OFFSET
1,1
COMMENTS
Consider k = 4 * x^2 + 1, where the largest square <= k^3 is (8 * x^3 + 3 * x)^2. Bunyakowsky's conjecture implies there are infinitely many primes of the form (4 * x^2 + 1)^3 + (8 * x^3 + 3 * x)^2, and so that the sequence is infinite. The first term of this form is 93389778901 = a(165), corresponding to x = 30. - Robert Israel, Jan 10 2024
LINKS
Robert Israel, Table of n, a(n) for n = 1..10000
MAPLE
g:= n -> n + floor(sqrt(n))^2:
select(isprime, map(g, [seq(i^3, i=1..1000)])); # Robert Israel, Jan 10 2024
MATHEMATICA
r[n_]:=n^3; f[n_]:=r[n]+Floor[Sqrt[r[n]]]^2; Select[Table[f[n], {n, 0, 6!}], PrimeQ[ # ]&]
PROG
(PARI) select(isprime, vector(1000, n, n^3+sqrtint(n^3)^2)) \\ Michel Marcus, Jan 10 2024
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Joseph Stephan Orlovsky, Apr 21 2010
STATUS
approved