OFFSET
1,1
COMMENTS
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.
LINKS
Eric Weisstein's World of Mathematics, Fibonacci n-Step Number.
PROG
(PARI) f(n) = poldisc('x^n-sum(k=0, n-1, 'x^k)); \\ A106273
isok(k) = my(x=abs(f(k))); ispseudoprime(x) || ispseudoprime(x/2^valuation(x, 2)); \\ Michel Marcus, Mar 26 2024
CROSSREFS
KEYWORD
nonn,hard,more
AUTHOR
T. D. Noe, May 02 2005
STATUS
approved