A336684 Irregular triangle in which row n lists residues k found in the sequence Lucas(i) mod n.

0, 0, 1, 0, 1, 2, 0, 1, 2, 3, 1, 2, 3, 4, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 7, 0, 1, 2, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4, 6, 7, 8, 9, 0, 1, 2, 3, 4, 7, 10, 1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 0, 1, 2, 3, 4, 5

Offset: 1

Author

Michael De Vlieger, Oct 07 2020

Keywords

• nonn
• tabf
• easy

Comments

For row n, it is sufficient to take the union of A000032(i) mod n for 0 <= i <= A106291(n - 1), since the Lucas numbers are cyclical mod n.

Row n contains the Lucas number k < n, and k such that (n + k) is a Lucas number.

Row n for n in A224482 is complete, i.e., it contains all residues k (mod n). This includes n that is a perfect power of 3.

Examples

			Row 1 contains 0 by convention.
Row 2 contains (0, 1) since the Lucas sequence contains both even and odd numbers.
Row 5 contains (1, 2, 3, 4) since the Lucas numbers mod 5 is {2,1,3,4,2,1} repeated; we are missing the residue 0.
Table begins as shown below, with residue k shown arranged in columns.
n    k (mod n)
--------------
1:   0
2:   0  1
3:   0  1  2
4:   0  1  2  3
5:      1  2  3  4
6:   0  1  2  3  4  5
7:   0  1  2  3  4  5  6
8:      1  2  3  4  5     7
9:   0  1  2  3  4  5  6  7  8
10:     1  2  3  4     6  7  8  9
11:  0  1  2  3  4        7       10
12:     1  2  3  4  5  6  7  8    10 11
13:     1  2  3  4  5  6  7  8  9 10 11 12
14:  0  1  2  3  4  5  6  7  8  9 10 11 12 13
15:     1  2  3  4        7          11       14
16:     1  2  3  4  5     7     9    11 12 13    15
...
		

Crossrefs

Cf. A000032, A066981, A106291, A223487. Analogous to A189768.

Programs

Mathematica

{Most@ #, #} &[Range[0, 1]]~Join~Array[Block[{w = {2, 1}}, Do[If[SequenceCount[w, {2, 1}] == 1, AppendTo[w, Mod[Total@ w[[-2 ;; -1]], #]], Break[]], {i, 2, Infinity}]; Union@ w] &, 12, 3] // Flatten

Formula

A066981(n) = length of row n.

A223487(n) = n - A066981(n) = number of residues missing from row n.

A224482(n) = rows n that have complete residue coverage, i.e., A066981(n) = n and A223487(n) = 0.