OFFSET
0,3
COMMENTS
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 ?
LINKS
FORMULA
PROG
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Antti Karttunen, Jul 08 2013
STATUS
approved