%I #8 Oct 01 2018 03:33:33
%S 0,0,0,2,10,16,32,42,66,112,130,192,240,266,322,416,522,560,682,770,
%T 816,962,1066,1232,1472,1600,1666,1802,1872,2016,2562,2730,2992,3082,
%U 3552,3650,3952,4266,4482,4816,5162,5280,5890,6016,6272,6402,7210,8066,8362,8512
%N Number of triples (a,b,c) with 0 < a < b < c < p and a + b + c == 0 mod p, where p = prime(n).
%C a(n) is even. (Proof. Each triple (a,b,c) with b < p/2 pairs uniquely with a triple (a',b',c') = (p-c,p-b,p-a) with b' > p/2.)
%H Fausto A. C. Cariboni, <a href="/A242089/b242089.txt">Table of n, a(n) for n = 1..1000</a>
%H Steven J. Miller, <a href="http://arxiv.org/pdf/1406.3558.pdf">Combinatorial and Additive Number Theory Problem Sessions</a>, arXiv:1406.3558 [math.NT]. See Nathan Kaplan's Problem 2014.1.4 on p. 30.
%F a(n) = 2*A242090(n).
%e For prime(4) = 7 there are 2 triples (a,b,c) with 0 < a < b < c < 7 and a + b + c == 0 mod 7, namely, 1+2+4 = 7 and 3+5+6 = 2*7, so a(4) = 2.
%t Table[ Length[ Reduce[ Mod[a + b + c, Prime[n]] == 0 && 0 < a < b < c < Prime[n], {a, b, c}, Integers]], {n, 40}]
%Y Cf. A242090.
%K nonn
%O 1,4
%A _Jonathan Sondow_, Jun 16 2014
%E a(41)-a(50) from _Fausto A. C. Cariboni_, Sep 30 2018