A377224 Number of ways to write n as x*(5*x+1) + y*(5*y+1)/2 + z*(5*z+1)/2, where x,y,z are integers with y*(5*y+1) <= z*(5*z+1).

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

Offset: 0

Author

Zhi-Wei Sun, Nov 13 2024

Keywords

• nonn

Comments

Conjecture 1: a(n) = 0 only for n = 1. Also, a(n) = 1 only for n = 0, 2, 3, 5, 7, 14, 16, 19, 37, 43, 58, 61, 79.

This has been verified for n <= 2*10^6.

Conjecture 2: Let N be the set of all nonnegative integers. Then

{x*(5*x+1) + y*(5*y+1)/2 + 5*z*(5*z+1)/2: x,y,z are integers} = N\{1,5},

{x*(5*x+1) + y*(5*y+1)/2 + 3*z*(5*z+1)/2: x,y,z are integers} = N\{1,5,32},

{x*(5*x+1) + y*(5*y+1)/2 + 2*z*(5*z+1): x,y,z are integers} = N\{1,5,70},

and

{x*(5*x+1)/2 + y*(5*y+1)/2 + z*(5*z+1)/2: x,y,z are integers} = N\{1,10,19,94}.

Conjecture 3: We have

{x*(5*x+3) + y*(5*y+3)/2 + 3*z*(5*z+3)/2: x,y,z are integers} = N\{31,77},

{x*(5*x+3) + y*(5*y+3)/2 + 5*z*(5*z+3): x,y,z are integers} = N\{10,16},

and

{x*(5*x+3)/2 + y*(5*y+3)/2 + 5*z*(5*z+3)/2: x,y,z are integers} = N\{3,15,29,44}.

Examples

			a(14) = 1 with 14 = 0*(5*0+1) + 1*(5*1+1)/2 + 2*(5*2+1)/2.
a(37) = 1 with 37 = (-1)*(5*(-1)+1) + (-2)*(5*(-2)+1)/2 + 3*(5*3+1)/2.
a(58) = 1 with 58 = (-2)*(5*(-2)+1) + (-1)*(5*(-1)+1)/2 + (-4)*(5*(-4)+1)/2.
a(79) = 1 with 79 = -4*(5*(-4)+1) + 0*(5*0+1)/2 + 1*(5*1+1)/2.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
• Zhi-Wei Sun, A result similar to Lagrange's theorem, J. Number Theory 162 (2016), 190-211.
• Zhi-Wei Sun, Universal sums of three quadratic polynomials, Sci. China Math. 63 (2020), 501-520.
• Zhi-Wei Sun, New results similar to Lagrange's four-square theorem, arXiv:2411.14308 [math.NT], 2024.

Crossrefs

Cf. A057569, A085787, A306383.

Programs

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={};Do[r=0;Do[If[SQ[40(n-x(5x+1)-y(5y+1)/2)+1],r=r+1],{x,-Floor[(Sqrt[20n+1]+1)/10],(Sqrt[20n+1]-1)/10},{y,-Floor[(Sqrt[20(n-x(5x+1))+1]+1)/10],Floor[(Sqrt[20(n-x(5x+1))+1]-1)/10]}];tab=Append[tab,r],{n,0,100}];Print[tab]