A296185 Numbers that are not the sum of 3 squares and an 8th power.

112, 240, 368, 496, 624, 752, 880, 1008, 1136, 1264, 1392, 1520, 1648, 1776, 1904, 2032, 2160, 2288, 2416, 2544, 2672, 2800, 2928, 3056, 3184, 3312, 3440, 3568, 3696, 3824, 3952, 4080, 4208, 4336, 4464, 4592, 4720, 4848, 4976, 5104, 5232, 5360, 5488, 5616

Offset: 1

Author

XU Pingya, Jan 13 2018

Keywords

• nonn

Comments

When m is in this sequence, 9m and m^9 are also in this sequence.

For nonnegative integers a, b, k, n, x, y, z and w, n = x^2 + y^2 + z^2 + w^8 if and only if n is not of the form 4^(4k + 2) * (8b + 7).

1. If n is not of the form 4^a * (8b + 7), then it follows from Legendre's three-square theorem that the equation x^2 + y^2 + z^2 + w^8 = n has a solution with w = 0.

2. If n = 4^a * (8b + 7), then (c, d and j are nonnegative integers):

(1) If a = 4k, then n - (2^k)^8 = 4^(4k) * (8b + 6), and the equation has a solution with w = 2^k.

(2) If a = 4k + 1, then n - (2^k)^8 = 4^(4k) * (32b + 27) is of the form 4^c * (8d + 3), and the equation has a solution with w = 2^k.

(3) If a = 4k + 3, then n - (2^(k + 1))^8 = 4^a * (8b + 3), and the equation has a solution with w = 2^(k + 1).

(4) If a = 4k + 2 and w = 2j + 1, then n == 0 (mod 8), w^8 == 1 (mod 8), and n - w^8 is number of the form 8c + 7. I.e., the equation does not have a solution with w odd.

If a = 4k + 2 and w = 4j + 2, then n - w^8 = 16 * (4^(4k) * (8b + 7) - 16 * (2j + 1)^8) = 4^4 * (4^(4k - 4) * (8b + 7) - (2j + 1)^8). When k = 0, n - w^8 is a number of the form 16 * (8c + 7); when k > 0, n - w^8 is a number of the form 4^4 * (8d + 7). Therefore, the equation does not have a solution with w = 4j + 2. Similarly, it can be proved that there is no solution with w = 4j.

Links

• Wikipedia, Legendre's three-square theorem

Crossrefs

Cf. A004215, A022552, A022557, A022561, A022566, A111151.

Programs

Mathematica

t1={};
Do[Do[If[x^2+y^2+z^2+w^8==n, AppendTo[t1,n]&&Break[]], {x,0,n^(1/2)}, {y,x,(n-x^2)^(1/2)}, {z,y,(n-x^2-y^2)^(1/2)}, {w,0,(n-x^2-y^2-z^2)^(1/8)}], {n,0,5700}];
t2={};
Do[If[FreeQ[t1,k]==True, AppendTo[t2,k]], {k,0,5700}];
t2

Python

from itertools import count, islice
def A296185_gen(): # generator of terms
    for k in count(0):
        r = 1<<((k<<1)+1<<2)
        yield from range(7*r,r*((r<<8)+7),r<<3)
A296185_list = list(islice(A296185_gen(),44)) # Chai Wah Wu, May 21 2025