%I #50 Apr 21 2023 12:47:38
%S 23,44,71,79,184,368,519,599,704,1136,1264,2944,4024,5888,8304,9584,
%T 11264,18176,20224,47104,64384,94208,132864,153344,180224,290816,
%U 323584,753664,1030144,1507328,2125824,2453504,2883584,4653056,5177344,12058624,16482304
%N Numbers that are not the sum of two squares and two fourth powers.
%C From _XU Pingya_, Feb 07 2018: (Start)
%C When n is a term, 16n is also. This can be proved as follows:
%C (1) If w is odd, then 16n - w^4 == 7 (mod 8), and it follows from Legendre's three-square theorem that the equation x^2 + y^2 + z^4 + w^4 = 16n has no solution (it is the same when x, y or z are odd numbers).
%C (2) If x, y, z and w are even numbers (x = 2a, y = 2b, z = 2c, w = 2d) such that x^2 + y^2 + z^4 + w^4 = 16n, then a^2 + b^2 = 4(n - c^4 - d^4). So there are integers u and v satisfying u^2 + v^2 = n - c^4 - d^4. i.e. u^2 + v^2 + c^4 + d^4 = n, which is a contradiction.
%C (End)
%C Conjecture: The set {a(n): n > 0} coincides with {16^k*m: k = 0, 1, 2, ... and m = 23, 44, 71, 79, 184, 519, 599, 4024}. - _Zhi-Wei Sun_, Jan 27 2022
%H Donovan Johnson, <a href="/A214891/b214891.txt">Table of n, a(n) for n = 1..52</a> (terms <= 4*10^9)
%H Zhi-Wei Sun, <a href="https://mathoverflow.net/questions/414791">On w^4+x^4+y^2+z^2 over a number field</a>, Question 414791 at MathOverflow, Jan. 27, 2022.
%o (PARI)
%o N=10^6; x='x+O('x^N);
%o S(e)=sum(j=0, ceil(N^(1/e)), x^(j^e));
%o v=Vec( S(4)^2 * S(2)^2 );
%o for(n=1,#v,if(!v[n],print1(n-1,", ")));
%Y Cf. A001481, A004999, A022549, A346643, A347865, A350857, A350860.
%K nonn
%O 1,1
%A _Joerg Arndt_, Jul 29 2012
%E a(29)-a(37) from _Donovan Johnson_, Jul 29 2012
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