OFFSET
1,9
COMMENTS
Conjecture: a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 16, 20, 22, 28, 42, 50.
We have verified a(n) > 0 for all n = 2..10^7.
See also A232174 for a similar conjecture.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Conjectures on representations involving primes, in: M. Nathanson (ed.), Combinatorial and Additive Number Theory II, Springer Proc. in Math. & Stat., Vol. 220, Springer, Cham, 2017, pp. 279-310. (See also arXiv:1211.1588 [math.NT], 2012-2017.)
EXAMPLE
a(2) = 1 since 2 = 1 + 1 with 1^3 + 2*1^2 = 3 prime.
a(3) = 1 since 3 = 1 + 2 with 1^3 + 3*2^2 = 13 prime.
a(11) = 1 since 11 = 5 + 6 with 5^3 + 11*6^2 = 521 prime.
a(20) = 1 since 20 = 3 + 17 with 3^3 + 20*17^2 = 5807 prime.
a(28) = 1 since 28 = 9 + 19 with 9^3 + 28*19^2 = 10837 prime.
a(42) = 1 since 42 = 19 + 23 with 19^3 + 42*23^2 = 29077 prime.
a(50) = 1 since 50 = 3 + 47 with 3^3 + 50*47^2 = 110477 prime.
MATHEMATICA
tab={}; Do[r=0; Do[If[PrimeQ[x^3+n(n-x)^2], r=r+1], {x, 1, n/2}]; tab=Append[tab, r], {n, 1, 90}]; Print[tab]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jun 03 2018
STATUS
approved