A270533 Number of ordered ways to write n = x^4 + x^3 + y^2 + z*(3z-1)/2, where x and y are nonnegative integers, and z is an integer.

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

Offset: 0

Author

Zhi-Wei Sun, Mar 18 2016

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 20, 22, 70, 87, 167, 252, 388, 562, 636, 658, 873, 2598, 14979, 18892, 20824.

(ii) Each n = 0,1,2,... can be written as x*P(x) + y^2 + z*(3z-1)/2 with x and y nonnegative integers, and z an integer, where P(x) is either of the polynomials x^3+2, x^3+3, x^3+2x+8, x^3+x^2+4x+2, x^3+x^2+7x+6.

(iii) Any nonnegative integer can be expressed as x*(x^3+3) + y*(5y+4) + z*(3z-1)/2, where x is an nonnegative integer, and y and z are integers.

See also A270516 for a similar conjecture.

Examples

			a(20) = 1 since 20 = 1^4 + 1^3 + 4^2 + (-1)*(3*(-1)-1)/2.
a(22) = 1 since 22 = 0^4 + 0^3 + 0^2 + 4*(3*4-1)/2.
a(873) = 1 since 873 = 5^4 + 5^3 + 11^2 + (-1)*(3*(-1)-1)/2.
a(2598) = 1 since 2598 =  4^4 + 4^3 + 4^2 + 39*(3*39-1)/2.
a(14979) = 1 since 14979 = 1^4 + 1^3 + 51^2 + 91*(3*91-1)/2.
a(18892) = 1 since 18892 = 3^4 + 3^3 + 137^2 + (-3)*(3*(-3)-1)/2.
a(20824) = 1 since 20824 = 1^4 + 1^3 + 115^2 + (-71)*(3*(-71)-1)/2.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
• Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 1367-1396.

Crossrefs

Cf. A000290, A000578, A000583, A001318, A262813, A262815, A262816, A262827, A262941, A262944, A262945, A262954, A262955, A262956, A270469, A270488, A270516.

Programs

Mathematica

pQ[x_]:=pQ[x]=IntegerQ[Sqrt[24x+1]]
Do[r=0;Do[If[pQ[n-y^2-x^3*(x+1)],r=r+1],{y,0,Sqrt[n]},{x,0,(n-y^2)^(1/4)}];Print[n," ",r];Continue,{n,0,90}]