A254629 Number of ways to write n as x^2 + y*(y+1) + z*(4*z+1) with x,y,z nonnegative integers.

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

Offset

0,7

Comments

Conjecture: (i) a(n) > 0 for all n, and a(n) > 1 for all n > 338.

(ii) If f(x) is one of the polynomials 3x^2+x, (7x^2+3x)/2, (9x^2-x)/2, (11x^2-7x)/2, (15x^2-7x)/2, (15x^2-11x)/2, then any nonnegative integer n can be written as x^2 + y*(y+1) + f(z) with x,y,z nonnegative integers.

We have proved that for each n = 0,1,... there are integers x,y,z such that n = x^2 + y*(y+1) + z*(4z+1).

It is known that {x^2+y*(y+1): x,y=0,1,...} = {x*(x+1)/2+y*(y+1)/2: x,y=0,1,...}.

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.

Example

a(103) = 1 since 103 = 8^2 + 0*1 + 3*(4*3+1).
a(122) = 1 since 122 = 9^2 + 1*2 +3*(4*3+1).
a(143) = 1 since 143 = 6^2 + 1*2 + 5*(4*5+1).
a(167) = 1 since 167 = 3^2 + 9*10 + 4*(4*4+1).
a(248) = 1 since 248 = 5^2 + 4*5 + 7*(4*7+1).
a(338) = 1 since 338 = 5^2 + 10*11 + 7*(4*7+1).

Mathematica

SQ[n_]:=IntegerQ[Sqrt[n]]
Do[r=0; Do[If[SQ[n-y(y+1)-z*(4z+1)], r=r+1], {y, 0, (Sqrt[4n+1]-1)/2}, {z, 0, (Sqrt[16(n-y(y+1))+1]-1)/8}];
Print[n, " ", r]; Continue, {n, 0, 100}]

Crossrefs

Cf. A000217, A000290, A007742, A254573, A254574, A254623.

Sequence in context: A342468 A324534 A248505 * A096248 A079109 A079099

Adjacent sequences: A254626 A254627 A254628 * A254630 A254631 A254632

Keyword

nonn

Author

Zhi-Wei Sun, Feb 03 2015

Status

approved