|
|
A119395
|
|
Number of nonnegative integer solutions to the equation x^2 + 3y^2 = n.
|
|
8
|
|
|
1, 1, 0, 1, 2, 0, 0, 1, 0, 1, 0, 0, 2, 1, 0, 0, 2, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 3, 0, 0, 1, 0, 0, 0, 0, 2, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 2, 2, 0, 0, 3, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 2, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 3, 0, 0, 1, 0, 1, 0, 0, 3, 0, 0, 0, 0, 0, 0, 2, 0, 1, 0, 0, 0, 1, 0, 0, 2, 0
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,5
|
|
COMMENTS
|
The number of integer solutions is given by A033716.
Records 1, 2, 3, 5, 6, 9, 12, 14, 18, ... occur at 0, 4, 28, 196, 364, 2548, 6916, 33124, 48412, ... - Antti Karttunen, Nov 20 2017
|
|
LINKS
|
|
|
FORMULA
|
For n > 0, a(n) = (A033716(n) + 2)/4 if n is a square or a triple of a square; otherwise a(n) = A033716(n)/4. Alternatively, a(n) = ceiling(A033716(n)/4).
G.f.: (1 + theta_3(q))*(1 + theta_3(q^3))/4, where theta_3() is the Jacobi theta function. - Ilya Gutkovskiy, Aug 01 2018
|
|
MATHEMATICA
|
QP = QPochhammer;
s = (QP[q^2]*QP[q^6])^5/(QP[q]*QP[q^3]*QP[q^4]*QP[q^12])^2 + O[q]^105;
|
|
PROG
|
(PARI) { A033716(n) = local(f, B); f=factorint(n); B=1; for(i=1, matsize(f)[1], if(f[i, 1]%3==1, B*=f[i, 2]+1); if(f[i, 1]%3==2, if(f[i, 2]%2, return(0)))); if(n%4, 2*B, 6*B) } { a(n) = ceil(A033716(n)/4) }
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|