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A300752
Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x + 3*y + 5*z a positive square, where x,y,z,w are nonnegative integers such that 3*x or y or z is a square.
11
1, 2, 2, 1, 1, 1, 1, 1, 2, 3, 2, 2, 2, 4, 2, 1, 5, 5, 2, 1, 2, 2, 2, 1, 4, 5, 2, 1, 4, 7, 1, 2, 5, 3, 2, 1, 3, 5, 4, 2, 8, 5, 1, 3, 5, 6, 1, 2, 4, 8, 5, 4, 2, 4, 4, 2, 6, 5, 5, 2, 1, 6, 4, 1, 8, 9, 6, 2, 3, 4, 1, 2, 6, 8, 5, 4, 5, 8, 2, 1
OFFSET
1,2
COMMENTS
Conjecture: a(n) > 0 for all n = 1,2,3,....
This is stronger than the author's 1-3-5 conjecture in A271518. See also A300751 for a similar conjecture stronger than the 1-3-5 conjecture.
In 2020, A. Machiavelo, R. Reis and N. Tsopanidis verified a(n) > 0 for n up to 1.05*10^11. - Zhi-Wei Sun, Oct 06 2020
LINKS
António Machiavelo, Rogério Reis, and Nikolaos Tsopanidis, Report on Zhi-Wei Sun's "1-3-5 conjecture" and some of its refinements, arXiv:2005.13526 [math.NT], 2020.
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], 2017-2018.
EXAMPLE
a(71) = 1 since 71 = 3^2 + 1^2 + 6^2 + 5^2 with 1 = 1^2 and 3 + 3*1 + 5*6 = 6^2.
a(248) = 1 since 248 = 10^2 + 2^2 + 0^2 + 12^2 with 0 = 0^2 and 10 + 3*2 + 5*0 = 4^2.
a(263) = 1 since 263 = 13^2 + 2^2 + 9^2 + 3^2 with 9 = 3^2 and 13 + 3*2 + 5*9 = 8^2.
a(808) = 1 since 808 = 12^2 + 14^2 + 18^2 + 12^2 with 3*12 = 6^2 and 12 + 3*14 + 5*18 = 12^2.
a(1288) = 1 since 1288 = 12^2 + 18^2 + 26^2 + 12^2 with 3*12 = 6^2 and 12 + 3*18 + 5*26 = 14^2.
a(3544) = 1 since 3544 = 14^2 + 34^2 + 16^2 + 44^2 with 16 = 4^2 and 14 + 3*34 + 5*16 = 14^2.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={}; Do[r=0; Do[If[(SQ[3(m^2-3y-5z)]||SQ[y]||SQ[z])&&SQ[n-(m^2-3y-5z)^2-y^2-z^2], r=r+1], {m, 1, (35n)^(1/4)}, {y, 0, Min[m^2/3, Sqrt[n]]}, {z, 0, Min[(m^2-3y)/5, Sqrt[n-y^2]]}]; tab=Append[tab, r], {n, 1, 80}]; Print[tab]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Mar 11 2018
STATUS
approved