A306592 Number of ways to write n as u^2 + 2*v^2 + 3*x^2 + 4*y^2 + 5*z^2, where u,v,x,y,z are generalized pentagonal numbers (A001318).

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

Offset: 0

Author

Zhi-Wei Sun, Feb 26 2019

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for any nonnegative integer n.

I'd like to call this the 1-2-3-4-5 conjecture. I have verified it for all n = 0..4*10^5.

It seems that a(n) = 1 only for n = 0,1,2.

Examples

			Let f(x) = x*(3x-1)/2. Then
a(2) = 1 with 2 = f(0)^2 + 2*f(1)^2 + 3*f(0)^2 + 4*f(0)^2 + 5*f(0)^2,
a(415) = 2 with 415 = f(-1)^2 + 2*f(3)^2 + 3*f(1)^2 + 4*f(2)^2 + 5*f(-1)^2 = f(3)^2 + 2*f(0)^2 + 3*f(2)^2 + 4*f(-2)^2 + 5*f(0)^2,
a(427) = 2 with 427 = f(-3)^2 + 2*f(2)^2 + 3*f(-2)^2 + 4*f(0)^2 + 5*f(1)^2 = f(-3)^2 + 2*f(1)^2 + 3*f(2)^2 + 4*f(0)^2 + 5*f(2)^2.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A000290, A001318.

Programs

Mathematica

f[x_]:=f[x]=(x(3x-1)/2)^2;
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
PQ[n_]:=PQ[n]=SQ[n]&&SQ[24Sqrt[n]+1];
tab={};Do[r=0;Do[If[PQ[n-5f[x]-4f[y]-3f[z]-2f[w]],r=r+1],{x,-Floor[(Sqrt[24Sqrt[n/5]+1]-1)/6],(Sqrt[24Sqrt[n/5]+1]+1)/6},{y,-Floor[(Sqrt[24Sqrt[(n-5f[x])/4]+1]-1)/6],(Sqrt[24Sqrt[(n-5f[x])/4]+1]+1)/6},{z,-Floor[(Sqrt[24Sqrt[(n-5f[x]-4f[y])/3]+1]-1)/6],(Sqrt[24Sqrt[(n-5f[x]-4f[y])/3]+1]+1)/6},{w,-Floor[(Sqrt[24Sqrt[(n-5f[x]-4f[y]-3f[z])/2]+1]-1)/6],(Sqrt[24Sqrt[(n-5f[x]-4f[y]-3f[z])/2]+1]+1)/6}];tab=Append[tab,r],{n,0,100}];Print[tab]