%I #5 Mar 13 2015 23:50:12
%S 209,517,3973,1081,1285,2893,13501,38579,105113,4897,12331
%N Least number k such that the polynomial x^n  x^(n1) ... 1 (mod k) has more than n distinct zeros.
%C This is the characteristic polynomial of the nstep Fibonacci and Lucas sequences. These terms produce the following number of distinct zeros: 4, 6, 8, 6, 8, 8, 10, 12, 15, 12, 18. The first 11 terms are semiprimes; the 12th term has 3 factors. For prime k, the polynomial can have at most n zeros.
%t Clear[x]; Table[poly = x^n  Sum[x^k, {k, 0, n  1}]; k = 1; While[cnt = 0; Do[If[Mod[poly, k] == 0, cnt++], {x, 0, k1}]; cnt <= n, k++]; k, {n, 2, 7}]
%Y Cf. A211671 (for prime k).
%K nonn,hard,more
%O 2,1
%A _T. D. Noe_, Apr 19 2012
