A349778 Number of ways to write n as x^2 + y^k + 2*z^m, where x,y,z are nonnegative integers with x >= y, and k and m belong to the set {2,3}.

4, 4, 8, 4, 8, 4, 4, 4, 4, 6, 8, 4, 8, 4, 2, 2, 7, 10, 12, 8, 9, 3, 4, 2, 5, 11, 10, 8, 8, 3, 1, 3, 7, 10, 11, 5, 12, 7, 7, 4, 5, 8, 8, 7, 8, 8, 2, 3, 4, 9, 11, 8, 18, 5, 11, 8, 4, 8, 11, 8, 7, 6, 3, 8, 7, 12, 12, 12, 11, 4, 7, 5, 10, 9, 11, 7, 11, 4, 3, 6, 11, 13, 17, 9, 10, 6, 5, 7, 7, 13, 13, 12, 5, 6, 3, 3, 5, 14, 12, 10, 18

Offset: 0

Author

Zhi-Wei Sun, Nov 29 2021

Keywords

• nonn

Comments

Conjecture 1: a(n) > 0 for all n >= 0, and a(n) = 1 only for n = 30, 120, 142.

We have verified this for all n <= 10^6.

Conjecture 2: Let S = {x^k: k = 2,3 and x = 0,1,2,...}, and let a be 3 or 4 or 5. Then any nonnegative integer can be written as x + 2*y + a*z, where x,y,z are elements of the set S.

Conjecture 3: Let T = {x^k: k = 2,3,4,... and x = 0,1,2,...}. If (b,c) is among the ordered pairs (1,2), (2,4), (2,5) and (3,2), then each n = 0,1,... can be written as x + b*y + c*z, where x and y are elements of T, and z is a square.

Examples

			a(3) = 4. In fact, 3 = 1^2 + 0^2 + 2*1^2 = 1^2 + 0^2 + 2*1^3 = 1^2 + 0^3 + 2*1^2 = 1^2 + 0^3 + 2*1^3 with 1 >= 0.
a(30) = 1 with 30 = 2^2 + 2^3 + 2*3^2 and 2 >= 2.
a(120) = 1 with 120 = 10^2 + 2^2 + 2*2^3 and 10 >= 2.
a(142) = 1 with 142 = 6^2 + 2^3 + 2*7^2 and 6 >= 2.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
• Zhi-Wei Sun, New Conjectures in Number Theory and Combinatorics (in Chinese), Harbin Institute of Technology Press, 2021.

Crossrefs

Cf. A000290, A000578, A001597, A002760, A275168, A275169, A349787.

Programs

Mathematica

tab={};Do[r=0;Do[If[IntegerQ[((n-x^2-y^k)/2)^(1/m)],r=r+1],{x,0,Sqrt[n]},{k,2,3},{y,0,Min[x,(n-x^2)^(1/k)]},{m,2,3}];tab=Append[tab,r],{n,0,100}];Print[tab]