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Numbers matched to polynomials divisible by x^2+x+1.
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%I #17 Nov 04 2014 02:08:32

%S 7,14,21,28,35,42,49,56,63,70,84,98,112,119,126,133,140,161,168,175,

%T 189,196,224,231,238,245,252,259,266,273,280,287,315,322,336,343,350,

%U 371,378,385,392,399,413,427,441,448,455,462,469,476,483,490,497

%N Numbers matched to polynomials divisible by x^2+x+1.

%C The polynomials having coefficients in {0,1} are enumerated as in A206074. The sequence A206717 shows the numbers of those satisfying p(n,r)=0, where r=(-1+i*sqrt(3))/2. Is every term divisible by 7?

%C Yes, because if p(n,x) is divisible by p(7,x) = 1 + x + x^2, then n = p(n,2) is divisible by p(7,2) = 7. Similarly, A000120(n) is divisible by 3, because A000120(n) = p(n,1) must be divisible by p(7,1) = 3. The least n such that n is divisible by 7 and A000120(n) is divisible by 3 but n is not in the sequence is 4683. - _Robert Israel_, Nov 03 2014

%H Robert Israel, <a href="/A206717/b206717.txt">Table of n, a(n) for n = 1..10000</a>

%e p(7,x) = 1 + x + x^2.

%e p(14,x) = x + x^2 + x^3.

%p filter:= proc(n) local L,P,x;

%p L:= convert(n,base,2);

%p P:= add(L[i]*x^(i-1),i=1..nops(L));

%p rem(P,x^2+x+1,x) = 0

%p end proc:

%p select(filter, 7*[$1..500]); # _Robert Israel_, Nov 03 2014

%t t = Table[IntegerDigits[n, 2], {n, 1, 3000}];

%t b[n_] := Reverse[Table[x^k, {k, 0, n}]]

%t p[n_, x_] := p[n, x] = t[[n]].b[-1 + Length[t[[n]]]]

%t TableForm[Table[{n, p[n, x], Factor[p[n, x]]},

%t {n, 1, 16}]]

%t u = {}; Do[n++; If[Simplify[(p[n, x] /. x -> (-1 + I*Sqrt[3])/2) == 0], AppendTo[u, n]], {n, 800}]

%t u (* A206717 *)

%t u/7 (* A206718 *)

%Y Cf. A000120, A206074, A206718.

%K nonn

%O 1,1

%A _Clark Kimberling_, Feb 11 2012