OFFSET
1,1
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.
PROG
(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, ", ")));
CROSSREFS
KEYWORD
nonn
AUTHOR
Joerg Arndt, Jul 29 2012
EXTENSIONS
a(29)-a(37) from Donovan Johnson, Jul 29 2012
STATUS
approved