Search: a223487 -id:a223487
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A336684
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Irregular triangle in which row n lists residues k found in the sequence Lucas(i) mod n.
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+10
1
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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
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OFFSET
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1,6
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COMMENTS
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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.
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LINKS
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FORMULA
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EXAMPLE
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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)
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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
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MATHEMATICA
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{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
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CROSSREFS
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KEYWORD
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nonn,tabf,easy
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AUTHOR
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STATUS
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approved
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A336685
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Sum of 2^k for residue k in among Lucas numbers mod n.
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+10
0
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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
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refs;
listen;
history;
text;
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OFFSET
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1,2
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COMMENTS
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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.
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LINKS
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FORMULA
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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).
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EXAMPLE
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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
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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
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MATHEMATICA
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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]
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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