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A290935
Number of ways to write 2*n+1 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that p = x^2 + 3*y^2 + 5*z^2 + 7*w^2 and p - 2 are twin prime.
6
2, 1, 4, 2, 1, 1, 4, 2, 2, 6, 2, 1, 6, 1, 2, 8, 5, 3, 7, 1, 4, 10, 3, 2, 9, 4, 8, 7, 5, 5, 11, 4, 7, 8, 4, 3, 10, 5, 6, 10, 7, 4, 16, 4, 9, 10, 2, 3, 11, 7, 5, 8, 3, 7, 13, 4, 4, 16, 2, 6, 15, 1, 4, 10, 6, 6, 13, 7, 2, 13, 8, 9, 15, 4, 12, 8, 7, 5, 7, 2, 9
OFFSET
0,1
COMMENTS
Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 1, 4, 5, 11, 13, 19, 61.
This refinement of Lagrange's four-square theorem implies the twin prime conjecture.
Below we list some similar conjectures:
(i) 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 p = 3*x^2 + 5*y^2 + 11*z^2 + 13*w^2 and p + 2 are twin prime.
(ii) Each positive odd integer can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that p = x + y + 3*z + 5*w and p + 2 (or p - 2) are twin prime.
(iii) Any positive odd integer can be can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x^3 + y^3 + z^3 + 3*w^3 is prime.
(iv) For each m = 1,2,3, any positive integer not divisible by 4 can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x^m + 2*y^m + 3*z^m + 4*w^m is prime.
(v) Let n be any positive integer not divisible by 4. 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^4 + 3*y^4 + 4*z^4 + 5*w^4 is prime. Also, we can write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that 3*x^5 + 4*y^5 + 5*z^6 + 6*w^5 is prime.
LINKS
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) = 2 since 2*0+1 = 0^2 + 0^2 + 1^2 + 0^2 with 0^2 + 3*0^2 + 5*1^2 + 7*0^2 = 5 and 5 - 2 = 3 twin prime, and 2*0+1 = 0^2 + 0^2 + 0^2 + 1^2 with 0^2 + 3*0^2 + 5*0^2 + 7*1^2 = 7 and 7 - 2 = 5 prime.
a(1) = 1 since 2*1+1 = 1^2 + 0^2 + 1^2 + 1^2 with 1^2 + 3*0^2 + 5*1^2 + 7*1^2 = 13 and 13 - 2 = 11 twin prime.
a(4) = 1 since 2*4+1 = 2^2 + 0^2 + 2^2 + 1^2 with 2^2 + 3*0^2 + 5*2^2 + 7*1^2 = 31 and 31 - 2 = 29 twin prime.
a(5) = 1 since 2*5+1 = 3^2 + 1^2 + 0^2 + 1^2 with 3^2 + 3*1^2 + 5*0^2 + 7*1^2 = 19 and 19 - 2 twin prime.
a(11) = 1 since 2*11+1 = 3^2 + 2^2 + 3^2 + 1^2 with 3^2 + 3*2^2 + 5*3^2 + 7*1^2 = 73 and 73 - 2 = 71 twin prime.
a(13) = 1 since 2*13+1 = 1^2 + 0^2 + 1^2 + 5^2 with 1^2 + 3*0^2 + 5*1^2 + 7*5^2 = 181 and 181 - 2 = 179 twin prime.
a(19) = 1 since 2*19+1 = 1^2 + 3^2 + 5^2 + 2^2 with 1^2 + 3*3^2 + 5*5^2 + 7*2^2 = 181 and 181 - 2 = 179 twin prime.
a(61) = 1 since 2*61+1 = 7^2 + 3^2 + 7^2 + 4^2 with 7^2 + 3*3^2 + 5*7^2 + 7*4^2 = 433 and 433 - 2 = 431 twin prime.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
TQ[p_]:=TQ[p]=PrimeQ[p]&&PrimeQ[p-2];
Do[r=0; Do[If[SQ[2n+1-x^2-y^2-z^2]&&TQ[x^2+3y^2+5z^2+7(2n+1-x^2-y^2-z^2)], r=r+1], {x, 0, Sqrt[2n+1]}, {y, 0, Sqrt[2n+1-x^2]}, {z, 0, Sqrt[2n+1-x^2-y^2]}]; Print[n, " ", r], {n, 0, 80}]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Aug 23 2017
STATUS
approved