A291191 Number of ways to write 2*n+1 as x^2 + y^2 + z^2 + w^2, where x,y,z,w are nonnegative integers with x <= y, z <= w and x + y < z + w such that 2^(x+y) + 2^(z+w) + 1 is prime.

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

Offset

1,4

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 2, 3, 6, 7, 11, 12, 16, 19, 20, 23, 26, 63, 75, 98.

(ii) Any integer n > 1 can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that 2^(x+y) + 2^(z+w) is a practical number (A005153).

(iii) Any positive odd integer can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that 2^(x+y) + 3^(z+w) is prime.

(iv) Any positive integer not divisible by 32 can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that 2^x + 3^y + 4^z is prime.

See also A291150 for a similar conjecture.

I have verified that a(n) > 0 for all n = 1..10^7. For example, a(6998538) > 0 since 2*6998538+1 = 122^2 + 220^2 + 208^2 + 3727^2 with 2^(122+220) + 2^(208+3727) + 1 = 2^342 + 2^3935 + 1 a prime of 1185 decimal digits. - Zhi-Wei Sun, Aug 23 2017

Links

Zhi-Wei Sun, Table of n, a(n) for n = 1..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(1) = 1 since 2*1+1 = 0^2 + 1^2 + 1^2 + 1^2 with 2^(0+1) + 2^(1+1) + 1 = 7 prime.
a(2) = 1 since 2*2+1 = 0^2 + 1^2 + 0^2 + 2^2 with 2^(0+1) + 2^(0+2) + 1 = 7 prime.
a(19) = 1 since 2*19+1 = 2^2 + 3^2 + 1^2 + 5^2 with 2^(2+3) + 2^(1+5) + 1 = 97 prime.
a(26) = 1 since 2*26+1 = 1^2 + 4^2 + 0^2 + 6^2 with 2^(1+4) + 2^(0+6) + 1 = 97 prime.
a(63) = 1 since 2*63+1 = 1^2 + 5^2 + 1^2 +10^2 with 2^(1+5) + 2^(1+10) + 1 = 2113 prime.
a(75) = 1 since 2*75+1 = 1^2 + 5^2 + 5^2 + 10^2 with 2^(1+5) + 2^(5+10) + 1 = 32833 prime.
a(98) = 1 since 2*98+1 = 6^2 + 6^2 + 2^2 + 11^2 with 2^(6+6) + 2^(2+11) + 1 = 12289 prime.

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0; Do[If[SQ[2n+1-x^2-y^2-z^2]&&x+y<z+Sqrt[2n+1-x^2-y^2-z^2]&&PrimeQ[2^(x+y)+2^(z+Sqrt[2n+1-x^2-y^2-z^2])+1], r=r+1], {x, 0, Sqrt[n]}, {y, x, Sqrt[2n+1-x^2]}, {z, 0, Sqrt[(2n+1-x^2-y^2)/2]}]; Print[n, " ", r], {n, 1, 100}]

Crossrefs

Cf. A000040, A000079, A000118, A000290, A005153, A271518, A281976, A290935, A291150.

Sequence in context: A122923 A113971 A109338 * A273866 A071202 A207379

Adjacent sequences: A291188 A291189 A291190 * A291192 A291193 A291194

Keyword

nonn

Author

Zhi-Wei Sun, Aug 20 2017

Status

approved