A214900 Number of ordered ways to represent n as the sum of three squares and one fourth power.

1, 4, 6, 4, 4, 9, 9, 3, 3, 9, 12, 9, 4, 7, 12, 6, 4, 15, 18, 10, 12, 18, 12, 3, 6, 18, 27, 19, 5, 18, 24, 6, 6, 18, 21, 18, 18, 18, 18, 9, 9, 30, 33, 13, 6, 27, 24, 6, 4, 16, 33, 27, 18, 24, 33, 12, 12, 27, 18, 18, 12, 24, 30, 12, 4, 30, 45, 21, 18, 33, 30, 6, 12, 21, 33, 34, 10, 27, 30, 6, 9, 40, 39, 24, 25, 33, 39, 18, 9

Offset: 0

Author

Joerg Arndt, Jul 29 2012

Keywords

• nonn

Comments

Different orderings of summands are counted, e.g., 1 = 1^2 + 0^2 + 0^4 + 0^4 = 0^2 + 1^2 + 0^4 + 0^4 = 0^2 + 0^2 + 1^4 + 0^4 = 0^2 + 0^2 + 0^4 + 1^4, so a(1)=4.

Conjecture: a(n) != 0, that is, all numbers are sums of three squares and one fourth power.

Crossrefs

Cf. A000925 (two squares), A002102 (three squares).

Programs

PARI

N=10^3;  x='x+O('x^N);
S(e)=sum(j=0, ceil(N^(1/e)), x^(j^e));
v=Vec( S(4)^1 * S(2)^3 )

Formula

G.f.: (Sum_{j>=0} x^(j^2))^3 * (Sum_{j>=0} x^(j^4)) (see PARI code).