A349957 Number of ways to write n as x^4 + y^2 + (z^2 + 11*16^w)/60, where x,y,z are nonnegative integers, and w is 0 or 1.

1, 2, 3, 5, 6, 4, 3, 3, 3, 4, 3, 4, 5, 3, 2, 2, 4, 4, 5, 9, 9, 3, 3, 4, 6, 5, 5, 9, 7, 4, 4, 6, 5, 2, 4, 8, 7, 3, 5, 7, 7, 4, 4, 4, 4, 4, 6, 9, 4, 3, 3, 9, 9, 4, 4, 5, 7, 2, 4, 4, 4, 2, 7, 7, 4, 3, 5, 12, 7, 3, 1, 6, 6, 4, 5, 8, 3, 1, 4, 5, 6, 3, 8, 14, 13, 6, 5, 5, 6, 6, 9, 8, 6, 3, 4, 8, 6, 6, 5, 12

Offset: 1

Author

Zhi-Wei Sun, Dec 06 2021

Keywords

• nonn

Comments

Conjecture 1: a(n) > 0 for all n > 0.

This has been verified for n <= 10^6. It seems that a(n) = 1 only for n = 1, 71, 78, 247, 542, 1258, 1907, 5225, 19798.

Conjecture 2: (i) If a is 1 or 3, then each n = 0,1,2,... can be written as a*x^8 + y^2 + (7*z^4 + w^2)/64 with x,y,z,w nonnegative integers.

(ii) If a is among 1,2,5, then each n = 0,1,2,... can be written as a*x^8 + y^2 + (11*z^4 + w^2)/60 with x,y,z,w nonnegative integers.

Conjecture 3: If (a,b, m) is among the triples (1,7,8), (1,11,12), (2,7,8), (3,11,12), (5,7,8), then each n = 0,1,2,... can be written as a*x^4 + y^2 + (b*z^6 + w^2)/m with x,y,z,w nonnegative integers.

Conjecture 4: (i) If F(x,y,z,w) is x^6 + y^2 + (5*z^4 + 3*w^2)/16 or 3x^6 + 2*y^2 + (11*z^4 + w^2)/60, then each n = 0,1,2,... can be written as F(x,y,z,w) with x,y,z,w nonnegative integers.

(ii) If (a,b,m) is among the triples (1,7,64), (1,11,12), (1,11,60), (2,1,25), (2,1,65), (2,11,4), (2,11,20), (2,11,60), (4,2,9), (4,7,64), (5,11,60), (6,1,10), then each n = 0,1,2,... can be written as a*x^6 + y^2 + (b*z^4 + w^2)/m with x,y,z,w nonnegative integers.

Examples

			a(1) = 1 with 1 = 0^4 + 0^2 + (7^2 + 11*16^0)/60.
a(16) = 2 with 16 = 0^4 + 0^2 + (28^2 + 11*16)/60 = 1^4 + 2^2 + (22^2 + 11*16)/60.
a(71) = 1 with 71 = 0^4 + 2^2 + (62^2 + 11*16)/60.
a(78) = 1 with 78 = 2^4 + 5^2 + (47^2 + 11*16^0)/60.
a(247) = 1 with 247 = 3^4 + 3^2 + (97^2 + 11*16^0)/60.
a(542) = 1 with 542 = 3^4 + 21^2 + (32^2 + 11*16)/60.
a(1258) = 1 with 1258 = 2^4 + 15^2 + (247^2 + 11*16^0)/60.
a(1907) = 1 with 1907 = 0^4 + 0^2 + (338^2 + 11*16)/60.
a(5225) = 1 with 5225 = 5^4 + 58^2 + (272^2 + 11*16)/60.
a(19798) = 1 with 19798 = 1^4 + 137^2 + (248^2 + 11*16)/60.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167--190.
• Zhi-Wei Sun, New Conjectures in Number Theory and Combinatorics (in Chinese), Harbin Institute of Technology Press, 2021.

Crossrefs

Cf. A000290, A000583, A001014, A001016, A349942, A349945, A349956.

Programs

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={};Do[r=0;Do[If[SQ[60(n-x^4-y^2)-11*16^z],r=r+1],{x,0,(n-1)^(1/4)},{y,0,Sqrt[n-1-x^4]},{z,0,1}];tab=Append[tab,r],{n,1,100}];Print[tab]