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A227351
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Permutation of nonnegative integers: map each number by lengths of runs of zeros in its Zeckendorf expansion shifted once left to the number which has the same lengths of runs (in the same order, but alternatively of runs of 0's and 1's) in its binary representation.
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3
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0, 1, 3, 7, 2, 15, 6, 4, 31, 14, 12, 8, 5, 63, 30, 28, 24, 13, 16, 9, 11, 127, 62, 60, 56, 29, 48, 25, 27, 32, 17, 19, 23, 10, 255, 126, 124, 120, 61, 112, 57, 59, 96, 49, 51, 55, 26, 64, 33, 35, 39, 18, 47, 22, 20, 511, 254, 252, 248, 125, 240, 121, 123, 224
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OFFSET
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0,3
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COMMENTS
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This permutation is based on the fact that by appending one extra zero to the right of Fibonacci number representation of n (aka "Zeckendorf expansion") and then counting the lengths of blocks of consecutive (nonleading) zeros we get bijective correspondence with compositions, and thus also with the binary representation of a unique n. See the chart below:
[shifted once left] zeros with same runs = a(n)
0: ......0 ......0 [] .....0 0
1: ......1 .....10 [1] .....1 1
2: .....10 ....100 [2] ....11 3
3: ....100 ...1000 [3] ...111 7
4: ....101 ...1010 [1,1] ....10 2
5: ...1000 ..10000 [4] ..1111 15
6: ...1001 ..10010 [2,1] ...110 6
7: ...1010 ..10100 [1,2] ...100 4
8: ..10000 .100000 [5] .11111 31
9: ..10001 .100010 [3,1] ..1110 14
10: ..10010 .100100 [2,2] ..1100 12
11: ..10100 .101000 [1,3] ..1000 8
12: ..10101 .101010 [1,1,1] ...101 5
13: .100000 1000000 [6] 111111 63
Are there any other fixed points after 0, 1, 6, 803, 407483 ?
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LINKS
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FORMULA
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This permutation effects following correspondences:
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PROG
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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