A283239 Number of ways to write n as x^2 + y^2 + z*(3*z-1)/2 with x,y,z integers such that x + 2*y is a square.

1, 2, 3, 2, 2, 5, 5, 5, 1, 2, 4, 2, 5, 3, 3, 4, 3, 7, 5, 3, 8, 3, 5, 3, 2, 6, 3, 7, 5, 3, 4, 7, 5, 3, 4, 6, 5, 3, 3, 7, 5, 5, 5, 1, 6, 7, 6, 4, 3, 2, 5, 5, 9, 6, 3, 7, 5, 7, 5, 4, 8, 5, 6, 4, 6, 5, 5, 8, 5, 6, 6, 7, 5, 5, 5, 7, 5, 6, 2, 4, 12

Offset

0,2

Comments

Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 8, 43, 84, 133, 253, 399, 488, 523, 803, 7369.

(ii) Any integer n > 1 can be written as p + x^2 + y^2 with p prime and x + 2*y (or x + 3*y) a square, where x is an integer and y is a nonnegative integer.

Note that those numbers z*(3*z-1)/2 with z integral are called generalized pentagonal numbers (A001318). By Theorem 1.7(ii) of the linked paper in Sci. China Math., each n = 0,1,2,... can be written as the sum of two squares and a generalized pentagonal number.

Ju. V. Linnik proved in 1960 that any sufficiently large integer can be expressed as the sum of a prime and two squares.

Links

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

Ju. V. Linnik, An asymptotic formula in an additive problem of Hardy-Littlewood, Izv. Akad. Nauk SSSR Ser. Mat., 24 (1960), 629-706 (Russian).

Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58 (2015), no. 7, 1367-1396.

Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175 (2017), 167-190.

Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017.

Example

a(8) = 1 since 8 = 1^2 + 0^2 + (-2)*(3*(-2)-1)/2 with 1 + 2*0 = 1^2.
a(43) = 1 since 43 = 1^2 + 4^2 + (-4)*(3*(-4)-1)/2 with 1 + 2*4 = 3^2.
a(84) = 1 since 84 = 7^2 + (-3)^2 + (-4)*(3*(-4)-1)/2 with 7 + 2*(-3) = 1^2.
a(133) = 1 since 133 = 4^2 + 0^2 + 9*(3*9-1)/2 with 4 + 2*0 = 2^2.
a(253) = 1 since 253 = (-13)^2 + 7^2 + 5*(3*5-1)/2 with (-13) + 2*7 = 1^2.
a(399) = 1 since 399 = 18^2 + (-7)^2 + (-4)*(3*(-4)-1)/2 with 18 + 2*(-7) = 2^2.
a(488) = 1 since 488 = 9^2 + 20^2 + (-2)*(3*(-2)-1)/2 with 9 + 2*20 = 7^2.
a(523) = 1 since 523 = 9^2 + 0^2 + (-17)*(3*(-17)-1)/2 with 9 + 2*0 = 3^2.
a(803) = 1 since 803 = (-17)^2 + 13^2 + (-15)*(3*(-15)-1)/2 with (-17) + 2*13 = 3^2.
a(7369) = 1 since 7369 = 0^2 + 72^2 + (-38)*(3*(-38)-1)/2 with 0 + 2*72 = 12^2.

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
PenQ[n_]:=PenQ[n]=SQ[24n+1];
Do[r=0; Do[If[PenQ[n-x^2-y^2], Do[If[SQ[(-1)^i*x+2(-1)^j*y], r=r+1], {i, 0, Min[x, 1]}, {j, 0, Min[y, 1]}]], {x, 0, Sqrt[n]}, {y, 0, Sqrt[n-x^2]}]; Print[n, " ", r]; Continue, {n, 0, 80}]

Crossrefs

Cf. A000040, A000290, A001318, A001481, A276415, A283170, A283196, A283204, A283205.

Sequence in context: A240230 A238689 A166454 * A318177 A128651 A093797

Adjacent sequences: A283236 A283237 A283238 * A283240 A283241 A283242

Keyword

nonn

Author

Zhi-Wei Sun, Mar 03 2017

Status

approved