[seqfan] Re: Odd behavior in a sequence
israel at math.ubc.ca
israel at math.ubc.ca
Wed Oct 21 03:10:37 CEST 2015
The solution of the recurrence with initial conditions a(0)=0, a(1)=1 is
a(n) = (s^n - t^n)/sqrt(17)
where s = (3+sqrt(17))/2 and t = (3-sqrt(17))/2 = -2/s. This works in any
field (not of characteristic 2 or 17) where 17 has a square root. Thus if p
is an odd prime and 17 is a quadratic residue mod p, say 17 == b^2 mod p,
we have a(n) == (s^n - t^n)/b mod p where s == (3+b)/2 and t == -2/s. Now
by Fermat's little theorem, a(n) is periodic in n with period at most p-1,
so in this case it's impossible to cover all the residues.
On the other hand, if 17 is not a quadratic residue mod p, you go to an
extension field, and it's possible that all residues will be covered.
However, this is not guaranteed. Thus for p = 683, s and t have order 396,
so at most 396 of the residues are covered (in fact only 177 are).
Note, by the way, that by quadratic reciprocity 17 is a quadratic residue
mod the odd prime p iff p is a quadratic residue mod 17, i.e. p == 1, 2, 4,
8, 9, 13, 15, or 16 mod 17.
Cheers,
Robert
On Oct 20 2015, David Wilson wrote:
>I'm looking at the linear recurrence
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>0, 1, 3, 11, 39, 139, .
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>with a(n) = 3a(n-1) + 2a(n-2) for n >= 2.
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> I was interested in the question, for which primes p do {a(n)} == Z (mod
> p)?
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>That is, for which primes p do all residues r appear in {a(n)} (mod p)
>(Forgive my notational abuse)?
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>We'll shorten that to "a covers p".
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>I expected this to have a simple answer, and it almost does.
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>It turns out that whether or not a covers p most of the time depends on p
>mod 17.
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>In general, a covers p if p == 0, 3, 5, 6, 7, 10, 11, 12, or 14 (mod 7).
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>However, there seem to be a smattering of primes p:
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>683, 1217, 2731, 11299, 48817, .
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>which, according to their residue mod 17, should be covered by a, but are
>not.
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>Very curious.
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