A305186 Number of invertible 4 X 4 matrices mod n.
1, 20160, 24261120, 1321205760, 116064000000, 489104179200, 27811094169600, 86586540687360, 1044361663787520, 2339850240000000, 41393302251840000, 32053931488051200, 610296923230525440, 560671658459136000, 2815842631680000000, 5674535530486824960
Offset: 1
Links
- Jianing Song, Table of n, a(n) for n = 1..10000
- C. J. Hillar and D. L. Rhea, Automorphisms of finite abelian groups, arXiv:math/0605185 [math.GR], 2006.
- C. J. Hillar and D. L. Rhea, Automorphisms of finite abelian groups, Amer. Math. Monthly 114 (2007), no 10, 917-923.
- J. Overbey, W. Traves and J. Wojdylo, On the Keyspace of the Hill Cipher, Cryptologia, Vol. 29, Iss. 1, 2005.
Crossrefs
Programs
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Mathematica
{1}~Join~Array[#^16*Product[(1 - 1/p^4) (1 - 1/p^3) (1 - 1/p^2) (1 - 1/p), {p, FactorInteger[#][[All, 1]]}] &, 12, 2] (* Michael De Vlieger, May 27 2018 *)
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PARI
a(n)=my(f=factor(n)[, 1]); n^16*prod(i=1, #f, (1-1/f[i]^4)*(1-1/f[i]^3)*(1-1/f[i]^2)*(1-1/f[i]))
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Python
from math import prod from sympy import factorint def A305186(n): return prod(p**((e<<3)-5<<1)*(p*(p*(p**3*(p**3*(p*(p-1)-1)+2)-1)-1)+1) for p,e in factorint(n).items()) # Chai Wah Wu, Mar 04 2025
Formula
Multiplicative with a(p^e) = (p - 1)*(p^2 - 1)*(p^3 - 1)*(p^4 - 1)*p^(16*e-10).
a(n) = n^16*Product_{primes p dividing n} (1 - 1/p^4)*(1 - 1/p^3)*(1 - 1/p^2)*(1 - 1/p).
Sum_{k=1..n} a(k) ~ c * n^17, where c = (1/17) * Product_{p prime} ((p^11 - p^9 - p^8 + 2*p^5 - p^2 - p + 1)/p^11) = 0.02958150406... . - Amiram Eldar, Oct 23 2022
Comments