A255191 Number of solutions to the equation n = x^2 + y*(y+1)/2 + z*(z+1)/2 (x,y,z=0,1,...) with x == Floor[sqrt(n)] (mod 2) and y <= z.

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

Offset

0,5

Comments

By Dickson's book in the reference, E. Lionnet observed in 1872 that any nonnegative integer can be written as the sum of a square and two triangular numbers.

We have a(n) > 0 for all n. This follows from Theorem 1(i) and Lemma 2 of Sun's reference in 2007. Note that if n = m*(m+1) with m a nonnegative integer then floor(sqrt(n)) = m.

References

L. E. Dickson, History of the Theory of Numbers, vol. II, AMS Chelsea Publ., 1999, p. 24.

Links

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

Zhi-Wei Sun, Mixed sums of squares and triangular numbers, Acta Arith. 127(2007), 103-113.

Zhi-Wei Sun, On universal sums a*x^2+b*y^2+f(z), a*T_x+b*T_y+f(z) and a*T_x+b*y^2+f(z), arXiv:1502.03056 [math.NT], 2015.

Example

a(14) = 1 since 14 = 1^2 + 2*3/2 + 4*5/2 with floor(sqrt(14)) == 1 (mod 2).
a(44) = 1 since 44 = 4^2 + 0*1/2 + 7*8/2 with floor(sqrt(44)) == 4 (mod 2).
a(63) = 1 since 63 = 5^2 + 4*5/2 + 7*8/2 with floor(sqrt(63)) == 5 (mod 2).

Mathematica

TQ[n_]:=IntegerQ[Sqrt[8n+1]]
Do[r=0; Do[If[Mod[Floor[Sqrt[n]]-x, 2]==0&&TQ[n-x^2-y*(y+1)/2], r=r+1], {x, 0, Sqrt[n]}, {y, 0, (Sqrt[4(n-x^2)+1]-1)/2}];
Print[n, " ", r]; Label[aa]; Continue, {n, 0, 100}]

Crossrefs

Cf. A000217, A000290, A254885.

Sequence in context: A374629 A145327 A374251 * A277818 A324607 A023512

Adjacent sequences: A255188 A255189 A255190 * A255192 A255193 A255194

Keyword

nonn

Author

Zhi-Wei Sun, Feb 16 2015

Status

approved