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A216066
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a(n) = card {cos((2^k)*Pi/(2*n-1)): k in N}.
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4
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1, 1, 2, 3, 3, 5, 6, 4, 4, 9, 6, 11, 10, 9, 14, 5, 5, 12, 18, 12, 10, 7, 12, 23, 21, 8, 26, 20, 9, 29, 30, 6, 6, 33, 22, 35, 9, 20, 30, 39, 27, 41, 8, 28, 11, 12, 10, 36, 24, 15, 50, 51, 12, 53, 18, 36, 14, 44, 12, 24, 55, 20, 50, 7, 7, 65, 18, 36, 34, 69, 46
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OFFSET
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1,3
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COMMENTS
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Essentially the same as A003558: a(n) is equal to the minimal value r in N for which either 2^r is congruent to 1 modulo 2*n-1 or 2^r is congruent to -1 modulo 2*n-1.
In view of Sharkovsky's Theorem numbers a(n) exert an essential influence on the chaotic nature (in the sense of Li and Yorke) of polynomials, for which the set {cos((2^k)*Pi/(2*n-1)): k in N} is a periodic cycle. For example from a(4) = 3 it follows (see Witula-Slota reference) that the set {c(1;7), c(2;7), c(4;7)}, where c(j;7) := cos(2*Pi*j/7), is a 3-element orbit of the polynomial p(x) = -x^3 + 2*x - 1 = -(x - c(1;9))*(x - c(2;9))*(x - c(4;9)), where c(j;9) := cos(2*Pi*j/9). "Period 3 implies chaos" of p(x) in the sense of Li and Yorke. Moreover from the Sharkovsky Theorem p(x) possesses cycle orbits of any positive lengths.
We note that A072451(n) is divisible by a(n) for every n in N (see Corollary 5.8 a) in Witula-Slota's paper - "whenever l(n)..." could be replaced by "whenever n..." in this Corollary). We have a(n) = A072451(n) for every n=1,...,20 except 9, 16 and 17 (a(9)=4, a(16)=a(17)=5, A072451(9)=8, A072451(16)=15 and A072451(17)=10).
The following fact (strongly than previously one) is also true: the value of the Carmichael lambda function for the argument 2*n-1, i.e., A002322(2*n-1) is divisible by a(n) for every n in N.
I want to formulate some problem: for which k in N there is a subsequence k,k in the sequence a(n)? We note that for k = 1,3,...,7 the answer is positive. Moreover, I am interesting for which k in N the equation a(n) = k has the infinite set of solutions n in N?
I observe that also A065457(n) is divisible by a(n) for every n in N and A002322(2*n+1) is divisible by A065457(n+1) for every n in N - but I don't know why these relations hold true. - Roman Witula, Sep 10 2012
If you write n letters in a line, for example n=5, abcde, and then put the last after the first, the second last after the second and so on, you will get aebdc. After this, you can apply the same transformation to the new string. Doing this transformation a(n) times will lead you eventually back to the original string; see the second PARI program. This idea is from Wolfgang Tomášek. - Robert Pfister, Sep 12 2013
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LINKS
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FORMULA
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EXAMPLE
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We have a(2)=1, a(3)=2, a(4)=3 and a(12)=11, a(11)=10, a(10)=9, and a(45)=11, a(46)=12, a(47)=10. Does exist some another k,l in N for which a(k)=p(l), a(k+1)=p(l+1), and a(k+2)=p(l+2), where p is a permutation on {l,l+1,l+2}?
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MATHEMATICA
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Suborder[k_, n_] := If[n > 1 && GCD[k, n] == 1, Min[MultiplicativeOrder[k, n, {-1, 1}]], 0];
a[n_] := If[n == 0, 1, Suborder[2, 2 n + 1]];
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PROG
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(PARI)
a(n) = {
my( g=Mod(2, 2*n-1), f=g );
for (r=1, 2*n+2,
if ( f == +1, return(r) );
if ( f == -1, return(r) );
f *= g;
);
}
a(n) = {
my( g = vectorsmall(n), e=vectorsmall(n, k, k), t );
my( ct = 1 );
\\ set g[] to the zip-permutation:
forstep ( k=1, n, 2, g[k] = k\2 + 1);
forstep ( k=2, n, 2, g[k] = n - k\2 + 1);
t = g;
while ( t != e, \\ until we hit identity
ct += 1;
t *= g; \\ t == g^ct
);
return( ct );
}
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CROSSREFS
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A003558 is essentially the same sequence except for the offset.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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