A115076 Number of 2 X 2 symmetric matrices over Z(n) having determinant 1.

1, 4, 6, 12, 30, 24, 42, 48, 54, 120, 110, 72, 182, 168, 180, 192, 306, 216, 342, 360, 252, 440, 506, 288, 750, 728, 486, 504, 870, 720, 930, 768, 660, 1224, 1260, 648, 1406, 1368, 1092, 1440, 1722, 1008, 1806, 1320, 1620, 2024, 2162, 1152, 2058, 3000

Offset

1,2

Comments

a(1)=1 because the matrix of all zeros has determinant 0, but 0=1 (mod 1).

Links

Andrew Howroyd, Table of n, a(n) for n = 1..2500

Formula

Multiplicative with a(2^1) = 4, a(2^e) = 3*2^(2*e-2) for e > 1, a(p^e) = (p+1)*p^(2*e-1) for p mod 4 == 1, a(p^e) = (p-1)*p^(2*e-1) for p mod 4 == 3. - Andrew Howroyd, Jul 04 2018

Sum_{k=1..n} a(k) ~ c * n^3, where c = (5/(2*Pi^2)) * A175647 * A243380 = 0.282098596071... . - Amiram Eldar, Aug 28 2023

Mathematica

Table[cnt=0; Do[m={{a, b}, {b, c}}; If[Det[m, Modulus->n]==1, cnt++ ], {a, 0, n-1}, {b, 0, n-1}, {c, 0, n-1}]; cnt, {n, 50}]
f[p_, e_] := If[Mod[p, 4] == 1, (p+1)*p^(2*e-1), (p-1)*p^(2*e-1)]; f[2, 1] = 4; f[2, e_] := 3*2^(2*e-2); a[n_] := Times @@ f @@@ FactorInteger[n]; a[1] = 1; Array[a, 100] (* Amiram Eldar, Aug 28 2023 *)

Prog

(PARI) a(n)={my(v=vector(n)); for(i=0, n-1, for(j=0, n-1, v[i*j%n+1]++)); sum(i=0, n-1, v[(i^2+1)%n+1])} \\ Andrew Howroyd, Jul 04 2018
(PARI) a(n)={my(f=factor(n)); prod(i=1, #f~, my(p=f[i, 1], e=f[i, 2]); p^(2*e-1)*if(p==2, if(e==1, 2, 3/2), if(p%4==1, p+1, p-1)))} \\ Andrew Howroyd, Jul 04 2018

Crossrefs

Cf. A000056 (order of the group SL(2, Z_n)), A175647, A243380.

Sequence in context: A215177 A294489 A344995 * A126259 A068570 A092320

Adjacent sequences: A115073 A115074 A115075 * A115077 A115078 A115079

Keyword

mult,nonn,easy

Author

T. D. Noe, Jan 12 2006

Status

approved