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A362917
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The part of n to the left of the decimal point in the Dekking-van-Loon-canonical base phi representation of n.
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4
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0, 1, 10, 11, 101, 1000, 1010, 1011, 10001, 10010, 10011, 10101, 100000, 100010, 100011, 100101, 101000, 101010, 101011, 1000001, 1000010, 1000011, 1000101, 1001000, 1001010, 1001011, 1010001, 1010010, 1010011, 1010101, 10000000
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OFFSET
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0,3
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COMMENTS
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The part to the right of the decimal point, reversed, is given by A341722, that is, it is the same as in the Bergman-canonical representation. I asked Jeffrey Shallit to confirm this, and he provided the following verification using the Walnut Theorem-Prover:
[Walnut]$ eval sloane "?msd_fib An,x1,x2,y1,y2 ($saka(n,x1,y1) & $dvl(n,x2,y2)) => $equal(y1,y2)":
(saka(n,x1,y1))&dvl(n,x2,y2))):54 states - 66ms
((saka(n,x1,y1))&dvl(n,x2,y2)))=>equal(y1,y2))):2 states - 25ms
(A n , x1 , x2 , y1 , y2 ((saka(n,x1,y1))&dvl(n,x2,y2)))=>equal(y1,y2)))):1 states - 81ms
Total computation time: 264ms.
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TRUE
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REFERENCES
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Dekking, Michel, and Ad van Loon. "On the representation of the natural numbers by powers of the golden mean." arXiv preprint arXiv:2111.07544 (2021); Fib. Quart. 61:2 (May 2023), 105-118.
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LINKS
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EXAMPLE
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The canonical base phi representations of the numbers 0 through 12 are:
0 = 0.
1 = 1.
2 = 10.01
3 = 11.01
4 = 101.01
5 = 1000.1001
6 = 1010.0001
7 = 1011.0001
8 = 10001.0001
9 = 10010.0101
10 = 10011.0101
11 = 10101.0101
12 = 100000.101001
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CROSSREFS
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KEYWORD
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nonn,base,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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