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A057300
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Binary counter with odd/even bit positions swapped; base-4 counter with 1's replaced by 2's and vice versa.
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30
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0, 2, 1, 3, 8, 10, 9, 11, 4, 6, 5, 7, 12, 14, 13, 15, 32, 34, 33, 35, 40, 42, 41, 43, 36, 38, 37, 39, 44, 46, 45, 47, 16, 18, 17, 19, 24, 26, 25, 27, 20, 22, 21, 23, 28, 30, 29, 31, 48, 50, 49, 51, 56, 58, 57, 59, 52, 54, 53, 55, 60, 62, 61, 63, 128, 130, 129, 131, 136, 138
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OFFSET
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0,2
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COMMENTS
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A self-inverse permutation of the integers.
a(n) = n if and only if n can be written as 3*Sum_{k>=0} d_i*4^k, where d_i is either 0 or 1. - Jon Perry, Oct 06 2012
In 1988 A. F. Sidorenko, see the Sidorenko reference, used this sequence as an example of a permutation of the set of positive integers with the property that if positive integers i, j, and k form a 3-term arithmetic progression then the corresponding terms a(i), a(j), and a(k) do not form an arithmetic progression.
In the terminology introduced in the Brown, Jungic, and Poelstra reference, the sequence does not contain "double 3-term arithmetic progressions".
It is not difficult to check that this sequence is with unbounded gaps, i.e., for any positive number m there is a natural number n such that a(n+1) - a(n) > m.
It is an open question if every sequence of integers with bounded gaps must contain a double 3-term arithmetic progression. This problem is equivalent to the well known additive square problem in infinite words: Is it true that any infinite word with a finite set of integers as its alphabet contains two consecutive blocks of the same length and the same sum? For more details about the additive square problem in infinite words see the following references: Ardal, et al.; Brown and Freedman; Freedman; Grytczuk; Halbeisen and Hungerbuhler, and Pirillo and Varricchio.
The sequence was attributed to Sidorenko in P. Hegarty's paper "Permutations avoiding arithmetic patterns". In his paper Hegarty characterized the countably infinite abelian groups for which there exists a bijection mapping arithmetic progressions to non-arithmetic progressions. This was further generalized by Jungic and Sahasrabudhe. (End)
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LINKS
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FORMULA
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Conjecture: a(2*n) = -2*a(n) + 5*n, a(2*n+1) = -2*a(n) + 5*n + 2. - Ralf Stephan, Oct 11 2003
a(4n+k) = 4a(n) + a(k), 0 <= k <= 3. - Jon Perry, Oct 06 2012
a(a(n)) = n.
a(n OR k)) = a(n) OR a(k), where OR is bitwise-or (A003986).
a(n XOR k) = a(n) XOR a(k), where XOR is bitwise exclusive-or (A003987).
a(n AND k) = a(n) AND a(k), where AND is bitwise-and (A004198).
(End)
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EXAMPLE
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a(31) = a(4*7+3) = 4*a(7) + a(3) = 4*11 + 3 = 47.
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MAPLE
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a:= proc(n) option remember; `if`(n=0, 0,
a(iquo(n, 4, 'r'))*4+[0, 2, 1, 3][r+1])
end:
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MATHEMATICA
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Table[FromDigits[IntegerDigits[n, 4]/.{1->2, 2->1}, 4], {n, 0, 70}] (* Harvey P. Dale, Aug 24 2017 *)
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PROG
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(PARI) A057300(n) = { my(t=1, s=0); while(n>0, if(1==(n%4), n++, if(2==(n%4), n--)); s += (n%4)*t; n >>= 2; t <<= 2); (s); }; \\ Antti Karttunen, Apr 14 2018
(C) uint32_t a(uint32_t n) { return ((n & 0x55555555) << 1) | ((n & 0xaaaaaaaa) >> 1); } /* Falk Hüffner, Jan 23 2022 */
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CROSSREFS
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KEYWORD
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easy,nonn,base
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AUTHOR
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STATUS
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approved
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