A300510 Number of ways to write n^2 as 4^k*(m^2+1) + x^2 + y^2, where m is 1 or 2, and k,x,y are nonnegative integers with x <= y.

0, 1, 2, 1, 3, 3, 3, 1, 4, 4, 5, 3, 5, 4, 6, 1, 3, 4, 5, 4, 7, 6, 5, 3, 8, 6, 6, 4, 5, 7, 7, 1, 5, 4, 11, 4, 7, 5, 6, 4, 6, 8, 5, 6, 12, 5, 5, 3, 6, 9, 8, 6, 7, 6, 10, 4, 7, 7, 6, 7, 5, 9, 9, 1, 8, 5, 10, 4, 9, 11, 9, 4, 11, 7, 12, 5, 8, 7, 7, 4

Offset: 1

Author

Zhi-Wei Sun, Mar 07 2018

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 1; in other words, for any integer n > 1 there is a nonnegative integer k such that either n^2 - 2*4^k or n^2 - 5*4^k can be written as the sum of two squares. Moreover, a(n) = 1 only for n = 2^k with k > 0.

This conjecture is stronger than the first conjecture in A300448. We have verified that a(n) > 0 for all n = 2..5*10^7.

Consider positive integers c not divisible by 4 such that for any integer n > 1 there is a nonnegative integer k for which n^2 - 2*4^k or n^2 - c*4^k can be written as the sum of two squares. Our computation for n up to 3*10^7 shows that the only candidates for values of c smaller than 160 are 5, 17, 18, 26, 29, 41, 45, 65, 74, 89, 98, 101, 113, 122, 125, 146, 149, 153. These numbers have the form 9^a*(3*b+2) with a and b nonnegative integers and the p-adic order of 3*b+2 is even for any prime p == 3 (mod 4). For n = 42211965 there is no nonnegative integer k such that n^2 - 2*4^k or n^2 - 162*4^k can be written as the sum of two squares.

Qing-Hu Hou at Tianjin Univ. reported that he had verified a(n) > 0 for n up to 10^9. - Zhi-Wei Sun, Mar 14 2018

Qing-Hu Hou found that 29, 65, 113 should be excluded from the candidates. In fact, for c = 29, 65, 113 there is no nonnegative integer k such that N(c)^2 - 2*4^k or N(c)^2 - c*4^k can be written as the sum of two squares, where N(29) = 51883659, N(65) = 56173837 and N(113) = 65525725. - Zhi-Wei Sun, Mar 23 2018

a(n) > 0 for 1 < n < 6*10^9. - Giovanni Resta, Jun 14 2019

Examples

			a(1) = 0 since 1^2 - 4^k*(m^2+1) < 0 for k = 0,1,2,... and m = 1, 2.
a(2) = 1 since 2^2 = 4^0*(1^2+1) + 1^2 + 1^2.
a(3) = 2 since 3^2 = 4^0*(2^2+1) + 0^2 + 2^2 = 4^1*(1^2+1) + 0^2 + 1^2.
		

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.

Crossrefs

Cf. A000118, A000302, A001481, A271518, A281976, A299924, A299537, A299794, A300362, A300396, A300448, A301452, A301471.

Programs

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[QQ[n^2-4^k*(m^2+1)],Do[If[SQ[n^2-4^k(m^2+1)-x^2],r=r+1],{x,0,Sqrt[(n^2-4^k(m^2+1))/2]}]],{m,1,2},{k,0,Log[4,n^2/(m^2+1)]}];tab=Append[tab,r],{n,1,80}];Print[tab]