A107751

[Artur]

If coefficient polynomials x^n+x+1 we will treated as binary digits
we will have
n=2  111 = 7
n=3  1011 =11
n=4 10011 =19
n=5 100011 =35
if polynomial is divisable by x^2+x+1 that mean that its binary number 
is divisable by 7 and this is true (see A046636 
<http://www.research.att.com/%7Enjas/sequences/A046636>):
a = {}; Do[m = CoefficientList[x^(3n + 2) + x + 1, x]; k = 
FromDigits[Reverse[m], 2]; AppendTo[a, k/7], {n, 1, 200}]; a

ARTUR JASINSKI




Robert Israel pisze:
> You may have noticed that (at least for n up to 200) whenever your 
> polynomial x^n+x+1 factors, one of the factors is x^2+x+1.  I don't
> know if this is true for all n, but if so there's doubtless a good
> reason for it.  In any case it's easy to see that for n >= 2, x^n+x+1
> is divisible by x^2+x+1 if and only if n == 2 mod 3.
>
> Robert Israel                                israel at math.ubc.ca
> Department of Mathematics        http://www.math.ubc.ca/~israel 
> University of British Columbia            Vancouver, BC, Canada
>
> On Sat, 13 Oct 2007, Artur wrote:
>
>> I would like to ask is somebody which understand A107751 and is able 
>> write Mathematica procedure. I was obtained that same sequence in 
>> completely another problem and connection is very unexpected  I want 
>> check that these two are that same. My procedure is on number of 
>> factors of  polynomials (x^n+x+1) :
>> Table[Length[FactorList[x^n + x + 1]] - 1, {n, 0, 200}]
>> {1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 
>> 1, 1,
>> 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 
>> 1, 2, 1,
>> 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 
>> 1, 1, 2,
>> 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 
>> 2, 1, 1,
>> 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 
>> 1, 2, 1,
>> 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 
>> 1, 1, 2,
>> 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 
>> 2, 1, 1,
>> 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2}
>>
>> I will be greatfull for procedure or explanation that these two 
>> different matters are that same
>>
>> Best wishes
>> Artur
>>
>
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