A254668 Number of ways to write n as the sum of a square, a second pentagonal number, and a hexagonal number.

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

Offset

0,2

Comments

Conjecture: a(n) > 0 for all n. Also, a(n) = 1 only for n = 0, 5, 13, 14, 20, 112, 125.

Compare this conjecture with the conjecture in A160324.

The conjecture that a(n) > 0 for all n = 0,1,2,... appeared in Conjecture 1.2(ii) of the author's JNT paper in the links. - Zhi-Wei Sun, Oct 03 2016

Links

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000

Zhi-Wei Sun, On universal sums of polygonal numbers, arXiv:0905.0635 [math.NT], 2009-2015.

Zhi-Wei Sun, On x(ax+1)+y(by+1)+z(cz+1) and x(ax+b)+y(ay+c)+z(az+d), J. Number Theory 171(2017), 275-283.

Example

a(20) = 1 since 20 = 2^2 + 3*(3*3+1)/2 + 1*(2*1-1).
a(112) = 1 since 112 = 7^2 + 6*(3*6+1)/2 + 2*(2*2-1).
a(125) = 1 since 125 = 5^2 + 8*(3*8+1)/2 + 0*(2*0-1).

Mathematica

SQ[n_]:=IntegerQ[Sqrt[n]]
Do[r=0; Do[If[SQ[n-y(3y+1)/2-z(2z-1)], r=r+1], {y, 0, (Sqrt[24n+1]-1)/6}, {z, 0, (Sqrt[8(n-y(3y+1)/2)+1]+1)/4}];
Print[n, " ", r]; Continue, {n, 0, 100}]

Crossrefs

Cf. A000290, A000384, A005449, A160324, A160325, A254661.

Sequence in context: A289641 A209312 A054715 * A306250 A145443 A192056

Adjacent sequences: A254665 A254666 A254667 * A254669 A254670 A254671

Keyword

nonn

Author

Zhi-Wei Sun, Feb 04 2015

Status

approved