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Number of triples (a,b,c) with 0 < a < b < c < p and a + b + c == 0 mod p, where 2*b < p = prime(n).
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%I #21 Feb 23 2019 14:48:43

%S 0,0,0,1,5,8,16,21,33,56,65,96,120,133,161,208,261,280,341,385,408,

%T 481,533,616,736,800,833,901,936,1008,1281,1365,1496,1541,1776,1825,

%U 1976,2133,2241,2408,2581,2640,2945,3008,3136,3201,3605,4033,4181,4256

%N Number of triples (a,b,c) with 0 < a < b < c < p and a + b + c == 0 mod p, where 2*b < p = prime(n).

%C Given a + b < 2*b < p and c < p it follows that a + b + c < 2*p then the condition reduces to a + b + c = p. - _Fausto A. C. Cariboni_, Sep 30 2018

%C Apparently a(n) = A030006(n) for 3 <= n <= 1000. - _Georg Fischer_, Oct 23 2018

%C Confirmed a(n) = A030006(n) for 3 <= n <= 4000. - _Fausto A. C. Cariboni_, Feb 23 2019

%H Fausto A. C. Cariboni, <a href="/A242090/b242090.txt">Table of n, a(n) for n = 1..4000</a>

%H Steven J. Miller, <a href="https://arxiv.org/abs/1406.3558">Combinatorial and Additive Number Theory Problem Sessions</a>, arXiv:1406.3558 [math.NT], 2014-2018. See Nathan Kaplan's Problem 2014.1.4 on p. 30.

%F a(n) = (1/2)*A242089(n).

%e For prime(4) = 7 there is 1 triple (a,b,c) with 0 < a < b < c < 7 and a+b+c == 0 mod 7, namely, 1+2+4 = 7, so a(4) = 1.

%t Table[(1/2) Length[ Reduce[ Mod[a + b + c, Prime[n]] == 0 && 0 < a < b < c < Prime[n], {a, b, c}, Integers]], {n, 40}]

%Y Cf. A030006, A242089.

%K nonn

%O 1,5

%A _Jonathan Sondow_, Jun 16 2014

%E a(41)-a(50) from _Fausto A. C. Cariboni_, Sep 30 2018