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%I #27 Jul 25 2023 07:34:47
%S 0,0,0,8,0,0,0,0,0,0,0,24,0,0,0,0,0,0,0,24,0,0,0,0,0,0,0,32,0,0,0,0,0,
%T 0,0,48,0,0,0,0,0,0,0,24,0,0,0,0,0,0,0,48,0,0,0,0,0,0,0,72,0,0,0,0,0,
%U 0,0,24,0,0,0,0,0,0,0,56,0,0,0,0,0,0,0,72,0,0,0,0,0,0,0,48,0,0,0,0,0,0,0,72
%N Number of ways of writing n as the sum of three odd squares.
%C Number of ways of writing n as the sum of the squares of three odd numbers (see example). Equals 8*A008437 because each summand can be the square of either a positive or negative odd number, and there are three summands, thus 2^3 = 8. - _Antti Karttunen_ & _Michel Marcus_, Jul 23 2018
%H Antti Karttunen, <a href="/A085121/b085121.txt">Table of n, a(n) for n = 0..65537</a>
%H J. E. Jones [Lennard-Jones] and A. E. Ingham, <a href="https://doi.org/10.1098/rspa.1925.0047">On the calculation of certain crystal potential constants and on the cubic crystal of least energy</a>, Proc. Royal Soc., A 107 (1925), 636-653 (see p. 650).
%F G.f.: (Sum_{n=-oo..oo} q^((2n+1)^2))^3.
%e a(3) = 8 because 3 = (+1)^2 + (+1)^2 + (+1)^2 = (-1)^2 + (+1)^2 + (+1)^2 = (+1)^2 + (-1)^2 + (+1)^2 = (+1)^2 + (+1)^2 + (-1)^2 = (-1)^2 + (-1)^2 + (+1)^2 = (-1)^2 + (+1)^2 + (-1)^2 = (+1)^2 + (-1)^2 + (-1)^2 = (-1)^2 + (-1)^2 + (-1)^2. - _Antti Karttunen_, Jul 23 2018
%o (PARI)
%o A008442(n) = if( n<1 || n%4!=1, 0, sumdiv(n, d, (d%4==1) - (d%4==3))); \\ From A008442.
%o A290081(n) = if(n%2,0,A008442(n/2));
%o A008437(n) = if((n<3)||!(n%2),0,my(s=0, k = sqrtint(n)); k -= ((1+k)%2); while(k>=1, s += A290081(n-(k*k)); k -= 2); (s));
%o A085121(n) = 8*A008437(n); \\ _Antti Karttunen_, Jul 22 2018
%Y Cf. A005875, A008437. The nonzero coefficients give A005878.
%K nonn
%O 0,4
%A _N. J. A. Sloane_, Apr 25 2004
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