A271237 Number of ordered ways to write n as u^3 + 2*v^3 + 3*x^3 + 4*y^3 + 5*z^3, where u, v, x, y and z are nonnegative integers.

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

Offset: 0

Author

Zhi-Wei Sun, Apr 02 2016

Keywords

• nonn

Comments

Conjecture: We have {u^3+a*v^3+b*x^3+c*y^3+d*z^3: u,v,x,y,z = 0,1,2,...} = {0,1,2,...} whenever (a,b,c,d) is among the following 32 quadruples: (1,2,2,3), (1,2,2,4), (1,2,3,4), (1,2,4,5), (1,2,4,6), (1,2,4,9), (1,2,4,10), (1,2,4,11), (1,2,4,18), (1,3,4,6), (1,3,4,9), (1,3,4,10), (2,2,4,5), (2,2,6,9), (2,3,4,5), (2,3,4,6), (2,3,4,7), (2,3,4,8), (2,3,4,9), (2,3,4,10), (2,3,4,12), (2,3,4,15), (2,3,4,18), (2,3,5,6), (2,3,6,12), (2,3,6,15), (2,4,5,6), (2,4,5,8), (2,4,5,9), (2,4,5,10), (2,4,6,7), (2,4,7,10).

In particular, this implies that a(n) > 0 for all n = 0,1,2,... We guess that a(n) = 1 only for n = 0, 1, 2, 18, 23, 79, 100.

If {m*u^3+a*v^3+b*x^3+c*y^3+d*z^3: u,v,x,y,z = 0,1,2,...} = {0,1,2,...} with 1 <= m <= a <= b <= c <= d, then m = 1, and we can show that (a,b,c,d) must be among the 32 quadruples listed in the conjecture (cf. Theorem 1.2 of the linked 2017 paper).

Conjecture verified for all the 32 quadruples up to 10^11. - Mauro Fiorentini, Jul 09 2023

It is known that there are exactly 54 quadruples (a,b,c,d) with 1 <= a <= b <= c <= d such that {a*w^2+b*x^2+c*y^2+d*z^2: w,x,y,z = 0,1,2,...} = {0,1,2,...}.

See also A271099 and A271169 for conjectures refining Waring's problem.

We also conjecture that if P(u,v,x,y,z) is one of the four polynomials u^6+v^3+2*x^3+4*y^3+5*z^3 and a*u^6+v^3+2*x^3+3*y^3+4*z^3 (a = 5,8,12) then any natural number can be written as P(u,v,x,y,z) with u,v,x,y,z nonnegative integers. - Zhi-Wei Sun, Apr 06 2016

Conjecture verified for all the 4 polynomials up to 10^11. - Mauro Fiorentini, Jul 09 2023

Examples

			a(2) = 1 since 2 = 0^3 + 2*1^3 + 3*0^3 + 4*0^3 + 5*0^3.
a(18) = 1 since 18 = 2^3 + 2*1^3 + 3*1^3 + 4*0^3 + 5*1^3.
a(23) = 1 since 23 = 0^3 + 2*2^3 + 3*1^3 + 4*1^3 + 5*0^3.
a(79) = 1 since 79 = 1^3 + 2*3^3 + 3*2^3 + 4*0^3 + 5*0^3.
a(100) = 1 since 100 = 2^3 + 2*1^3 + 3*3^3 + 4*1^3 + 5*1^3.
		

References

• S. Ramanujan, On the expression of a number in the form a*x^2 + b*y^2 + c*z^2 + d*w^2, Proc. Cambridge Philos. Soc. 19(1917), 11-21.

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
• L. E. Dickson, Quaternary quadratic forms representing all integers, Amer. J. Math. 49(1927), 39-56.
• Zhi-Wei Sun, A result similar to Lagrange's theorem, J. Number Theory 162(2016), 190-211.
• Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120.

Crossrefs

Cf. A000578, A002804, A001014, A271099, A271169.

Programs

Mathematica

CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)]
Do[r=0;Do[If[CQ[n-5z^3-4y^3-3x^3-2v^3],r=r+1],{z,0,(n/5)^(1/3)},{y,0,((n-5z^3)/4)^(1/3)},{x,0,((n-5z^3-4y^3)/3)^(1/3)},{v,0,((n-5z^3-4y^3-3x^3)/2)^(1/3)}];Print[n," ",r];Continue,{n,0,100}]