) = A250469(a(A269380(2n+1))).

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A269388 Permutation of natural numbers: a(1) = 0, after which, a(2n) = 1 + 2*a(n), a(2n+1) = 2 * a(A269380(n)).

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A269386 Permutation of nonnegative integers: a(1) = 0, a(2) = 1, a(2n) = 2*a(n), a(2n+1) = 1 + 2*a(A269380(2n+1)).

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A269355 Permutation of natural numbers: a(n) = A269380(A250469(n)).

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A269357 Permutation of natural numbers: a(1) = 1, a(2n) = 2*a(n), a(2n+1) = A269380(A250469(2n+1)).

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A255127 Ludic array: square array A(row,col), where row n lists the numbers removed at stage n in the sieve which produces Ludic numbers. Array is read by antidiagonals A(1,1), A(1,2), A(2,1), A(1,3), A(2,2), A(3,1), ...

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A269379 a(1) = 1; for n > 1, a(n) = A255127(A260738(n)+1, A260739(n)).

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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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A302026 Permutation of natural numbers mapping "Ludic factorization" to ordinary factorization: a(1) = 1, a(2n) = 2*a(n), a(A269379(n)) = A003961(a(n)).

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A269388 Permutation of natural numbers: a(1) = 0, after which, a(2n) = 1 + 2a(n), a(2n+1) = 2 a(A269380(n)).

A269386 Permutation of nonnegative integers: a(1) = 0, a(2) = 1, a(2n) = 2a(n), a(2n+1) = 1 + 2a(A269380(2n+1)).