OFFSET
1,3
COMMENTS
Conjecture: a(n) > 0 for all n > 0. Moreover, any positive square n^2 can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w integers and y even such that x^2 - (3*y)^2 = 4^k for some k = 0,1,2,....
We have verifed this for all n = 1..10^7.
Compare this conjecture with the conjectures in A299537.
As 3*A001353(n)^2 + 1 = A001075(n)^2, the conjecture in A300441 implies that any positive square can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w integers such that x^2 - 3*y^2 = 4^k for some k = 0,1,2,....
See also A301391 for a similar conjecture.
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], 2017-2018.
EXAMPLE
a(1) = 1 since 1^2 = 1^2 + 0^2 + 0^2 + 0^2 with 1^2 - (3*0)^2 = 4^0.
a(5) = 1 since 5^2 = 4^2 + 0^2 + 0^2 + 3^2 with 4^2 - (3*0)^2 = 4^2.
a(7) = 1 since 7^2 = 2^2 + 0^2 + 3^2 + 6^2 with 2^2 - (3*0)^2 = 4^1.
a(31) = 3 since 31^2 = 10^2 + 2^2 + 4^2 + 29^2 with 10^2 - (3*2)^2 = 4^3, and 31^2 = 20^2 + 4^2 + 4^2 + 23^2 = 20^2 + 4^2 + 16^2 + 17^2 with 20^2 - (3*4)^2 = 4^4.
MATHEMATICA
f[n_]:=f[n]=FactorInteger[n];
g[n_]:=g[n]=Sum[Boole[Mod[Part[Part[f[n], i], 1]-3, 4]==0&&Mod[Part[Part[f[n], i], 2], 2]==1], {i, 1, Length[f[n]]}]==0;
QQ[n_]:=QQ[n]=n==0||(n>0&&g[n]);
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={}; Do[r=0; Do[If[SQ[4^k+9y^2]&&QQ[n^2-4^k-10y^2], Do[If[SQ[n^2-(4^k+10y^2)-z^2], r=r+1], {z, 0, Sqrt[(n^2-4^k-10y^2)/2]}]], {k, 0, Log[2, n]}, {y, 0, Sqrt[(n^2-4^k)/10]}]; tab=Append[tab, r], {n, 1, 80}]; Print[tab]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Mar 19 2018
STATUS
approved