A253663 - OEIS

%I #29 Jun 02 2024 13:28:21

%S 0,0,1,7,17,38,78,127,196,296,410,564,738,958,1220,1514,1848,2235,

%T 2686,3175,3719,4365,5007,5758,6568,7442,8415,9477,10597,11779,13100,

%U 14459,15954,17566,19231,21029,22916,24930,27030,29293,31616,34103,36732,39459

%N Number of positive solutions to x^2+y^2+z^2 <= n^2.

%C Whereas A000604 counts solutions where x>=0, y>=0, z>=0, this sequence counts solutions where x>0, y>0, z>0.

%H Alois P. Heinz, <a href="/A253663/b253663.txt">Table of n, a(n) for n = 0..1000</a>

%F a(n) = A211639(n^2).

%F a(n) = [x^(n^2)] (theta_3(x) - 1)^3/(8*(1 - x)), where theta_3() is the Jacobi theta function. - _Ilya Gutkovskiy_, Apr 17 2018

%F Comment from _N. J. A. Sloane_, Jun 02 2024 (Start)

%F The one-dimensional lattice {n: n an integer} , which graphically looks like

%F    ...o   o   o   o   o   o   ...

%F has theta series 1 + 2 q + 2 q^4 + 2 q^9 + 2 q^16 + ... = sum {n=-oo..oo} q^(n^2),

%F and that power series is called theta_3(q).

%F   Raising it to the power 3 counts points with x^2+y^2+z^2 = k.

%F Dividing it by 1-x gives the partial sums, which basically is what this sequence is.

%F So a first approximation to a theta series for the sequence is theta_3(q)^8/(1-q).

%F Subtracting 1 and dividing by 8 is because here we only want positive solutions.

%F (End)

%e a(4)=17 counts the following solutions (x,y,z): (1,1,1), (2,2,2), three permutations of (1,1,2), three permutations of (1,1,3), three permutations of (1,2,2), and six permutations of (1,2,3).

%o (Sage)

%o [len([(x,y,z) for x in [1..n] for y in [1..n] for z in [1..n] if x^2+y^2+z^2<=n^2]) for n in [0..43]] # _Tom Edgar_, Jan 07 2015

%Y Cf. A000604.

%K nonn

%O 0,4

%A _R. J. Mathar_, Jan 07 2015