OFFSET
1,2
COMMENTS
For odd k, Lucas(5*k)/Lucas(k) = A124297(k)*A124296(k) is the Aurifeuillian factorization into "plus" and "minus" halves; this sequence gives the k with prime plus half. Cf. A190781.
11 = Lucas(5) divides A124297(k) for k == 1, 9 (mod 10) and A124296(k) for k == 3, 7 (mod 10), so every term is == 3 or 7 (mod 10) apart from k = 1 and the multiples of 5.
5*a(n) is a term of A061443. That sequence is larger: it also contains 5*k when only the primitive part of A124297(k) is prime, as at k = 9, where A124297(9) = 5951 = 11*541.
A124297(k) - 1 = 5*F(k)*(F(k) + 1) factors by construction, and 5*(F(k) + 1) exceeds sqrt(A124297(k)), so a fully factored algebraic part supports an N-1 primality proof.
a(1)-a(13) correspond to proven primes, a(12) and a(13) with complete Lucas N-1 certificates (witnesses 2 and 11); a(14)-a(18) correspond to probable primes.
a(19) > 10^5. - Michael S. Branicky, Jul 22 2026
REFERENCES
J. Brillhart, D. H. Lehmer and J. L. Selfridge, New primality criteria and factorizations of 2^m +- 1, Mathematics of Computation, 29 (1975), 620-647.
EXAMPLE
PROG
(PARI) isok(k) = (k % 2) && ispseudoprime(5*fibonacci(k)^2 + 5*fibonacci(k) + 1);
select(isok, [1..1000])
CROSSREFS
KEYWORD
nonn,hard,more
AUTHOR
Dillon Ryan, Jul 20 2026
EXTENSIONS
a(16)-a(18) from Michael S. Branicky, Jul 20 2026
STATUS
approved