A301471 Number of ways to write n^2 as x^2 + 2*y^2 + 3*2^z with x,y,z nonnegative integers.

0, 1, 2, 1, 3, 4, 3, 1, 5, 4, 4, 4, 5, 4, 10, 1, 4, 7, 4, 4, 10, 4, 3, 4, 6, 6, 11, 4, 7, 10, 6, 1, 9, 5, 7, 7, 7, 6, 12, 4, 6, 12, 7, 4, 14, 4, 8, 4, 3, 8, 10, 6, 8, 13, 6, 4, 16, 8, 7, 10, 7, 6, 14, 1, 7, 11, 6, 5, 16, 9, 5, 7, 7, 7, 18, 6, 7, 14, 6, 4

Offset: 1

Author

Zhi-Wei Sun, Mar 21 2018

Keywords

• nonn

Comments

Square Conjecture: a(n) > 0 for all n > 1. Moreover, for any integer n > 3 we can write n^2 as x^2 + 2*y^2 + 3*2^z, where x,y,z are nonnegative integers with y even and z > 1.

It is known that a positive integer n has the form x^2 + 2*y^2 with x and y integers if and only if the p-adic order of n is even for any prime p == 5 or 7 (mod 8).

See also A301472 for the list of positive integers not of the form x^2 + 2*y^2 + 3*2^z with x,y,z nonnegative integers.

If n^2 = x^2 + 2*y^2 + 3*2^z with x,y,z nonnegative integers, then it is easy to see that x is not divisible by 3.

The Square Conjecture implies that for each n = 1,2,3,... we can write 3*n^2 as x^2 + 2*y^2 + 2^z with x,y,z nonnegative integers. In fact, if (3*n)^2 = u^2 + 2*v^2 + 3*2^z with u,v,z integers and z >= 0, then u^2 == v^2 (mod 3) and thus we may assume u == v (mod 3) without loss of generality, hence 3*n^2 = (u^2+2*v^2)/3 + 2^z = x^2 + 2*y^2 + 2^z with x = (u+2*v)/3 and y = (u-v)/3 integers.

On March 25, 2018 Qing-Hu Hou at Tianjin Univ. finished his verification of the Square Conjecture for n <= 4*10^8. Then I used Hou's program to verify the conjecture for n <= 5*10^9. - Zhi-Wei Sun, Apr 10 2018

I have found a counterexample to the Square Conjecture, namely a(5884015571) = 0. Note that 5884015571 is the product of the three primes 7, 17 and 49445509. - Zhi-Wei Sun, Apr 15 2018

Examples

			a(2) = 1 with 2^2 = 1^2 + 2*0^2 + 3*2^0.
a(3) = 2 with 3^2 = 2^2 + 2*1^2 + 3*2^0 = 1^2 + 2*1^2 + 3*2^1.
a(4) = 1 with 4^2 = 2^2 + 2*0^2 + 3*2^2.
a(1131599953) = 1 with 1131599953^2 = 316124933^2 + 2*768304458^2 + 3*2^6.
a(5884015571) = 0 since there are no nonnegative integers x,y,z such that  x^2 + 2*y^2 + 3*2^z = 5884015571^2.
		

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. A000079, A000290, A002479, A299924, A299537, A299794, A300219, A300362, A300396, A300510, A301376, A301391, A301452, A301472, A301479, A301579, A301640, A302641.

Programs

Mathematica

f[n_]:=f[n]=FactorInteger[n];
g[n_]:=g[n]=Sum[Boole[(Mod[Part[Part[f[n],i],1],8]==5||Mod[Part[Part[f[n],i],1],8]==7)&&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^2-3*2^k],Do[If[SQ[n^2-3*2^k-2x^2],r=r+1],{x,0,Sqrt[(n^2-3*2^k)/2]}]],{k,0,Log[2,n^2/3]}];tab=Append[tab,r],{n,1,80}];Print[tab]