A366091 a(n) is the number of ways to write n = i^2 + 2*j^2 + 3*k^2 with i,j,k >= 0.

1, 1, 1, 2, 2, 1, 2, 1, 1, 3, 0, 2, 4, 1, 2, 2, 2, 1, 3, 2, 2, 4, 2, 1, 2, 2, 0, 4, 3, 2, 5, 2, 1, 3, 2, 2, 7, 2, 2, 5, 0, 2, 0, 2, 4, 4, 3, 1, 4, 3, 3, 5, 3, 2, 7, 1, 2, 6, 0, 3, 6, 2, 2, 4, 2, 2, 6, 3, 2, 4, 3, 3, 3, 2, 0, 7, 5, 2, 6, 3, 2, 8, 2, 2, 11, 2, 5, 2, 2, 3, 0, 4, 3, 7, 3, 2, 2, 3, 3

Offset

0,4

Links

Robert Israel, Table of n, a(n) for n = 0..10000

Formula

G.f. (1 + theta_3(0,z)) * (1 + theta_3(0,z^2)) * (1 + theta_3(0,z^3))/8 where theta_3 is a Jacobi theta function.

Example

a(9) = 3 because 9 = 3^2 + 2*0^2 + 3*0^2 = 1^2 + 2*2^2 + 3*0^2 = 2^2 + 2*1^2 + 3*1^2.

Maple

g:= (1+JacobiTheta3(0, z))*(1+JacobiTheta3(0, z^2))*(1+JacobiTheta3(0, z^3))/8:
S:= series(g, z, 101):
seq(coeff(S, z, j), j=0..100);

Prog

(Python)
from itertools import count
from sympy.ntheory.primetest import is_square
def A366091(n):
    c = 0
    for k in count(0):
        if (a:=3*k**2)>n:
            break
        for j in count(0):
            if (b:=a+(j**2<<1))>n:
                break
            if is_square(n-b):
                c += 1
    return c # Chai Wah Wu, Sep 29 2023

Crossrefs

Cf. A028594 (allows any integer i,j,k), A055042 (a(n) = 0)

Sequence in context: A332278 A182596 A087775 * A089955 A352942 A180312

Adjacent sequences: A366088 A366089 A366090 * A366092 A366093 A366094

Keyword

nonn

Author

Robert Israel, Sep 28 2023

Status

approved