A291150 Number of ways to write 2*n+1 as x^2 + y^2 + z^2 + 4*w^2, where x,y,z,w are nonnegative integers with x <= y <= z such that 2^x + 2^y + 2^z + 1 is prime.

1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 3, 2, 1, 1, 3, 2, 3, 2, 2, 1, 6, 1, 3, 2, 3, 2, 5, 2, 3, 3, 5, 2, 5, 1, 6, 5, 6, 2, 6, 1, 5, 1, 5, 4, 8, 3, 4, 2, 2, 2, 8, 2, 6, 4, 3, 2, 4, 1, 5, 4, 7, 3, 7, 2, 7, 4, 5, 3, 10, 1, 7, 4, 5, 2, 13, 4, 6, 5, 5

Offset: 0

Author

Zhi-Wei Sun, Aug 19 2017

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 1, 2, 3, 5, 7, 11, 14, 15, 21, 23, 35, 41, 43, 59, 71, 309, 435.

(ii) Any positive even number not divisibly by 8 and other than 6 and 14 can be written as x^2 + y^2 + z^2 + w^2, where w is a positive odd integer, and x,y,z are nonnegative integers with 2^x + 2^y + 2^z + 1 prime.

(iii) Let n be a positive integer. If n is not divisible by 8, then n can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that 2^x + 2^y + 1 is prime. If n is not a multiple of 2^7 = 128, then we can write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that 2^x + 2^y - 1 is prime.

(iv) Let n be a positive integer. If n is not divisible by 8, then n can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that 2^x + 2*2^y + 3*2^z + 4*2^w - 1 is prime. If n is not a multiple of 2^8 = 256, then we can write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that 2^x + 2*2^y +3*2^z + 4*2^w + 1 is prime.

I have verified that a(n) > 0 for all n = 0..10^7. - Zhi-Wei Sun, Aug 23 2017

Examples

			a(0) = 1 since 2*0+1 = 0^2 + 0^2 + 1^2 + 4*0^2 with 2^0 + 2^0 + 2^1 + 1 = 5 prime.
a(14) = 1 since 2*14+1 = 2^2 + 3^2 + 4^2 + 4*0^2 with 2^2 + 2^3 + 2^4 + 1 = 29 prime.
a(35) = 1 since 2*35+1 = 1^2 + 3^2 + 5^2 + 4*3^2 with 2^1 + 2^3 + 2^5 + 1 = 43 prime.
a(43) = 1 since 2*43+1 = 1^2 + 5^2 + 5^2 + 4*3^2 with 2^1 + 2^5 + 2^5 + 1 = 67 prime.
a(59) = 1 since 2*59+1 = 1^2 + 3^2 + 3^2 + 4*5^2 with 2^1 + 2^3 + 2^3 + 1 = 19 prime.
a(71) = 1 since 2*71+1 = 1^2 + 5^2 + 9^2 + 4*3^2 with 2^1 + 2^5 + 2^9 + 1 = 547 prime.
a(309) = 1 since 2*309+1 = 5^2 + 13^2 + 13^2 + 4*8^2 with 2^5 + 2^13 + 2^13 + 1 = 16417 prime.
a(435) = 1 since 2*435+1 = 13^2 + 13^2 + 23^2 + 4*1^2 with 2^13 + 2^13 + 2^23 + 1 = 8404993 prime.
		

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.

Crossrefs

Cf. A000040, A000079, A000118, A000290, A271518, A281976, A290935, A291191.

Programs

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0;Do[If[SQ[(2n+1-x^2-y^2-z^2)/4]&&PrimeQ[2^x+2^y+2^z+1],r=r+1],{x,0,Sqrt[(2n+1)/3]},{y,x,Sqrt[(2n+1-x^2)/2]},{z,y,Sqrt[2n+1-x^2-y^2]}];Print[n," ",r],{n,0,80}]