|
%I #29 Aug 26 2021 19:15:54
%S 1,4,4,0,0,0,0,0,4,8,0,0,0,0,0,0,4,0,0,0,0,0,0,0,0,0,0,4,8,0,0,0,0,0,
%T 0,8,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,4,0,0,0,0,0,0,0,0,0,4,8,0,0,
%U 0,0,0,0,8,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,8,0,0,0,0,0,0,0,0,0,0,0,0,0
%N Number of integer pairs (x,y) satisfying |x|^3 + |y|^3 = n, -n <= x,y <= n.
%C Cube variant of A004018.
%C Obviously, a(n) must be 4*k, for k >= 0, n > 0. - _Altug Alkan_, Apr 09 2016
%C From _Robert Israel_, Jan 26 2017: (Start)
%C a(k^3*n) >= a(n) for k >= 1.
%C a(n) >= 16 for n in A001235.
%C a(A011541(n)) >= 8*n. (End)
%H Robert Israel, <a href="/A175362/b175362.txt">Table of n, a(n) for n = 0..10000</a>
%F G.f.: ( 1 + 2 * Sum_{j>=1} x^(j^3) )^2.
%F a(n^3) = 4 for n > 0. - _Altug Alkan_, Apr 09 2016
%F a(n) = 4*Sum_{k=1..floor(n^(1/3))} A010057(n - k^3), for n > 0. - _Daniel Suteu_, Aug 15 2021
%e a(2) = 4 counts (x,y) = (-1,1), (1,1), (-1,-1) and (1,-1).
%e a(9) = 8 counts (x,y) = (-2,-1), (-2,1), (-1,-2), (-1,2), (1,-2), (1,2), (2,-1) and (2,1).
%p N:= 200: # to get a(0)..a(N)
%p G:= (1+2*add(x^(j^3),j=1..floor(N^(1/3))))^2:
%p seq(coeff(G,x,j),j=0..N); # _Robert Israel_, Jan 26 2017
%o (PARI) a(n) = if(n==0, 1, 4*sum(k=1, sqrtnint(n, 3), ispower(n - k^3, 3))); \\ _Daniel Suteu_, Aug 16 2021
%Y Cf. A001235, A010057, A011541, A025446, A025455, A025464, A025468, A121980.
%K nonn
%O 0,2
%A _R. J. Mathar_, Apr 24 2010
%E Invalid claim that belonged to A004018 removed by _R. J. Mathar_, Apr 24 2010
|
