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

0, 0, 0, 2, 10, 16, 32, 42, 66, 112, 130, 192, 240, 266, 322, 416, 522, 560, 682, 770, 816, 962, 1066, 1232, 1472, 1600, 1666, 1802, 1872, 2016, 2562, 2730, 2992, 3082, 3552, 3650, 3952, 4266, 4482, 4816, 5162, 5280, 5890, 6016, 6272, 6402, 7210, 8066, 8362, 8512

Offset: 1

Author

Jonathan Sondow, Jun 16 2014

Keywords

• nonn
• easy

Comments

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.)

Examples

			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.
		

Links

• Fausto A. C. Cariboni, Table of n, a(n) for n = 1..1000
• Steven J. Miller, Combinatorial and Additive Number Theory Problem Sessions, arXiv:1406.3558 [math.NT], 2014-2023. See Nathan Kaplan's Problem 2014.1.4 on p. 30.

Crossrefs

Cf. A069905, A242090.

Programs

Mathematica

Table[ Length[ Reduce[ Mod[a + b + c, Prime[n]] == 0 && 0 < a < b < c < Prime[n], {a, b, c}, Integers]], {n, 40}]

PARI

a(n) = 2 * round((prime(n) - 3)^2/12) \\ David A. Corneth, May 27 2025

Formula

a(n) = 2*A242090(n).

a(n) = 2*A069905(prime(n)-3) = 2 * round((prime(n) - 3)^2/12). - David A. Corneth, May 27 2025

Extensions

a(41)-a(50) from Fausto A. C. Cariboni, Sep 30 2018