OFFSET
1,5
COMMENTS
Conjecture: a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 3, 4, 9, 17, 21, 25, 33, 49, 109, 169, 189, 361, 841, 961, 12769, 19321.
See also A276825 for a similar conjecture involving cubes, and some comments on x^2 + P_2.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
EXAMPLE
a(2) = 1 since 2 = ((3-1)/2)^2 + 1 with 3 prime.
a(3) = 1 since 3 = ((3-1)/2)^2 + 2 with 3 and 2 both prime.
a(4) = 1 since 4 = ((3-1)/2)^2 + 3 with 3 prime.
a(9) = 1 since 9 = ((5-1)/2)^2 + 5 with 5 prime.
a(17) = 1 since 17 = ((5-1)/2)^2 + 13 with 5 and 13 both prime.
a(21) = 1 since 21 = ((5-1)/2)^2 + 17 with 5 and 17 both prime.
a(25) = 1 since 25 = ((5-1)/2)^2 + 3*7 with 5, 3 and 7 all prime.
a(33) = 1 since 33 = ((5-1)/2)^2 + 29 with 5 and 29 both prime.
a(49) = 1 since 49 = ((13-1)/2)^2 + 13 with 13 prime.
a(109) = 1 since 109 = ((13-1)/2)^2 + 73 with 13 and 73 both prime.
a(169) = 1 since 169 = ((13-1)/2)^2 + 7*19 with 13, 7 and 19 all prime.
a(189) = 1 since 189 = ((5-1)/2)^2 + 5*37 with 5 and 37 both prime.
a(361) = 1 since 361 = ((37-1)/2)^2 + + 37 with 37 prime.
a(841) = 1 since 841 = ((37-1)/2)^2 + 11*47 with 37, 11 and 47 all prime.
a(961) = 1 since 961 = ((61-1)/2)^2 + 61 with 61 prime.
a(12769) = 1 since 12769 = ((109-1)/2)^2 + 59*167 with 109, 59 and 167 all prime.
a(19321) = 1 since 19321 = ((277-1)/2)^2 + 277 with 277 prime.
MATHEMATICA
PP[n_]:=PP[n]=PrimeQ[Sqrt[n]]||(SquareFreeQ[n]&&Length[FactorInteger[n]]<=2)
Do[r=0; Do[If[PP[n-((Prime[k]-1)/2)^2], r=r+1; If[r>1, Goto[aa]]], {k, 2, PrimePi[2*Sqrt[n]+1]}]; Print[n, " ", r];
Label[aa]; If[Mod[n, 50000]==0, Print[n]]; Continue, {n, 10^5, 1000000}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Sep 20 2016
STATUS
approved