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

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%I A336684 #7 Feb 21 2021 10:56:07
%S A336684 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,
%T A336684 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,
%U A336684 1,2,3,4,5,6,7,8,9,10,11,12,0,1,2,3,4,5
%N A336684 Irregular triangle in which row n lists residues k found in the sequence Lucas(i) mod n.
%C A336684 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.
%C A336684 Row n contains the Lucas number k < n, and k such that (n + k) is a Lucas number.
%C A336684 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.
%F A336684 A066981(n) = length of row n.
%F A336684 A223487(n) = n - A066981(n) = number of residues missing from row n.
%F A336684 A224482(n) = rows n that have complete residue coverage, i.e., A066981(n) = n and A223487(n) = 0.
%e A336684 Row 1 contains 0 by convention.
%e A336684 Row 2 contains (0, 1) since the Lucas sequence contains both even and odd numbers.
%e A336684 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.
%e A336684 Table begins as shown below, with residue k shown arranged in columns.
%e A336684 n    k (mod n)
%e A336684 --------------
%e A336684 1:   0
%e A336684 2:   0  1
%e A336684 3:   0  1  2
%e A336684 4:   0  1  2  3
%e A336684 5:      1  2  3  4
%e A336684 6:   0  1  2  3  4  5
%e A336684 7:   0  1  2  3  4  5  6
%e A336684 8:      1  2  3  4  5     7
%e A336684 9:   0  1  2  3  4  5  6  7  8
%e A336684 10:     1  2  3  4     6  7  8  9
%e A336684 11:  0  1  2  3  4        7       10
%e A336684 12:     1  2  3  4  5  6  7  8    10 11
%e A336684 13:     1  2  3  4  5  6  7  8  9 10 11 12
%e A336684 14:  0  1  2  3  4  5  6  7  8  9 10 11 12 13
%e A336684 15:     1  2  3  4        7          11       14
%e A336684 16:     1  2  3  4  5     7     9    11 12 13    15
%e A336684 ...
%t A336684 {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
%Y A336684 Cf. A000032, A066981, A106291, A223487. Analogous to A189768.
%K A336684 nonn,tabf,easy
%O A336684 1,6
%A A336684 _Michael De Vlieger_, Oct 07 2020