A115226 Order of the group of invertible 3 X 3 symmetric matrices over Z(n).
1, 28, 468, 1792, 12400, 13104, 100548, 114688, 341172, 347200, 1609300, 838656, 4453488, 2815344, 5803200, 7340032, 22713088, 9552816, 44563284, 22220800, 47056464, 45060400, 141587908, 53673984, 193750000, 124697664, 248714388, 180182016, 574288624, 162489600
Offset: 1
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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Mathematica
Table[cnt=0; Do[m={{a, b, c}, {b, d, e}, {c, e, f}}; If[Det[m, Modulus->n]>0 && MatrixQ[Inverse[m, Modulus->n]], cnt++ ], {a, 0, n-1}, {b, 0, n-1}, {c, 0, n-1}, {d, 0, n-1}, {e, 0, n-1}, {f, 0, n-1}]; cnt, {n, 2, 20}] f[p_, e_] := p^(6*e - 4)*(p^3 - 1)*(p - 1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 10 2020 *)
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PARI
a(n) = {my(f = factor(n)); prod(i = 1, #f~, f[i,1]^(6*f[i,2] - 4)*(f[i,1]^3 - 1)*(f[i,1] - 1));} \\ Amiram Eldar, Nov 05 2022
Formula
For prime p, a(p) = (p^3-1)*(p-1)*p^2.
Multiplicative with a(p^e) = p^(6*e - 4)*(p^3 - 1)*(p - 1). - Amiram Eldar, Sep 10 2020
Sum_{k>=1} 1/a(k) = Product_{primes p} (1 + p^4/((p-1)^3 * (p^2+p+1)^2 * (p^3+1))) = 1.03859354030263389220782701124174403591851545785245128014455467710993780757... - Vaclav Kotesovec, Sep 20 2020
Sum_{k=1..n} a(k) ~ c * n^7, where c = (1/7) * Product_{p prime} (1 - 1/p^2 - 1/p^4 + 1/p^5) = 0.08230753362... . - Amiram Eldar, Nov 05 2022
Extensions
More terms from Amiram Eldar, Sep 10 2020
Comments