A272979 Number of ways to write n as x^2 + 2*y^2 + 3*z^3 + 4*w^4 with x,y,z,w nonnegative integers.

1, 1, 1, 2, 3, 2, 3, 3, 3, 4, 2, 3, 4, 3, 1, 3, 4, 1, 3, 3, 2, 3, 4, 2, 3, 5, 3, 4, 4, 3, 4, 4, 4, 4, 4, 2, 7, 5, 2, 4, 6, 4, 3, 4, 3, 3, 4, 3, 4, 2, 3, 6, 3, 3, 5, 5, 2, 7, 5, 1, 5, 6, 3, 1, 6, 2, 5, 5, 5, 4, 5

Offset: 0

Author

Zhi-Wei Sun, Jul 13 2016

Keywords

• nonn

Comments

Conjecture: For positive integers a,b,c,d, any natural number can be written as a*x^2 + b*y^2 + c*z^3 + d*w^4 with x,y,z,w nonnegative integers, if and only if (a,b,c,d) is among the following 49 quadruples: (1,2,1,1), (1,3,1,1), (1,6,1,1), (2,3,1,1), (2,4,1,1), (1,1,2,1), (1,4,2,1), (1,2,3,1), (1,2,4,1), (1,2,12,1), (1,1,1,2), (1,2,1,2), (1,3,1,2), (1,4,1,2), (1,5,1,2), (1,11,1,2), (1,12,1,2), (2,4,1,2), (3,5,1,2), (1,1,4,2), (1,1,1,3), (1,2,1,3), (1,3,1,3), (1,2,4,3), (1,2,1,4), (1,3,1,4), (2,3,1,4), (1,1,2,4), (1,2,2,4), (1,8,2,4), (1,2,3,4), (1,1,1,5), (1,2,1,5), (2,3,1,5), (2,4,1,5), (1,3,2,5), (1,1,1,6), (1,3,1,6), (1,1,2,6), (1,2,1,8), (1,2,4,8), (1,2,1,10), (1,1,2,10), (1,2,1,11), (2,4,1,11), (1,2,1,12), (1,1,2,13), (1,2,1,14),(1,2,1,15).

See also A262824, A262827, A262857 and A273917 for similar conjectures.

Examples

			a(0) = 1 since 0 = 0^2 + 2*0^2 + 3*0^3 + 4*0^4.
a(1) = 1 since 1 = 1^2 + 2*0^2 + 3*0^3 + 4*0^4.
a(2) = 1 since 2 = 0^2 + 2*1^2 + 3*0^3 + 4*0^4.
a(14) = 1 since 14 = 3^2 + 2*1^2 + 3*1^3 + 4*0^4.
a(17) = 1 since 17 = 3^2 + 2*2^2 + 3*0^3 + 4*0^4.
a(59) = 1 since 59 = 3^2 + 2*5^2 + 3*0^3 + 4*0^4.
a(63) = 1 since 63 = 3^2 + 2*5^2 + 3*0^2 + 4*1^4.
a(287) = 1 since 287 = 11^2 + 2*9^2 + 3*0^2 + 4*1^4.
		

Links

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

Crossrefs

Cf. A000290, A000578, A000583, A262824, A262827, A262857, A270969, A273429, A273915, A273917.

Programs

Mathematica

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