OFFSET
1,1
COMMENTS
This permutation consists of just two cycles, both infinite.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Joe Buhler and R. L. Graham, Juggling Drops and Descents, Amer. Math. Monthly, 101, (no. 6) 1994, 507 - 519.
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 2|z|+1, with inverse g(z) = z/2 if z even else (1-z)/2. Then a(n) = f(g(n)-2).
G.f.: x*(x^6-x^5+4*x^4-4*x^3-x^2-2*x+5) / ((x-1)^2*(x+1)). - Colin Barker, Feb 18 2013
a(n) = -4*(-1)^n+n for n>4. a(n) = a(n-1)+a(n-2)-a(n-3) for n>7. - Colin Barker, Mar 07 2014
MATHEMATICA
CoefficientList[Series[(x^6 - x^5 + 4 x^4 - 4 x^3 - x^2 - 2 x + 5)/((x - 1)^2 (x + 1)), {x, 0, 100}], x] (* Vincenzo Librandi, Mar 08 2014 *)
PROG
(PARI) Vec(x*(x^6-x^5+4*x^4-4*x^3-x^2-2*x+5)/((x-1)^2*(x+1)) + O(x^100)) \\ Colin Barker, Mar 07 2014
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Antti Karttunen, Oct 19 2001
STATUS
approved