cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A057300 Binary counter with odd/even bit positions swapped; base-4 counter with 1's replaced by 2's and vice versa.

Original entry on oeis.org

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
Offset: 0

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Author

Marc LeBrun, Aug 24 2000

Keywords

Comments

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
From Veselin Jungic, Mar 03 2015: (Start)
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)

Examples

			a(31) = a(4*7+3) = 4*a(7) + a(3) = 4*11 + 3 = 47.
		

Crossrefs

Sequences used in definitions of this sequence: A000695, A059905, A059906.
Sequences with similar definitions: A057301, A126006, A126007, A126008, A163241, A163327.
A003986, A003987, A004198, A053985, A054240 are used to express relationships between sequence terms.

Programs

  • C
    #include 
    uint32_t a(uint32_t n) { return ((n & 0x55555555) << 1) | ((n & 0xaaaaaaaa) >> 1); } /* Falk Hüffner, Jan 23 2022 */
  • Maple
    a:= proc(n) option remember; `if`(n=0, 0,
          a(iquo(n, 4, 'r'))*4+[0, 2, 1, 3][r+1])
        end:
    seq(a(n), n=0..69);  # Alois P. Heinz, Jan 25 2022
  • Mathematica
    Table[FromDigits[IntegerDigits[n,4]/.{1->2,2->1},4],{n,0,70}] (* Harvey P. Dale, Aug 24 2017 *)
  • 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
    

Formula

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(n) = A000695(A059906(n)) + 2*A000695(A059905(n)). - Antti Karttunen, Apr 14 2018
From Peter Munn, Dec 10 2019: (Start)
a(a(n)) = n.
a(A000695(m) + 2*A000695(n)) = 2*A000695(m) + A000695(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).
a(A054240(n,k)) = A054240(a(n), a(k)). (End)
a(n) = 5*n/4 - 3*A053985(2*n)/8. - Alan Michael Gómez Calderón, May 20 2025