A262815 Number of ordered ways to write n as x^3 + y*(y+1)/2 + z*(3*z+1)/2, where x, y and z are nonnegative integers.

1, 2, 2, 3, 2, 1, 2, 2, 4, 3, 3, 4, 1, 3, 2, 3, 5, 3, 5, 1, 1, 3, 3, 4, 2, 3, 3, 3, 4, 6, 6, 3, 2, 3, 2, 4, 6, 6, 3, 2, 3, 3, 4, 5, 8, 2, 3, 3, 5, 3, 2, 5, 3, 3, 3, 7, 3, 4, 4, 3, 3, 3, 5, 8, 2, 4, 3, 5, 7, 4, 7, 4, 5, 3, 6, 1, 4, 4, 6, 7, 4, 8, 5, 1, 4, 7, 7, 4, 4, 5, 2, 3, 5, 10, 6, 4, 2, 1, 3, 5, 7

Offset

0,2

Comments

Conjecture: a(n) > 0 for all n >= 0, and a(n) = 1 only for n = 0, 5, 12, 19, 20, 75, 83, 97, 117.

Conjecture verified up to 10^11. - Mauro Fiorentini, Jul 20 2023

See also A262813 and A262816 for similar conjectures.

By Theorem 1.7(i) in the linked paper, each natural number can be written as the sum of a triangular number, an even square and a generalized pentagonal number.

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.

Example

a(0) = 1 since 0 = 0^3 + 0*1/2 + 0*(3*0+1)/2.
a(5) = 1 since 5 = 0^3 + 2*3/2 + 1*(3*1+1)/2.
a(12) = 1 since 12 = 0^3 + 4*5/2 + 1*(3*1+1)/2.
a(19) = 1 since 19 = 1^3 + 2*3/2 + 3*(3*3+1)/2.
a(20) = 1 since 20 = 2^3 + 4*5/2 + 1*(3*1+1)/2.
a(75) = 1 since 75 = 2^3 + 4*5/2 + 6*(3*6+1)/2.
a(83) = 1 since 83 = 0^3 + 3*4/2 + 7*(3*7+1)/2.
a(97) = 1 since 97 = 3^3 + 10*11/2 + 3*(3*3+1)/2.
a(117) = 1 since 117 = 0^3 + 13*14/2 + 4*(3*4+1)/2.

Mathematica

TQ[n_]:=IntegerQ[Sqrt[8n+1]]
Do[r=0; Do[If[TQ[n-x^3-z(3z+1)/2], r=r+1], {x, 0, n^(1/3)}, {z, 0, (Sqrt[24(n-x^3)+1]-1)/6}]; Print[n, " ", r]; Continue, {n, 0, 100}]

Crossrefs

Cf. A000217, A000578, A005449, A160325, A262813, A262816.

Sequence in context: A368798 A253141 A100890 * A076494 A217721 A071862

Adjacent sequences: A262812 A262813 A262814 * A262816 A262817 A262818

Keyword

nonn

Author

Zhi-Wei Sun, Oct 03 2015

Status

approved