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Number of nonnegative solutions to x^3 + y^3 + z^3 <= n^3.
5

%I #35 Feb 01 2025 14:57:21

%S 1,4,11,30,66,115,200,302,441,619,829,1085,1395,1771,2200,2666,3228,

%T 3843,4564,5351,6185,7143,8158,9349,10526,11934,13375,14896,16652,

%U 18381,20370,22411,24629,26963,29406,32101,34840,37766,40920,44164,47587,51200

%N Number of nonnegative solutions to x^3 + y^3 + z^3 <= n^3.

%H David A. Corneth, <a href="/A224215/b224215.txt">Table of n, a(n) for n = 0..1500</a>

%H David A. Corneth, <a href="/A224215/a224215_1.gp.txt">Pari prog</a>

%F a(n) = [x^(n^3)] (1/(1 - x))*(Sum_{k>=0} x^(k^3))^3. - _Ilya Gutkovskiy_, Apr 20 2018

%e For n=1, the four solutions are {0,0,0}, {0,0,1}, {0,1,0} and {1,0,0}, so a(1)=4.

%o (Python)

%o for a in range(99):

%o n = a*a*a

%o k = 0

%o for x in range(99):

%o s = x*x*x

%o if s>n: break

%o for y in range(99):

%o sy = s + y*y*y

%o if sy>n: break

%o for z in range(99):

%o sz = sy + z*z*z

%o if sz>n: break

%o k+=1

%o print(k, end=',')

%o (PARI) a(n) = n++; p = Pol((1/(1 - x))*sum(k=0, n, x^(k^3))^3 + O(x^(n^3))); polcoeff(p, (n-1)^3); \\ _Michel Marcus_, Apr 21 2018

%o (PARI) \\ See PARI link. _David A. Corneth_, May 22 2018

%Y Cf. A224214.

%K nonn,changed

%O 0,2

%A _Alex Ratushnyak_, Apr 01 2013