cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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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), ...

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

Views

Author

Antti Karttunen, Feb 22 2015

Keywords

Comments

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

Examples

			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
...

Links

Crossrefs

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).

Programs

  • Mathematica
    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 *)
  • Python
    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
  • Scheme
    (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)))))))))

Formula

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)

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.

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Jul 30 2015

Keywords

Comments

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

Links

Crossrefs

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.

Programs

  • Scheme
    (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.

Formula

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

Extensions

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

Views

Author

Antti Karttunen, Mar 31 2018

Keywords

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

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)

A276580 Square array A(n,k) = A276570(A255127(n,k)), numbers in Ludic array reduced by the first element of each row. 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

0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 3, 0, 0, 0, 4, 3, 0, 0, 0, 0, 0, 1, 4, 8, 0, 0, 0, 4, 1, 8, 3, 16, 0, 0, 0, 0, 4, 5, 2, 0, 7, 0, 0, 0, 4, 2, 9, 3, 9, 18, 17, 0, 0, 0, 0, 2, 2, 0, 10, 12, 11, 2, 0, 0, 0, 4, 0, 2, 8, 2, 17, 7, 23, 31, 0, 0, 0, 0, 3, 6, 7, 3, 11, 24, 0, 6, 6, 0, 0, 0, 4, 3, 3, 4, 0, 22, 20, 23, 10, 30, 16, 0
Offset: 2

Views

Author

Antti Karttunen, Sep 13 2016

Keywords

Comments

The starting offset of the sequence giving the terms of square array is 2, to tally with the indexing used in A255127. The row and column indices both start from 1.
Row 4 seems to have a period of 8: [0, 3, 3, 1, 1, 4, 2, 2], while row 5 (A276577) seems to have a period of 48.

Examples

			The top left 17 x 15 corner of the array:
   n   A003309(n+1) = A255127(n,1).
   |   |
   |   | |           A255127(n,k) modulo A003309(n+1)
   v   v |
   1   2 |0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0
   2   3 |0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0
   3   5 |0,  4,  0,  4,  0,  4,  0,  4,  0,  4,  0,  4,  0,  4,  0,  4,  0
   4   7 |0,  3,  3,  1,  1,  4,  2,  2,  0,  3,  3,  1,  1,  4,  2,  2,  0
   5  11 |0,  0,  4,  8,  5,  9,  2,  2,  6,  3,  7,  0,  4,  4,  1,  5,  9
   6  13 |0,  8,  3,  2,  3,  0,  8,  7,  4, 12,  4, 12,  7,  4,  3, 11, 12
   7  17 |0, 16,  0,  9, 10,  2,  3,  0,  7, 10,  0,  3,  8, 13, 12,  0,  5
   8  23 |0,  7, 18, 12, 17, 11, 22, 14,  2,  7,  1, 14,  2, 17,  7, 18, 10
   9  25 |0, 17, 11,  7, 24, 20,  8,  8,  6, 21, 19, 15,  3,  3, 16, 16, 14
  10  29 |0,  2, 23,  0, 23, 25,  0, 25,  0, 25,  2,  2,  0,  2,  6,  8,  8
  11  37 |0, 31,  6, 10, 30, 36,  3, 21, 17, 29, 16, 14, 22, 34,  1, 13, 27
  12  41 |0,  6, 30, 38, 25, 37,  4, 16,  5, 21,  0, 12, 22, 40, 29,  6, 24
  13  43 |0, 16, 42,  9, 35, 38,  7, 31,  8,  1, 25,  6, 24,  5, 35, 26,  1
  14  47 |0, 26, 13, 45, 32, 23, 10, 40, 19, 14, 21,  8, 28, 15,  0, 40, 25
  15  53 |0, 42, 31, 14, 11,  0,  1,  0, 50, 31, 20, 29, 12, 11, 20,  1, 51

Links

Crossrefs

Cf. A003309, A255127, A272565, A276570.
Column 1, Rows 1 & 2: A000004.
Column 2: A276576.
Row 5: A276577.

