A306225 Number of ways to write n as w + x^5 + pen(y) + pen(z), where w is 0 or 1, and x,y,z are integers with x >= w and pen(y) < pen(z), and where pen(m) denotes the pentagonal number m*(3*m-1)/2.

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

Offset

1,2

Comments

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

We have verified a(n) > 0 for all n = 1..2*10^6. The conjecture implies that the set A = {x^5 + pen(y): x = 0,1,2,... and y is an integer} is an additive basis of order two (i.e., the sumset A + A coincides with {0,1,2,...}).

See also A306227 for a similar conjecture.

Links

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

Example

a(11) = 1 with 11 = 1 + 1^5 + pen(-1) + pen(-2).
a(1000) = 1 with 1000 = 0 + 2^5 + pen(8) + pen(-24).
a(5104) = 1 with 5104 = 1 + 3^5 + pen(-3) + pen(57).
a(8196) = 1 with 8196 = 0 + 2^5 + pen(48) + pen(-56).
a(9537) = 1 with 9537 = 1 + 6^5 + pen(17) + pen(30).
a(15049) = 1 with 15049 = 0 + 6^5 + pen(-44) + pen(54).
a(16775) = 1 with 16775 = 1 + 5^5 + pen(-17) + pen(94).

Mathematica

PenQ[n_]:=PenQ[n]=IntegerQ[Sqrt[24n+1]];
tab={}; Do[r=0; Do[If[PenQ[n-x-y^5-z(3z-1)/2], r=r+1], {x, 0, Min[1, (n-1)/2]}, {y, x, (n-1-x)^(1/5)}, {z, -Floor[(Sqrt[12(n-1-x-y^5)+1]-1)/6], (Sqrt[12(n-1-x-y^5)+1]+1)/6}]; tab=Append[tab, r], {n, 1, 100}]; Print[tab]

Crossrefs

Cf. A000584, A001318, A270920, A306227.

Sequence in context: A288677 A187757 A286529 * A373249 A077199 A145390

Adjacent sequences: A306222 A306223 A306224 * A306226 A306227 A306228

Keyword

nonn

Author

Zhi-Wei Sun, Jan 30 2019

Status

approved