

A065164


Permutation t>t+1 of Z, folded to N.


6



2, 4, 1, 6, 3, 8, 5, 10, 7, 12, 9, 14, 11, 16, 13, 18, 15, 20, 17, 22, 19, 24, 21, 26, 23, 28, 25, 30, 27, 32, 29, 34, 31, 36, 33, 38, 35, 40, 37, 42, 39, 44, 41, 46, 43, 48, 45, 50, 47, 52, 49, 54, 51, 56, 53, 58, 55, 60, 57, 62, 59, 64, 61, 66, 63, 68, 65, 70, 67, 72, 69, 74
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OFFSET

1,1


COMMENTS

Corresponds to simple periodic asynchronic site swap pattern ...111111... (tossing one ball from hand to hand forever).
This permutation consists of a single infinite cycle.
This is, starting at a(2) = 4, the same as the "increasing oscillating sequence" shown in Proposition 3.1, p.7 and plotted in the right of figure 1, of Vatter. The same paper, p.4, cites Comtet and uses without giving the Anumber of A003319. Abstract: We prove that there are permutation classes (hereditary properties of permutations) of every growth rate (StanleyWilf limit) at least lambda = approx 2.48187, the unique real root of x^52x^42x^22x1, thereby establishing a conjecture of Albert and Linton.  Jonathan Vos Post, Jul 18 2008


REFERENCES

Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 819.


LINKS

Table of n, a(n) for n=1..72.
Michael H. Albert, Robert Brignall, Vincent Vatter, Large infinite antichains of permutations, arXiv:1212.3346 [math.CO], 2012.
Joe Buhler and R. L. Graham, Juggling Drops and Descents, Amer. Math. Monthly, 101, (no. 6) 1994, 507  519.
Jay Pantone, Vincent Vatter, Growth rates of permutation classes: categorization up to the uncountability threshold, arXiv:1605.04289 [math.CO], 2016.
Vincent Vatter, Permutation classes of every growth rate (a.k.a. StanleyWilf limit) above 2.48187.., arXiv:0807.2815 [math.CO], 20082009.
Vincent Vatter, Permutation classes, arXiv:1409.5159 [math.CO], 2014.
Index entries for sequences that are permutations of the natural numbers
Index entries for linear recurrences with constant coefficients, signature (1,1,1).


FORMULA

Let f: Z > N be given by f(z) = 2z if z>0 else 2z+1, with inverse g(z) = z/2 if z even else (1z)/2. Then a(n) = f(g(n)+1).
a(n) = n + 2*(1^n) for n > 1.  Frank Ellermann, Feb 12, 2002
a(n) = 2*na(n1)1, n>2.  Vincenzo Librandi, Dec 07 2010, corrected by R. J. Mathar, Dec 07 2010
From Colin Barker, Feb 18 2013: (Start)
a(n) = a(n1) + a(n2)  a(n3) for n>4.
G.f.: x*(3*x^35*x^2+2*x+2) / ((x1)^2*(x+1)). (End)


MAPLE

ss1 := [seq(PerSS(n, 1), n=1..120)]; PerSS := (n, c) > Z2N(N2Z(n)+c);
N2Z := n > ((1)^n)*floor(n/2); Z2N := z > 2*abs(z)+`if`((z < 1), 1, 0);


CROSSREFS

Row 1 of A065167. Obtained by composing permutations A014681 and A065190. Inverse permutation: A065168.
Sequence in context: A225679 A081879 A066248 * A138124 A128860 A019680
Adjacent sequences: A065161 A065162 A065163 * A065165 A065166 A065167


KEYWORD

nonn,easy


AUTHOR

Antti Karttunen, Oct 19 2001


STATUS

approved



