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

0,2

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.

Links

Table of n, a(n) for n=0..88.

Formula

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

Prog

(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 )

Crossrefs

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

Sequence in context: A161965 A256419 A256424 * A362817 A078385 A137444

Adjacent sequences: A214897 A214898 A214899 * A214901 A214902 A214903

Keyword

nonn

Author

Joerg Arndt, Jul 29 2012

Status

approved