A290472 Number of ways to write 6*n+1 as x^2 + 3*y^2 + 7*z^2, where x is a positive integer, and y and z are nonnegative integers.

1, 1, 1, 2, 1, 1, 2, 3, 3, 1, 1, 4, 1, 3, 1, 9, 1, 1, 2, 4, 3, 3, 3, 5, 1, 4, 2, 6, 3, 6, 1, 4, 2, 3, 1, 7, 3, 3, 3, 6, 2, 3, 2, 15, 2, 5, 2, 4, 2, 2, 7, 6, 3, 6, 2, 11, 3, 7, 3, 6, 4, 5, 2, 11, 4, 3, 1, 7, 3, 2, 4, 17, 2, 3, 3, 8, 2, 5, 7, 9, 4, 4, 2, 13, 1, 13, 1, 5, 4, 3, 4, 6, 7, 7, 3, 10, 4, 6, 3, 20, 3

Offset: 0

Author

Zhi-Wei Sun, Aug 03 2017

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 1, 2, 4, 5, 9, 10, 12, 14, 16, 17, 24, 30, 34, 66, 84, 86, 116, 124, 152, 286.

We also conjecture that {6n+5: n = 0,1,2,...} is a subset of {2x^2+3y^2+5z^2: x,y,z are nonnegative integers with y > 0}.

See A286885 for more similar conjectures.

In support of the first conjecture, a(n) > 1 for 286 < n <= 10^7. - Charles R Greathouse IV, Aug 04 2017

Examples

			a(4) = 1 since 6*4+1 = 5^2 + 3*0^2 + 7*0^2.
a(5) = 1 since 6*5+1 = 2^2 + 3*3^2 + 7*0^2.
a(9) = 1 since 6*9+1 = 6^2 + 3*2^2 + 7*1^2.
a(116) = 1 since 6*116+1 = 9^2 + 3*14^2 + 7*2^2.
a(124) = 1 since 6*124+1 = 21^2 + 3*8^2 + 7*4^2.
a(152) = 1 since 6*152+1 = 19^2 + 3*10^2 + 7*6^2.
a(286) = 1 since 6*286+1 = 11^2 + 3*14^2 + 7*12^2.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
• Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58 (2015), 1367-1396.
• Zhi-Wei Sun, On universal sums x(ax+b)/2+y(cy+d)/2+z(ez+f)/2, arXiv:1502.03056 [math.NT], 2015-2017.
• Hai-Liang Wu and Zhi-Wei Sun, Some universal quadratic sums over the integers, arXiv:1707.06223 [math.NT], 2017.

Crossrefs

Cf. A000290, A286885, A286944, A287616, A290342.

Programs

Mathematica

SQ[n_]:=SQ[n]=n>0&&IntegerQ[Sqrt[n]];
Do[r=0;Do[If[SQ[6n+1-3y^2-7z^2],r=r+1],{y,0,Sqrt[(6n+1)/3]},{z,0,Sqrt[(6n+1-3y^2)/7]}];Print[n," ",r],{n,0,100}]

PARI

a(n)=my(s=6*n+1,t); sum(z=0,sqrtint((s-1)\7), t=s-7*z^2; sum(y=0,sqrtint((t-1)\3), issquare(t-3*y^2))) \\ Charles R Greathouse IV, Aug 03 2017

PARI

first(n)=my(v=vector(n+1),mx=6*n+1,s,t,u); for(x=1,sqrtint(mx), s=x^2; for(y=0,sqrtint((mx-s)\3), t=s+3*y^2; for(z=0,sqrtint((mx-t)\7), u=t+7*z^2; if(u%6==1, v[u\6+1]++)))); v \\ Charles R Greathouse IV, Aug 03 2017