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

1,3

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.

Links

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

Example

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.

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]

Crossrefs

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

Sequence in context: A164898 A100801 A194310 * A359455 A363853 A272231

Adjacent sequences: A306224 A306225 A306226 * A306228 A306229 A306230

Keyword

nonn

Author

Zhi-Wei Sun, Jan 30 2019

Status

approved