A306614 Number of ways to write n as w^2 + (x*(x+1)/2)^2 + (y*(3y+1)/2)^2 + (z*(5z+1)/2)^2, where w,x,y,z are integers with w > 0 and x >= 0.

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

Offset: 1

Author

Zhi-Wei Sun, Feb 28 2019

Keywords

• nonn

Comments

Conjecture 1 (Big 1-3-5 Conjecture): a(n) > 0 for any positive integer n not divisible by 8.

This is stronger than Lagrange's four-square theorem, but different from the 1-3-5 conjecture in A271518. I'd like to call it "the Big 1-3-5 Conjecture". I have verified a(n) > 0 for any positive integer n < 3*10^6 not divisible by 8.

Conjecture 2: If c is 5 or 7, then any positive integer n not divisible by 8 can be written as w^2 + (x*(x+1)/2)^2 + (y(3*y+1)/2)^2 + (z*(c*z+3)/2)^2, where w,x,y,z are integers with w > 0.

Conjecture 3: If a positive integer n is neither 76 nor divisible by 8, then it can be written as w^2 + (x*(3x+1)/2)^2 + (y*(5y+1)/2)^2 + (z*(5z+3)/2)^2, where w,x,y,z are integers with w > 0.

Conjecture 4: Each n = 0,1,2,... can be written as 2*w^2 + (x*(3x+1)/2)^2 + (y*(5y+1)/2)^2 + (z*(5z+3)/2)^2 with w,x,y,z integers.

Examples

			a(3) = 1 with 3 = 1^2 + (1*(1+1)/2)^2 + ((-1)*(3*(-1)+1)/2)^2 + (0*(5*0+1)/2)^2.
a(4) = 1 with 4 = 2^2 + (0*(0+1)/2)^2 + (0*(3*0+1)/2)^2 + (0*(5*0+1)/2)^2.
a(7) = 1 with 7 = 1^2 + (1*(1+1)/2)^2 + ((-1)*(3*(-1)+1)/2)^2 + ((-1)*(5*(-1)+1)/2)^2.
a(28) = 1 with 28 = 3^2 + (2*(2+1)/2)^2 + ((-1)*(3*(-1)+1)/2)^2 + (1*(5*1+1)/2)^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.

Crossrefs

Cf. A000118, A000217, A000290, A001318, A057569, A271518.

Programs

Mathematica

t[x_]:=t[x]=x(x+1)/2; p[x_]:=p[x]=x(3x+1)/2; q[x_]:=q[x]=x(5x+1)/2; SQ[n_]:=SQ[n]=n>0&&IntegerQ[Sqrt[n]];
tab={};Do[r=0;Do[If[SQ[n-t[x]^2-p[y]^2-q[z]^2],r=r+1],{x,0,(Sqrt[8*Sqrt[n-1]+1]-1)/2},{y,-Floor[(Sqrt[24*Sqrt[n-1-t[x]^2]+1]+1)/6],(Sqrt[24*Sqrt[n-1-t[x]^2]+1]-1)/6}, {z,-Floor[(Sqrt[40*Sqrt[n-1-t[x]^2-p[y]^2]+1]+1)/10],(Sqrt[40*Sqrt[n-1-t[x]^2-p[y]^2]+1]-1)/10}];tab=Append[tab,r],{n,1,100}];Print[tab]