OFFSET
0,2
COMMENTS
Conjecture: (i) a(n) > 0 for all n, and a(n) = 1 only for n = 0, 3, 6, 23, 44.
(ii) Any positive integer relatively prime to 6 can be written as (p-1)^2 + q^2 + 3*r^2, where p is prime, and q and r are integers.
(iii) Let n be any nonnegative integer. Then 4*n+1 can be written as x^2 + y^2 + z^2, where x,y,z are nonnegative integers with x + y + 2*z prime; 4*n+2 can be written as x^2 + y^2 + z^2, where x,y,z are nonnegative integers with x + 3*y prime. Also, we can write 4*n+3 as 2*x^2 + y^2 + z^2, where x,y,z are nonnegative integers with 2*x+1 prime.
(iv) Let n be any nonnegative integer. Then we can write 4*n+1 as x^2 + y^2 + z^2 with x,y,z integers such that x^2 + 5*y^2 + 7*z^2 prime, and write 4*n+2 as x^2 + y^2 + z^2 with x,y,z integers such that x^2 + 2*y^2 + 5*z^2 prime. Also, we can write 8*n+3 as x^2 + y^2 + z^2 with x,y,z integers such that 2*x^2 + 4*y^2 + 5*z^2 is prime.
Note that by the Gauss-Legendre theorem any positive integer not of the form 4^k*(8*m+7) (k,m = 0,1,2,...) can be written as the sum of three squares.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190.
Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017.
EXAMPLE
a(0) = 1 since 4*0+1 = (2-1)^2 + 0^2 + 0^2 with 2 prime.
a(3) = 1 since 4*3+1 = (3-1)^2 + 0^2 + 3^2 with 3 prime.
a(6) = 1 since 4*6+1 = (5-1)^2 + 0^2 + 3^2 with 5 prime.
a(23) = 1 since 4*23+1 = (3-1)^2 + 5^2 + 8^2 with 3 prime.
a(44) = 1 since 4*44+1 = (3-1)^2 + 2^2 + 13^2 with 3 prime.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0; Do[If[PrimeQ[p]&&SQ[4n+1-(p-1)^2-q^2], r=r+1], {p, 2, Sqrt[4n+1]+1}, {q, 0, Sqrt[(4n+1-(p-1)^2)/2]}]; Print[n, " ", r], {n, 0, 80}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Aug 17 2017
STATUS
approved