A214891 Numbers that are not the sum of two squares and two fourth powers.

23, 44, 71, 79, 184, 368, 519, 599, 704, 1136, 1264, 2944, 4024, 5888, 8304, 9584, 11264, 18176, 20224, 47104, 64384, 94208, 132864, 153344, 180224, 290816, 323584, 753664, 1030144, 1507328, 2125824, 2453504, 2883584, 4653056, 5177344, 12058624, 16482304

Offset: 1

Author

Joerg Arndt, Jul 29 2012

Keywords

• nonn

Comments

From XU Pingya, Feb 07 2018: (Start)

When n is a term, 16n is also. This can be proved as follows:

(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).

(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.

(End)

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

Links

• Donovan Johnson, Table of n, a(n) for n = 1..52 (terms <= 4*10^9)
• Zhi-Wei Sun, On w^4+x^4+y^2+z^2 over a number field, Question 414791 at MathOverflow, Jan. 27, 2022.

Crossrefs

Cf. A001481, A004999, A022549, A346643, A347865, A350857, A350860.

Programs

PARI

N=10^6;  x='x+O('x^N);
S(e)=sum(j=0, ceil(N^(1/e)), x^(j^e));
v=Vec( S(4)^2 * S(2)^2 );
for(n=1,#v,if(!v[n],print1(n-1,", ")));

Extensions

a(29)-a(37) from Donovan Johnson, Jul 29 2012