cp's OEIS Frontend

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

%I A296185 #47 May 21 2025 18:47:30
%S A296185 112,240,368,496,624,752,880,1008,1136,1264,1392,1520,1648,1776,1904,
%T A296185 2032,2160,2288,2416,2544,2672,2800,2928,3056,3184,3312,3440,3568,
%U A296185 3696,3824,3952,4080,4208,4336,4464,4592,4720,4848,4976,5104,5232,5360,5488,5616
%N A296185 Numbers that are not the sum of 3 squares and an 8th power.
%C A296185 When m is in this sequence, 9m and m^9 are also in this sequence.
%C A296185 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).
%C A296185 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.
%C A296185 2. If n = 4^a * (8b + 7), then (c, d and j are nonnegative integers):
%C A296185 (1) If a = 4k, then n - (2^k)^8 = 4^(4k) * (8b + 6), and the equation has a solution with w = 2^k.
%C A296185 (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.
%C A296185 (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).
%C A296185 (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.
%C A296185 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.
%H A296185 Wikipedia, <a href="https://en.wikipedia.org/wiki/Legendre%27s_three-square_theorem">Legendre's three-square theorem</a>
%t A296185 t1={};
%t A296185 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}];
%t A296185 t2={};
%t A296185 Do[If[FreeQ[t1,k]==True, AppendTo[t2,k]], {k,0,5700}];
%t A296185 t2
%o A296185 (Python)
%o A296185 from itertools import count, islice
%o A296185 def A296185_gen(): # generator of terms
%o A296185     for k in count(0):
%o A296185         r = 1<<((k<<1)+1<<2)
%o A296185         yield from range(7*r,r*((r<<8)+7),r<<3)
%o A296185 A296185_list = list(islice(A296185_gen(),44)) # _Chai Wah Wu_, May 21 2025
%Y A296185 Cf. A004215, A022552, A022557, A022561, A022566, A111151.
%K A296185 nonn
%O A296185 1,1
%A A296185 _XU Pingya_, Jan 13 2018