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Search: id:A154435
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| A154435 |
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Permutation of non-negative integers induced by Lamplighter group generating wreath recursion, variant 3: a = s(b,a), b = (a,b), starting from the state a. |
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+0
14
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| 0, 1, 3, 2, 6, 7, 5, 4, 13, 12, 14, 15, 10, 11, 9, 8, 26, 27, 25, 24, 29, 28, 30, 31, 21, 20, 22, 23, 18, 19, 17, 16, 53, 52, 54, 55, 50, 51, 49, 48, 58, 59, 57, 56, 61, 60, 62, 63, 42, 43, 41, 40, 45, 44, 46, 47, 37, 36, 38, 39, 34, 35, 33, 32, 106, 107, 105, 104, 109, 108
(list; graph; listen)
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OFFSET
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0,3
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COMMENT
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This permutation is induced by the third Lamplighter group generating wreath recursion a = s(b,a), b = (a,b) (i.e. binary transducer, where s means that the bits at that state are toggled: 0 <-> 1) given on page 104 of Bondarenko, Grigorchuk, et al. paper, starting from the active (swapping) state a and rewriting bits from the second most significant bit to the least significant end.
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LINKS
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A. Karttunen, Table of n, a(n) for n = 0..2047
Index entries for sequences that are permutations of the natural numbers
R. I. Grigorchuk and A. Zuk, The lamplighter group as a group generated by a 2-state automaton and its spectrum, Geometriae Dedicata, vol. 87 (2001), no. 1-3, pp. 209--244.
Bondarenko, Grigorchuk, Kravchenko, Muntyan, Nekrashevych, Savchuk, Sunic, Classification of groups generated by 3-state automata over a 2-letter alphabet, pp. 8--9 & 103.
S. Wolfram, R. Lamy, Discussion on the NKS Forum
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EXAMPLE
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475 = 111011011 in binary. Starting from the second most significant bit and, as we begin with the swapping state a, we complement the bits up to and including the first zero encountered and so the beginning of the binary expansion is complemented as 1001....., then, as we switch to the inactive state b, the following bits are kept same, again up to and including the first zero encountered, after which the binary expansion is 1001110.., after which we switch again to the active state (state a), which complements the two rightmost 1's and we obtain the final answer 100111000, which is 312's binary representation, thus a(475)=312.
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PROGRAM
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(MIT Scheme:) (define (A154435 n) (if (< n 2) n (let loop ((maskbit (A072376 n)) (state 1) (z 1)) (if (zero? maskbit) z (let ((dombit (modulo (floor->exact (/ n maskbit)) 2))) (cond ((= 0 dombit) (loop (floor->exact (/ maskbit 2)) (- 1 state) (+ z z (modulo (- state dombit) 2)))) (else (loop (floor->exact (/ maskbit 2)) state (+ z z (modulo (- state dombit) 2))))))))))
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CROSSREFS
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Inverse: A154436. a(n) = A059893(A154437(A059893(n))) = A054429(A006068(A054429(n))). Corresponds to A122301 in the group of Catalan bijections. Cf. also A153141-A153142, A154439-A154448, A072376.
Sequence in context: A153142 A154447 A003188 this_sequence A006042 A100280 A092745
Adjacent sequences: A154432 A154433 A154434 this_sequence A154436 A154437 A154438
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KEYWORD
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nonn,base
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AUTHOR
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Antti Karttunen (His-Firstname.His-Surname(AT)gmail.com), Jan 17 2009
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