A262827 Number of ordered ways to write n as w^2 + x^3 + y^4 + 2*z^4, where w, x, y and z are nonnegative integers.

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

Offset: 0

Author

Zhi-Wei Sun, Oct 03 2015

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n >= 0. Also, a(n) = 1 only for the following 41 values of n: 0, 14, 15, 21, 22, 23, 62, 71, 78, 87, 136, 216, 405, 437, 448, 477, 535, 583, 591, 623, 671, 696, 885, 950, 1046, 1135, 1206, 1208, 1248, 1317, 2288, 2383, 2543, 3167, 3717, 3974, 6847, 7918, 8328, 9096, 21935.

We have verified that a(n) > 0 for all n = 0..10^7.

Conjecture verified up to 10^11. - Mauro Fiorentini, Jul 07 2023

We also conjecture that if f(w,x,y,z) is one of the 8 polynomials 2w^2+x^3+4y^3+z^4, w^2+x^3+2y^3+c*z^3 (c = 3,4,5,6) and w^2+x^3+2y^3+d*z^4 (d = 1,3,6) then each n = 0,1,2,... can be written as f(w,x,y,z) with w,x,y,z nonnegative integers. - Zhi-Wei Sun, Dec 30 2017

Conjecture verified up to 10^11 for all 8 polynomials. - Mauro Fiorentini, Jul 07 2023

Examples

			a(14) = 1 since 14 = 2^2 + 2^3 + 0^4 + 2*1^4.
a(87) = 1 since 87 = 2^2 + 0^3 + 3^4 + 2*1^4.
a(216) = 1 since 216 = 0^2 + 6^3 + 0^4 + 2*0^4.
a(405) = 1 since 405 = 18^2 + 0^3 + 3^4 + 2*0^4.
a(1248) = 1 since 1248 = 31^2 + 5^3 + 0^4 + 2*3^4.
a(1317) = 1 since 1317 = 23^2 + 1^3 + 5^4 + 2*3^4.
a(2288) = 1 since 2288 = 44^2 + 4^3 + 4^4 +2*2^4.
a(2383) = 1 since 2383 = 1462 + 9^3 + 6^4 + 2*3^4.
a(2543) = 1 since 2543 = 50^2 + 3^3 + 2^4 + 2*0^4.
a(3167) = 1 since 3167 = 54^2 + 2^3 + 3^4 + 2*3^4.
a(3717) = 1 since 3717 = 18^2 + 15^3 + 2^4 + 2*1^4.
a(3974) = 1 since 3974 = 39^2 + 13^3 + 4^4 + 2*0^4.
a(6847) = 1 since 6847 = 52^2 + 15^3 + 4^4 + 2*4^4.
a(7918) = 1 since 7918 = 46^2 + 10^3 + 0^4 + 2*7^4.
a(8328) = 1 since 8328 = 42^2 + 1^3 + 9^4 + 2*1^4.
a(9096) = 1 since 9096 = 44^2 + 18^3 + 6^4 + 2*2^4.
a(21935) = 1 since 21935 = 66^2 + 26^3 + 1^4 + 2*1^4.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
• Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190. (See Remark 1.1.)
• Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120. (See Theorem 1.1 and Conjecture 1.1.)

Crossrefs

Cf. A000290, A000578, A000583, A262813, A262815, A262816, A262824.

Programs

Mathematica

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