A306227 Number of ways to write n as w + x^4 + y*(y+1)/2 + z*(z+1)/2, where w is 0 or 1, and x, y, z are nonnegative integers with x >= w and y < z.

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

Offset: 1

Author

Zhi-Wei Sun, Jan 30 2019

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 0. In other words, any positive integer can be written as the sum of two fourth powers one of which is 0 or 1, and two distinct triangular numbers.

We have verified a(n) > 0 for all n = 1..10^6. The conjecture implies that the set S = {x^4 + y*(y+1)/2: x,y = 0,1,2,...} is an additive basis of order two (i.e., the sumset S + S coincides with {0,1,2,...}).

See also A306225 for a similar conjecture.

Examples

			a(1) = 1 with 1 = 0 + 0^4 + 0*1/2 + 1*2/2.
a(2) = 1 with 2 = 0 + 1^4 + 0*1/2 + 1*2/2.
a(14) = 1 with 14 = 0 + 1^4 + 2*3/2 + 4*5/2.
a(3774) = 1 with 3774 = 1 + 5^4 + 52*53/2 + 59*60/2.
a(7035) = 1 with 7035 = 0 + 3^4 + 48*49/2 + 107*108/2.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000217, A000583, A262813, A262941, A262954, A262956, A262959, A270566, A306225.

Programs

Mathematica

TQ[n_]:=TQ[n]=IntegerQ[Sqrt[8n+1]];
tab={};Do[r=0;Do[If[TQ[n-x-y^4-z(z+1)/2],r=r+1],{x,0,Min[1,(n-1)/2]},{y,x,(n-1-x)^(1/4)},{z,0,(Sqrt[4(n-1-x-y^4)+1]-1)/2}];tab=Append[tab,r],{n,1,100}];Print[tab]