A343397 Number of ways to write n as 2^x + [y^2/3] + [z^2/4] with x,y,z positive integers, where [.] is the floor function.

0, 1, 2, 3, 4, 4, 5, 5, 8, 5, 9, 5, 8, 8, 6, 9, 9, 10, 8, 11, 10, 10, 9, 9, 14, 8, 8, 10, 12, 11, 6, 14, 13, 10, 12, 13, 15, 11, 13, 9, 20, 6, 12, 17, 13, 13, 10, 11, 17, 12, 11, 13, 15, 14, 9, 13, 13, 14, 11, 18, 11, 15, 7, 12, 22, 13, 14, 17, 17, 11, 15, 13, 24, 16, 9, 17, 15, 15, 14, 18

Offset: 1

Author

Zhi-Wei Sun, Apr 13 2021

Keywords

• nonn

Comments

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

We have verified a(n) > 0 for all n = 2..2*10^6.

Conjecture verified up to 10^10. - Giovanni Resta, Apr 14 2021

Examples

			a(2) = 1 with 2 = 2^1 + [1^2/3] + [1^2/4].
a(3) = 2 with 3 = 2^1 + [1^2/3] + [2^2/4] = 2^1 + [2^2/3] + [1^2/4].
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Zhi-Wei Sun, Natural numbers represented by [x^2/a] + [y^2/b] + [z^2/c], arXiv:1504.01608 [math.NT], 2015.

Crossrefs

Cf. A000079, A000290, A343326, A343368, A343384, A343387, A343391, A343411.

Programs

Mathematica

PowQ[n_]:=PowQ[n]=n>1&&IntegerQ[Log[2,n]];
tab={};Do[r=0;Do[If[PowQ[n-Floor[x^2/3]-Floor[y^2/4]],r=r+1],{x,1,Sqrt[3n-1]},{y,1,Sqrt[4(n-Floor[x^2/3]-1)+1]}];tab=Append[tab,r],{n,1,80}];Print[tab]