A223487 -id:A223487 - cp's OEIS frontend

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

Michael De Vlieger, Oct 07 2020

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.

			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
...

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

{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

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.

A336685 Sum of 2^k for residue k in among Lucas numbers mod n.

Original entry on oeis.org

1, 3, 7, 15, 30, 63, 127, 190, 511, 990, 1183, 3582, 8190, 16383, 18590, 47806, 131070, 247967, 298911, 854686, 1453502, 2423967, 8362495, 10366142, 31738014, 67100670, 134217727, 262073758, 302254239, 609175710, 1779923167, 3133061822, 4962151582, 16855148990

Offset: 1

Michael De Vlieger, Oct 07 2020

Row n of A336684 compactified as a binary number.

a(n) contains even numbers whereas A336683 (pertaining to the Fibonacci sequence) is strictly odd, since 0 is a Fibonacci number but not a Lucas number.

			a(1) = 1 by convention.
a(2) = 3 = 2^0 + 2^1, since the Lucas sequence contains both even and odd numbers.
a(5) = 30 = 2^1 + 2^2 + 2^3 + 2^4, since the Lucas numbers mod 5 is {2,1,3,4,2,1} repeated, and we are missing 0, leaving the exponents of 2 as shown.
Binary equivalents of first terms:
   n    a(n)   a(n) in binary
   --------------------------
    1      1                 1
    2      3                11
    3      7               111
    4     15              1111
    5     30             11110
    6     63            111111
    7    127           1111111
    8    190          10111110
    9    511         111111111
   10    990        1111011110
   11   1183       10010011111
   12   3582      110111111110
   13   8190     1111111111110
   14  16383    11111111111111
   15  18590   100100010011110
   16  47806  1011101010111110
   ...

Cf. A000032, A066981, A106291, A223487, A336684. Analogous to A336683.

Total /@ {Most@ #, #} &[2^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}]; Total[2^Union@ w]] &, 32, 3]

a(3^j) = 2^(3^j+1) - 1 for all j.
A066981(n) = binary weight of a(n).
A223487(n) = n - A066981(n) = number of zeros in the binary expansion of a(n).
a(m) = 2^(m+1) - 1 for m = A224482(n).

cp's OEIS Frontend

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

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