Programs

A271419 If n is a ludic number, a(n)=0; if n is not a ludic number, a(n) is the ludic number that rejects n from the ludic number sieve.

Original entry on oeis.org

0, 0, 0, 2, 0, 2, 0, 2, 3, 2, 0, 2, 0, 2, 3, 2, 0, 2, 5, 2, 3, 2, 0, 2, 0, 2, 3, 2, 0, 2, 7, 2, 3, 2, 5, 2, 0, 2, 3, 2, 0, 2, 0, 2, 3, 2, 0, 2, 5, 2, 3, 2, 0, 2, 11, 2, 3, 2, 7, 2, 0, 2, 3, 2, 5, 2, 0, 2, 3, 2, 0, 2, 13, 2, 3, 2, 0, 2, 5, 2, 3, 2, 0, 2, 7
Offset: 1

Views

Author

Max Barrentine, Apr 07 2016

Keywords

Links

Crossrefs

Cf. A003309, A192607, A192490, A255127, A260738, A271420, A272565.
Cf. A264940 (analogous version for lucky numbers).

Programs

Formula

a(n) = (1-A192490(n)) * A272565(n). - Antti Karttunen, Sep 11 2016

A276440 a(n) = greatest ludic number (A003309) that divides n.

Original entry on oeis.org

1, 2, 3, 2, 5, 3, 7, 2, 3, 5, 11, 3, 13, 7, 5, 2, 17, 3, 1, 5, 7, 11, 23, 3, 25, 13, 3, 7, 29, 5, 1, 2, 11, 17, 7, 3, 37, 2, 13, 5, 41, 7, 43, 11, 5, 23, 47, 3, 7, 25, 17, 13, 53, 3, 11, 7, 3, 29, 1, 5, 61, 2, 7, 2, 13, 11, 67, 17, 23, 7, 71, 3, 1, 37, 25, 2, 77, 13, 1, 5, 3, 41, 83, 7, 17, 43, 29, 11, 89, 5, 91, 23, 3, 47
Offset: 1

Views

Author

Antti Karttunen, Sep 11 2016

Keywords

Examples

			a(19) = 1 as 19 is not a ludic number, but it is a prime, thus only ludic number that divides it is the very first one A003309(1) = 1.
a(589) = 1 also as 589 = 19*31 and both 19 and 31 are in A192505.

Links

Crossrefs

Cf. A003309, A192505, A272565, A276568, A276569.
Differs from A006530 for the first time at n=19.

Programs

  • Scheme
    (define (A276440 n) (let loop ((k 1) (mt 1)) (let ((t (A003309 k))) (cond ((> t n) mt) ((zero? (modulo n t)) (loop (+ 1 k) t)) (else (loop (+ 1 k) mt))))))

A274568 Irregular triangle read by rows in which the n-th row (n>1) lists distinct "factors" of n for the sieve described in A262775.

Original entry on oeis.org

2, 0, 2, 3, 7, 2, 3, 2, 9, 2, 3, 13, 2, 7, 3, 2, 15, 2, 3, 19, 2, 9, 3, 7, 2, 21, 2, 3, 27, 2, 13, 3, 2, 7, 9, 2, 3, 33, 2, 15, 3, 2, 7, 2, 3, 37, 2, 9, 19, 3, 13, 2, 39, 2, 3, 7, 49, 2, 21, 3, 2, 9, 15, 2, 3, 7, 2, 3, 2, 13, 27, 51, 2, 3, 55, 2, 7, 9, 3, 19
Offset: 0

Views

Author

Max Barrentine, Jun 28 2016

Keywords

Comments

This sequence is analogous to the prime factors of n, but for the sifted set A262775. If n is equivalent to some term A262775(k) mod A262775(k+1), then row n will include the term A262775(k+1). Row 1 is zero.

Examples

			Row 2 is 2,3 because 2==0 mod 2 and A262775(0)=0, A262775(1)=2, and 2==2 mod 3 and A262775(1)=2, A262775(2)=3.

Links

Crossrefs

Cf. A262775, A027748, A264940, A272565.
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