%I #2 Mar 30 2012 17:34:25
%S 2,1,1,2,1,1,1,1,1,0,1,1,1,0,0,1,1,1,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,
%T 0,0,1,1,1,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,0,1
%N Triangle of coefficients from a polynomial recursion for Galois field GF(2^n) polynomials: p(x,n)=(x+1)*p(x,n-1)-x*p(x,n-2); or f(x,n)=x^n+x+1;.
%C Row sums are:
%C {3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3}
%C The result is very dependent on the two initial polynomials.
%D Taylor L. Booth, Sequential Machines and Automata Theory, John Wiley and Sons, Inc., 1967, Appendix I
%F p(x,0)=2+x;p(x,1)=1+2*x; p(x,n)=(x+1)*p(x,n-1)-x*p(x,n-2); or f(x,n)=x^n+x+1;
%e {2, 1},
%e {1, 2},
%e {1, 1, 1},
%e {1, 1, 0, 1},
%e {1, 1, 0, 0, 1},
%e {1, 1, 0, 0, 0, 1},
%e {1, 1, 0, 0, 0, 0, 1},
%e {1, 1, 0, 0, 0, 0, 0, 1},
%e {1, 1, 0, 0, 0, 0, 0, 0, 1},
%e {1, 1, 0, 0, 0, 0, 0, 0, 0, 1},
%e {1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1}
%t p[x, 0] = 2 + x; p[x, 1] = 1 + 2*x; p[x_, n_] := p[x, n] = (x + 1)*p[x, n - 1] - x*p[x, n - 2]; Table[ExpandAll[p[x, n]], {n, 0, 10}]; a = Table[CoefficientList[p[x, n], x], {n, 0, 10}] Flatten[a]
%Y Cf. A057764, A103204.
%K nonn,uned,tabl
%O 1,1
%A _Roger L. Bagula_, Apr 05 2008