

A206717


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


2



7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 84, 98, 112, 119, 126, 133, 140, 161, 168, 175, 189, 196, 224, 231, 238, 245, 252, 259, 266, 273, 280, 287, 315, 322, 336, 343, 350, 371, 378, 385, 392, 399, 413, 427, 441, 448, 455, 462, 469, 476, 483, 490, 497
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OFFSET

1,1


COMMENTS

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?
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


LINKS



EXAMPLE

p(7,x) = 1 + x + x^2.
p(14,x) = x + x^2 + x^3.


MAPLE

filter:= proc(n) local L, P, x;
L:= convert(n, base, 2);
P:= add(L[i]*x^(i1), i=1..nops(L));
rem(P, x^2+x+1, x) = 0
end proc:


MATHEMATICA

t = Table[IntegerDigits[n, 2], {n, 1, 3000}];
b[n_] := Reverse[Table[x^k, {k, 0, n}]]
p[n_, x_] := p[n, x] = t[[n]].b[1 + Length[t[[n]]]]
TableForm[Table[{n, p[n, x], Factor[p[n, x]]},
{n, 1, 16}]]
u = {}; Do[n++; If[Simplify[(p[n, x] /. x > (1 + I*Sqrt[3])/2) == 0], AppendTo[u, n]], {n, 800}]


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



