A279725 Number of 3 X 3 matrices having all terms in {0,1,...,n} with |det| = 1.

0, 168, 2022, 15090, 53160, 196962, 409956, 1096368, 2062140, 4070796, 6674010, 12603174, 18410352, 31642836, 45306438, 67301682, 93747984, 142196892, 183799392, 267038772, 342684960, 458663640, 582535842, 793793994, 963867732, 1266864846, 1550198598, 1957887150, 2357651670, 3015489714

Offset: 0

Author

Indranil Ghosh, Jan 04 2017

Keywords

• nonn

Comments

a(n) is always even.

a(n) mod 6 = 0.

Examples

			For n=2, a few of the possible matrices are [0,0,1,0,1,0,1,0,0], [0,0,1,0,1,0,1,0,1], [0,0,1,0,1,0,1,0,2], [1,0,0,0,1,1,2,0,1], [1,0,0,0,1,1,2,1,0], [1,0,0,0,1,1,2,1,2], [2,2,1,2,1,2,1,0,2], [2,2,1,2,1,2,1,1,0], [2,2,1,2,1,2,1,1,1], [2,2,1,2,1,2,1,2,0], .... There are 2022 possibilities.
Here each of the matrices is defined as M=[a,b,c,d,e,f,g,h,i] where a=M[1][1], b=M[1][2], c=M[1][3], d=M[2][1], e=M[2][2], f=M[2][3], g=M[3][1], h=M[3][2] and i=M[3][3].
So, for n=2, a(n)=2022.
		

Links

• Robin Visser, Table of n, a(n) for n = 0..35
• Eric Weisstein's World of Mathematics, Unimodular Matrix
• Wikipedia, Unimodular Matrix

Crossrefs

Cf. A210000.

Programs

Sage

import itertools
def a(n):
    ans, W = 0, itertools.product(range(n+1), repeat=9)
    for w in W:
        if abs(Matrix(ZZ, 3, 3, w).det())==1: ans += 1
    return ans  # Robin Visser, May 01 2025

Extensions

More terms from Robin Visser, May 01 2025