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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.

A054639 Queneau numbers: numbers n such that the Queneau-Daniel permutation {1, 2, 3, ..., n} -> {n, 1, n-1, 2, n-2, 3, ...} is of order n.

Original entry on oeis.org

1, 2, 3, 5, 6, 9, 11, 14, 18, 23, 26, 29, 30, 33, 35, 39, 41, 50, 51, 53, 65, 69, 74, 81, 83, 86, 89, 90, 95, 98, 99, 105, 113, 119, 131, 134, 135, 146, 155, 158, 173, 174, 179, 183, 186, 189, 191, 194, 209, 210, 221, 230, 231, 233, 239
Offset: 1

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Author

Gilles Esposito-Farese (gef(AT)cpt.univ-mrs.fr), May 17 2000

Keywords

Comments

The troubadour Arnaut Daniel composed sestinas based on the permutation 123456 -> 615243, which cycles after 6 iterations.
Roubaud quotes the number 141, but the corresponding Queneau-Daniel permutation is only of order 47 = 141/3.
This appears to coincide with the numbers n such that a type-2 optimal normal basis exists for GF(2^n) over GF(2). But are these two sequences really the same? - Joerg Arndt, Feb 11 2008
The answer is Yes - see Theorem 2 of the Dumas reference. [Jean-Guillaume Dumas (Jean-Guillaume.Dumas(AT)imag.fr), Mar 20 2008]
From Peter R. J. Asveld, Aug 17 2009: (Start)
a(n) is the n-th T-prime (Twist prime). For N >= 2, the family of twist permutations is defined by
p(m,N) == +2m (mod 2N+1) if 1 <= m < k = ceiling((N+1)/2),
p(m,N) == -2m (mod 2N+1) if k <= m < N.
N is T-prime if p(m,N) consists of a single cycle of length N.
The twist permutation is the inverse of the Queneau-Daniel permutation.
N is T-prime iff p=2N+1 is a prime number and exactly one of the following three conditions holds;
(1) N == 1 (mod 4) and +2 generates Z_p^* (the multiplicative group of Z_p) but -2 does not,
(2) N == 2 (mod 4) and both +2 and -2 generate Z_p^*,
(3) N == 3 (mod 4) and -2 generate Z_p^* but +2 does not. (End)
The sequence name says the permutation is of order n, but P. R. J. Asveld's comment says it's an n-cycle. Is there a proof that those conditions are equivalent for the Queneau-Daniel permutation? (They are not equivalent for any arbitrary permutation; e.g., (123)(45)(6) has order 6 but isn't a 6-cycle.) More generally, I have found that for all n <= 9450, (order of Queneau-Daniel permutation) = (length of orbit of 1) = A003558(n). Does this hold for all n? - David Wasserman, Aug 30 2011

Examples

			For N=6 and N=7 we obtain the permutations (1 2 4 5 3 6) and (1 2 4 7)(3 6)(5): 6 is T-prime, but 7 is not. - _Peter R. J. Asveld_, Aug 17 2009
		

References

  • Raymond Queneau, Note complémentaire sur la Sextaine, Subsidia Pataphysica 1 (1963), pp. 79-80.
  • Jacques Roubaud, Bibliothèque Oulipienne No 65 (1992) and 66 (1993).

Crossrefs

Not to be confused with Queneau's "s-additive sequences", see A003044.
A005384 is a subsequence.
Union of A163782 (Josephus_2-primes) and A163781 (dual Josephus_2-primes); also the union of A163777 (Archimedes_0-primes) and A163778 (Archimedes_1-primes); also the union of A071642/2 (shuffle primes) and A163776/2 (dual shuffle primes). - Peter R. J. Asveld, Aug 17 2009
Cf. A216371, A003558 (for which a(n) == n).

Programs

  • Maple
    QD:= proc(n) local i;
      if n::even then map(op,[seq([n-i,i+1],i=0..n/2-1)])
      else map(op, [seq([n-i,i+1],i=0..(n-1)/2-1),[(n+1)/2]])
      fi
    end proc:
    select(n -> GroupTheory:-PermOrder(Perm(QD(n)))=n, [$1..1000]); # Robert Israel, May 01 2016
  • Mathematica
    a[p_] := Sum[Cos[2^n Pi/((2 p + 1) )], {n, 1, p}];
    Select[Range[500],Reduce[a[#] == -1/2, Rationals] &] (* Gerry Martens, May 01 2016 *)
  • PARI
    is(n)=
    {
        if (n==1, return(1));
        my( m=n%4 );
        if ( m==4, return(0) );
        my(p=2*n+1, r=znorder(Mod(2,p)));
        if ( !isprime(p), return(0) );
        if ( m==3 && r==n, return(1) );
        if ( r==2*n, return(1) ); \\ r == 1 or 2
        return(0);
    }
    for(n=1,10^3, if(is(n),print1(n,", ")) );
    \\ Joerg Arndt, May 02 2016

Formula

a(n) = (A216371(n)-1)/2. - L. Edson Jeffery, Dec 18 2012
a(n) >> n log n, and on the Bateman-Horn-Stemmler conjecture a(n) << n log^2 n. I imagine a(n) ≍ n log n, and numerics suggest that perhaps a(n) ~ kn log n for some constant k (which seems to be around 1.122). - Charles R Greathouse IV, Aug 02 2023