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A242089 Number of triples (a,b,c) with 0 < a < b < c < p and a + b + c == 0 mod p, where p = prime(n).

%I A242089 #16 May 28 2025 00:58:19
%S A242089 0,0,0,2,10,16,32,42,66,112,130,192,240,266,322,416,522,560,682,770,
%T A242089 816,962,1066,1232,1472,1600,1666,1802,1872,2016,2562,2730,2992,3082,
%U A242089 3552,3650,3952,4266,4482,4816,5162,5280,5890,6016,6272,6402,7210,8066,8362,8512
%N A242089 Number of triples (a,b,c) with 0 < a < b < c < p and a + b + c == 0 mod p, where p = prime(n).
%C A242089 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 A242089 Fausto A. C. Cariboni, <a href="/A242089/b242089.txt">Table of n, a(n) for n = 1..1000</a>
%H A242089 Steven J. Miller, <a href="https://arxiv.org/abs/1406.3558">Combinatorial and Additive Number Theory Problem Sessions</a>, arXiv:1406.3558 [math.NT], 2014-2023. See Nathan Kaplan's Problem 2014.1.4 on p. 30.
%F A242089 a(n) = 2*A242090(n).
%F A242089 a(n) = 2*A069905(prime(n)-3) = 2 * round((prime(n) - 3)^2/12). - _David A. Corneth_, May 27 2025
%e A242089 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 A242089 Table[ Length[ Reduce[ Mod[a + b + c, Prime[n]] == 0 && 0 < a < b < c < Prime[n], {a, b, c}, Integers]], {n, 40}]
%o A242089 (PARI) a(n) = 2 * round((prime(n) - 3)^2/12) \\ _David A. Corneth_, May 27 2025
%Y A242089 Cf. A069905, A242090.
%K A242089 nonn,easy
%O A242089 1,4
%A A242089 _Jonathan Sondow_, Jun 16 2014
%E A242089 a(41)-a(50) from _Fausto A. C. Cariboni_, Sep 30 2018