A306606 Number of ways to write n as w^2 + T(x)^2 + Pen(y)^2 + 2*Pen(z)^2, where w,x,y,z are integers with w > 0 and x >= 0, and T(m) = m*(m+1)/2 and Pen(m) = m*(3m-1)/2.

1, 2, 2, 3, 4, 3, 3, 3, 3, 6, 5, 4, 7, 5, 2, 5, 5, 6, 6, 3, 4, 5, 1, 2, 5, 6, 5, 5, 6, 3, 3, 2, 2, 6, 5, 3, 9, 7, 3, 6, 4, 5, 8, 3, 6, 7, 2, 4, 7, 6, 8, 9, 9, 6, 6, 5, 3, 9, 9, 6, 11, 7, 4, 10, 5, 6, 13, 5, 5, 9, 2, 5, 7, 7, 7, 10, 7, 3, 5, 3, 3, 6, 6, 5, 11, 6, 6, 9, 4, 6, 11, 3, 5, 9, 2, 5, 4, 4, 7, 11

Offset: 1

Author

Zhi-Wei Sun, Feb 28 2019

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 0. In other words, any positive integer n can be written as s^2 + t^2 + u^2 + 2*v^2, where s is a positive integer, t is a triangular number, u and v are generalized pentagonal numbers.

I have verified a(n) > 0 for all n = 1..5*10^6.

It is well known that any positive odd integer can be written as x^2 + y^2 + 2*z^2 with x,y,z integers.

See also A306614 for similar conjectures.

Examples

			a(1) = 1 with 1 = 1^2 + T(0)^2 + Pen(0)^2 + 2*Pen(0)^2.
a(23) = 1 with 23 = 4^2 + T(1)^2 + Pen(-1)^2 + 2*Pen(1)^2.
a(335) = 1 with 335 = 18^2 + T(2)^2 + Pen(0)^2 + 2*Pen(1)^2.
a(3695) = 1 with 3695 = 53^2 + T(7)^2 + Pen(-1)^2 + 2*Pen(-2)^2.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017.

Crossrefs

Cf. A000217, A000290, A001318, A275344, A306614.

Programs

Mathematica

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