A343400 Number of ways to write n as 3^x + [y^2/3] + [z^2/4], where [.] is the floor function, x is a nonnegative integer, and y and z are positive integers.

1, 2, 3, 4, 4, 5, 4, 6, 4, 9, 5, 7, 8, 6, 9, 6, 7, 9, 7, 6, 9, 7, 8, 7, 7, 10, 6, 9, 11, 9, 12, 8, 9, 14, 5, 13, 11, 8, 11, 11, 7, 13, 9, 12, 11, 9, 9, 11, 8, 12, 11, 11, 11, 6, 16, 4, 11, 12, 11, 13, 12, 6, 10, 9, 8, 17, 8, 12, 11, 10, 8, 10, 12, 10, 8, 11, 12, 12, 13, 7

Offset: 1

Author

Zhi-Wei Sun, Apr 13 2021

Keywords

• nonn

Comments

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

We have verified a(n) > 0 for all n = 1..2*10^6.

The first indices n for which a(n) = 0 are 4051736, 7479656, 8592680, 9712160, 14039792, 16726256, 24914510. - Giovanni Resta, Apr 14 2021

Examples

			a(2) = 2 with 2 = 3^0 + [1^2/3] + [2^2/4] = 3^0 + [2^2/3] + [1^2/4].
a(2942) = 2 with 2942 = 3^1 + [93^2/3] + [15^2/4] = 3^7 + [44^2/3] + [21^2/4].
a(627662) = 5 with 627662 - 3^0 = [330^2/3] + [1538^2/4] = [1042^2/3] + [1031^2/4] = [1318^2/3] + [441^2/4] = [1328^2/3] + [399^2/4] = [1352^2/3] + [271^2/4].
a(1103096) = 3 with 1103096 = 3^1 + [260^2/3] + [2079^2/4] = 3^1 + [508^2/3] + [2017^2/4] = 3^9 + [328^2/3] + [2047^2/4].
a(1694294) = 3 with 1694294 = 3^8 + [860^2/3] + [2401^2/4] = 3^8 + [928^2/3] + [2367^2/4] = 3^13 + [112^2/3] + [619^2/4].
		

Links

• Zhi-Wei Sun, Natural numbers represented by [x^2/a] + [y^2/b] + [z^2/c], arXiv:1504.01608 [math.NT], 2015.

Crossrefs

Cf. A000290, A000244, A343387, A343391, A343397.

Programs

Mathematica

PowQ[n_]:=PowQ[n]=IntegerQ[Log[3,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]