A297970 Numbers that are not the sum of 3 squares and a nonnegative 7th power.

112, 240, 368, 496, 624, 752, 880, 1008, 1136, 1264, 1392, 1520, 1648, 1776, 1904, 2032, 2160

Offset: 1

Author

XU Pingya, Jan 10 2018

Keywords

• nonn
• fini
• full

Comments

The last term in this sequence is 2160. The reasons are as follows (let b, c, d, i, j, k, m, r, s, t, w, x, y and z be nonnegative integers).

For the Diophantine equation x^2 + y^2 + z^2 + w^7 = m:

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

(2) 8b + 7 - 1^7 = 8b + 6. Then m = 8b + 7, the equation has a solution with w = 1.

(3) 4 * (8b + 7) - 1^7 = (8 * (4b + 3) + 3) = 8d + 3. Then m = 4 * (8b + 7), the equation has a solution with w = 1.

(4) For b >= 17, 16 * (8b + 7) - 3^7 = 8 * (16 * (b - 17) + 12) + 5 = 8i + 5. Then m = 16 * (8b + 7) and b >= 17, the equation has a solution with w = 3.

(5) 4^3 * (8b + 7) - 2^7 = 4^3 * (8b + 5). Then m = 4^3 * (8b + 7), the equation has a solution with w = 2. And 4^3 * (8b + 7) - 3^7 = 8 * (4^3 * (b - 4) + 38) + 5 = 8j + 5. Then m = 4^3 * (8b + 7) and b >= 4, the equation has a solution with w = 3.

(6) 4^4 * (8b + 7) - 2^7 = 4^3 * (8 * (4b + 3) + 3) = 4^3 * (8k + 3). 4^4 * (8b + 7) - 3^7 = 8 * (256b - 217) + 3 = 8r + 3. Then m = 4^4 * (8b + 7), the equation has a solution with w = 2 and when b > 0, the equation has a solution with w = 3.

(7) When c >= 5, 4^c * (8b + 7) - 2^7 = 4^3 * (8 * (b * 4^(c - 3) + 14 * 4^(c - 5) + 5) = 4^3 * (8s + 5). 4^c * (8b + 7) - 3^7 = 8 * (b * 4^(c - 3) + 14 * 4^(c - 3) - 273) + 3 = 8t + 3. Then n = 4^c * (8b + 7), the equation has solutions with w = 2 and 3.

In short, except for the 17 numbers in the sequence, every nonnegative integer can be represented as the sum of 3 squares and a nonnegative 7th power.

Links

• Wikipedia, Legendre's three-square theorem

Crossrefs

Finite subsequence of A004215 and A296185.

Cf. A004771, A022552, A022557, A022561, A022566, A111151.

Programs

Mathematica

t1={};
Do[Do[If[x^2+y^2+z^2+w^7==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)^(1/7)}], {n,0,3000}];
t2={};
Do[If[FreeQ[t1,k]==True, AppendTo[t2,k]], {k,0,3000}];
t2

Formula

a(n) = 128n - 16 = 16 * A004771(n - 1), 1 <= n <= 17.