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), ...

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

Author

Antti Karttunen, Feb 22 2015

Keywords

• nonn
• tabl
• look

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

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

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)