OFFSET
1,1
COMMENTS
Equivalently, numbers k such that there exists a prime of the form k^6 - m^3. Proof: Let d = k^2 - m. Then m = k^2 - d, so k^6 - m^3 = k^6 - (k^2 - d)^3 = k^6 - (k^6 - 3*k^4*d + 3*k^2*d^2 - d^3) = d*(3*k^4 - 3*k^2*d + d^2), which cannot be prime unless d = 1, i.e., k^6 - m^3 = 3*k^4 - 3*k^2 + 1.
CROSSREFS
KEYWORD
nonn
AUTHOR
Jon E. Schoenfield, Jul 12 2024
STATUS
approved