A011785 Number of 3 X 3 matrices whose determinant is 1 mod n.
1, 168, 5616, 43008, 372000, 943488, 5630688, 11010048, 36846576, 62496000, 212427600, 241532928, 810534816, 945955584, 2089152000, 2818572288, 6950204928, 6190224768, 16934047920, 15998976000, 31621943808, 35687836800
Offset: 1
Links
- T. D. Noe, Table of n, a(n) for n = 1..1000
- Geoffrey Critzer, Combinatorics of Vector Spaces over Finite Fields, Master's thesis, Emporia State University, 2018.
Crossrefs
Programs
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Mathematica
a[n_] := (n^9*Times @@ Function[p, (1 - 1/p^3)*(1 - 1/p^2)*(1 - 1/p)] /@ FactorInteger[n][[All, 1]])/EulerPhi[n]; a[1] = 1; Array[a, 30] (* Jean-François Alcover, Mar 21 2017 *)
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PARI
a(n) = n^9*prod(k=2, n, if (!isprime(k) || (n % k), 1, (1-1/k^3)*(1-1/k^2)*(1-1/k)))/eulerphi(n); \\ Michel Marcus, Jun 30 2015
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Python
from math import prod from sympy import factorint def A011785(n): return prod(p**((e<<3)-5)*(p**2*(p*(p-1)*(p+1)-1)+1) for p,e in factorint(n).items()) # Chai Wah Wu, Mar 04 2025
Formula
Multiplicative with a(p^e) = p^(8*e-5)*(p^3 - 1)*(p^2 - 1). - Vladeta Jovovic, Nov 18 2001
For a formula see A064767.
a(n) = A064767(n)/phi(n). - Jianing Song, Nov 24 2018
Sum_{k>=1} 1/a(k) = Product_{primes p} (1 + p^5/((p-1)^3 * (p+1)^2 * (p^2 + p + 1) * (p^6 + p^4 + p^2 + 1))) = 1.0061577672748872278355775942508642214184417621389767880397578015151659965... - Vaclav Kotesovec, Sep 19 2020
Sum_{k=1..n} a(k) ~ c * n^9, where c = (1/9) * Product_{p prime} (1 - (p^3 + p^2 -1)/p^6) = 0.08630488937... . - Amiram Eldar, Oct 23 2022
Extensions
More terms from John W. Layman, Feb 16 2001
Further terms from Vladeta Jovovic, Oct 29 2001
Comments