A242090 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).

0, 0, 0, 1, 5, 8, 16, 21, 33, 56, 65, 96, 120, 133, 161, 208, 261, 280, 341, 385, 408, 481, 533, 616, 736, 800, 833, 901, 936, 1008, 1281, 1365, 1496, 1541, 1776, 1825, 1976, 2133, 2241, 2408, 2581, 2640, 2945, 3008, 3136, 3201, 3605, 4033, 4181, 4256

Offset

1,5

Comments

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

Apparently a(n) = A030006(n) for 3 <= n <= 1000. - Georg Fischer, Oct 23 2018

Confirmed a(n) = A030006(n) for 3 <= n <= 4000. - Fausto A. C. Cariboni, Feb 23 2019

Links

Fausto A. C. Cariboni, Table of n, a(n) for n = 1..4000

Steven J. Miller, Combinatorial and Additive Number Theory Problem Sessions, arXiv:1406.3558 [math.NT], 2014-2018. See Nathan Kaplan's Problem 2014.1.4 on p. 30.

Formula

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

Example

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.

Mathematica

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

Crossrefs

Cf. A030006, A242089.

Sequence in context: A126695 A365316 A001082 * A030006 A229849 A357276

Adjacent sequences: A242087 A242088 A242089 * A242091 A242092 A242093

Keyword

nonn

Author

Jonathan Sondow, Jun 16 2014

Extensions

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

Status

approved