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A115224 Number of 3 X 3 symmetric matrices over Z(n) having determinant 1.

%I A115224 #30 Sep 20 2020 03:49:25
%S A115224 1,28,234,896,3100,6552,16758,28672,56862,86800,160930,209664,371124,
%T A115224 469224,725400,917504,1419568,1592136,2475738,2777600,3921372,4506040,
%U A115224 6435814,6709248,9687500,10391472,13817466,15015168,20510308,20311200,28628190,29360128,37657620
%N A115224 Number of 3 X 3 symmetric matrices over Z(n) having determinant 1.
%H A115224 Amiram Eldar, <a href="/A115224/b115224.txt">Table of n, a(n) for n = 1..10000</a> (terms 1..4885 from Enrique Pérez Herrero)
%F A115224 a(1)=1 because the matrix of all zeros has determinant 0, but 0=1 (mod 1).
%F A115224 For prime p, a(p) = (p^3-1)*p^2.
%F A115224 Multiplicative with a(p^e) = (p^3-1)*p^(5e-3).
%F A115224 a(n) = A011785(n)/A000056(n).
%F A115224 a(n) = A059376(n^2)/n. - _Enrique Pérez Herrero_, Sep 14 2010
%F A115224 a(n) = n^2*A059376(n). Dirichlet g.f.: zeta(s-5)/zeta(s-2). - _R. J. Mathar_, Feb 27 2011
%F A115224 Sum_{k=1..n} a(k) ~ 15*n^6 / Pi^4. - _Vaclav Kotesovec_, Feb 07 2019
%F A115224 Sum_{k>=1} 1/a(k) = Product_{primes p} (1 + p^3/(1 - p^3 - p^5 + p^8)) = 1.04172462829914219180789244796430293454403616906393417764614215669994022537... - _Vaclav Kotesovec_, Sep 20 2020
%t A115224 Table[cnt=0; Do[m={{a, b, c}, {b, d, e}, {c, e, f}}; If[Det[m, Modulus->n]==1, 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}]
%t A115224 JordanTotient[n_,k_:1] := DivisorSum[n,#^k*MoebiusMu[n/# ]&]/;(n>0)&&IntegerQ[n]; A115224[n_IntegerQ] := JordanTotient[n^2,3]/n; Table[A115224[n], {n,100}] (* _Enrique Pérez Herrero_, Sep 14 2010 *)
%t A115224 f[p_, e_] := (p^3 - 1)*p^(5*e - 3); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 50] (* _Amiram Eldar_, Sep 15 2020 *)
%Y A115224 Cf. A000056 (order of the group SL(2, Z_n)), A011785 (number of 3 X 3 matrices whose determinant is 1 mod n, i.e. order of SL(3, Z_n)).
%K A115224 mult,nonn
%O A115224 1,2
%A A115224 _T. D. Noe_, Jan 16 2006