%I #24 Mar 26 2024 05:31:48
%S 2,3,4,5,6,7,21,26,99,158,405
%N Numbers k for which the absolute value of the discriminant of the polynomial x^k - x^(k-1) - ... - x - 1 is a prime times 2^m for some m >= 0.
%C This polynomial is the characteristic polynomial of the Fibonacci and Lucas k-step recursions. Are the k-step recursions different -- in some way -- for the values of k that yield a prime*2^m discriminant? No other k < 10000.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Fibonaccin-StepNumber.html">Fibonacci n-Step Number</a>.
%o (PARI) f(n) = poldisc('x^n-sum(k=0, n-1, 'x^k)); \\ A106273
%o isok(k) = my(x=abs(f(k))); ispseudoprime(x) || ispseudoprime(x/2^valuation(x, 2)); \\ _Michel Marcus_, Mar 26 2024
%Y Cf. A106273 (discriminant of the polynomial x^n - x^(n-1) - ... - x - 1).
%K nonn,hard,more
%O 1,1
%A _T. D. Noe_, May 02 2005