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%I #122 Sep 08 2022 08:46:20
%S 1,3,5,4,9,11,9,5,12,12,7,23,8,20,29,6,33,35,20,39,41,28,12,36,15,51,
%T 53,36,44,24,20,7,65,36,69,60,42,15,20,52,81,83,9,60,89,60,40,95,12,
%U 99,84,66,105,28,18,37,113,30,92,119,81,36,25,8,36,131,22,135,20,30,47,60,48,116,132,100,51,155
%N Order of the "inside-out" permutation on 2n+1 letters.
%C The "inside-out" permutation (closely related to the Mongean shuffle, see A019567) sends (t_1, t_2, ..., t_{2n+1}) to (t_{n+1}, t_{n+2}, t_{n}, t_{n+3}, t_{n-1}, ..., t_1). For n = 0, 1, 2, 3, this is (1), (2,3,1), (3,4,2,5,1), (4,5,3,6,2,7,1), whose orders are respectively 1,3,5,4.
%C This is the odd bisection of A238371 and also the odd bisection of A003558 (see Joseph L. Wetherell's comment below).
%H Robert Israel, <a href="/A294673/b294673.txt">Table of n, a(n) for n = 0..10000</a>
%H N. J. A. Sloane, <a href="/A294673/a294673.txt">Table of n, a(n) for n = 0..32683</a> (computed using Robert Israel's Maple program)
%F The permutation sends i (1 <= i <= 2n+1) to p(i) = n + 1 + f(i), where f(i) = (-1)^i*ceiling((i-1)/2).
%F a(n) = minimal k>0 such that p^k() = p^0().
%F a((A163778(n)-1)/2) = A163778(n). - _Andrew Howroyd_, Nov 11 2017.
%F From _Joseph L. Wetherell_, Nov 14 2017: (Start)
%F a(n) is equal to the order of multiplication-by-2 acting on the set of nonzero elements in (Z/(4n+3)Z), modulo the action of +-1. To be precise, identify i=1,2,...,2*n+1 with the odd representatives J=1,3,...,4*n+1 of this set, via the map J = 2*i-1. It is not hard to show that the induced permutation on the set of J values is given on integer representatives by J -> (4*n+3+J)/2 if i=(J+1)/2 is even and J -> (4*n+3-J)/2 if i=(J+1)/2 is odd. It follows that this induces the permutation J -> +-J/2 (mod 4*n+3), from which we immediately see that the order is as stated.
%F Note that the order of 2 acting on (Z/(4n+3)Z)/{+-1} is the same as the order of either 2 or -2 acting on (Z/(4n+3)Z), depending on which of these is a quadratic residue modulo 4n+3. Thus an equivalent (and often easier) way to compute a(n) is as the order of -2*(-1)^n acting on (Z/(4n+3)Z).
%F Among other things, the lower and upper bounds log_2(n) + 2 < a(n) <= 2*n+1 follow immediately.
%F (End)
%F It appears that the upper bound a(n) = 2n+1 occurs iff 2n+1 belongs to A163778 or equivalently iff n belongs to A294434. This almost (but not quite) follows from the above comments by _Andrew Howroyd_ and _Joseph L. Wetherell_. - _N. J. A. Sloane_, Nov 16 2017
%e For n=2: Iterating the "inside-out" permutation of a string of length 2n+1=5:
%e 12345
%e 34251
%e 25413
%e 41532
%e 53124
%e 12345
%e ...
%e which has order a(2) = 5.
%p f:= proc(n)
%p ilcm(op(map(nops,convert(map(op, [[n+1],seq([n+1+i,n+1-i],i=1..n)]),disjcyc))))
%p end proc:
%p map(f, [$0..100]); # _Robert Israel_, Nov 09 2017
%t a[n_] := MultiplicativeOrder[-2(-1)^n, 4n+3];
%t a /@ Range[0, 100] (* _Jean-François Alcover_, Apr 07 2020 *)
%o (PARI)
%o Follow(s,f)={my(t=f(s),k=1); while(t>s, k++; t=f(t)); if(s==t, k, 0)}
%o CyclePoly(n,x)={my(p=0); for(i=1, 2*n+1, my(l=Follow(i, j->n+1+(-1)^j*ceil((j-1)/2) )); if(l,p+=x^l)); p}
%o a(n)={my(p=CyclePoly(n,x), m=1); for(i=1,poldegree(p),if(polcoeff(p,i), m=lcm(m,i))); m} \\ _Andrew Howroyd_, Nov 08 2017
%o (PARI) a(n)=znorder(Mod(if(n%2,2,-2),4*n+3)) \\ See Wetherell formula; _Charles R Greathouse IV_, Nov 15 2017
%o (Magma)
%o f:=func<n| Order(Sym(2*n+1)![n+1+(-1)^i*Ceiling((i-1)/2): i in [1..2*n+1]]) >;
%o [f(n): n in [0..100]]; \\ _Joseph L. Wetherell_, Nov 12 2017
%o (Magma)
%o [Order(Integers(4*n+3)!-2*(-1)^n): n in [0..100]];
%o \\ _Joseph L. Wetherell_, Nov 15 2017
%Y Cf. A003558, A019567, A163778, A238371, A294434.
%K nonn,look
%O 0,2
%A _P. Michael Hutchins_, Nov 06 2017
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