A235384 Number of involutions in the group Aff(Z/nZ).
2, 4, 6, 6, 8, 8, 16, 10, 12, 12, 24, 14, 16, 24, 28, 18, 20, 20, 36, 32, 24, 24, 64, 26, 28, 28, 48, 30, 48, 32, 52, 48, 36, 48, 60, 38, 40, 56, 96, 42, 64, 44, 72, 60, 48, 48, 112, 50, 52, 72, 84, 54, 56, 72, 128, 80, 60, 60, 144, 62, 64, 80, 100, 84
Offset: 2
Examples
Since 18 = 2*3^2, we get a(18) = 2*(3^2+1) = 20. Since 120 = 2^3*3*5, we have a(120) = (4+2^2+2^3)*(3+1)*(5+1) = 384.
Links
- Alois P. Heinz, Table of n, a(n) for n = 2..10000
- K. K. A. Cunningham, Tom Edgar, A. G. Helminck, B. F. Jones, H. Oh, R. Schwell and J. F. Vasquez, On the Structure of Involutions and Symmetric Spaces of Dihedral Groups, Note di Mat., Volume 34, No. 2, 2014.
Programs
-
Maple
a:= n-> add(`if`(irem(k^2, n)=1, igcd(n, k+1), 0), k=1..n-1): seq(a(n), n=2..100); # Alois P. Heinz, Jan 20 2014
-
Mathematica
a[n_] := Sum[If[Mod[k^2, n] == 1, GCD[n, k+1], 0], {k, 1, n-1}]; Table[a[n], {n, 2, 100}] (* Jean-François Alcover, Mar 24 2014, after Alois P. Heinz *) f[p_, e_] := p^e + 1; f[2, 1] = 2; f[2, 2] = 6; f[2, e_] := 3*2^(e - 1) + 4; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100, 2] (* Amiram Eldar, Dec 05 2022 *)
-
PARI
A034448(n,f=factor(n))=factorback(vector(#f~,i,f[i,1]^f[i,2]+1)) a(n)=my(m=valuation(n,2)); if(m==0,1,m==1,2,m==2,6,4+3<<(m-1))*A034448(n>>m) \\ Charles R Greathouse IV, Jul 29 2016
-
Sage
def a(n): L=list(factor(n)) if L[0][0]==2: m=L[0][1] L.pop(0) else: m=0 order=prod([x[0]^x[1]+1 for x in L]) if m==1: order=2*order elif m==2: order=6*order elif m>=3: order=(4+2^(m-1)+2^m)*order return order [a(i) for i in [2..100]]
-
Sage
def b(n): sum = 0 for a in [x for x in range(n) if ((x^2) % n == 1)]: sum += gcd(a+1,n) return sum [b(i) for i in [2..100]]
Formula
Suppose n = 2^m*p_1^(r_1)*p_2^(r_2)*...*p_k^(r_k) where each p_i>2 is prime, r_i>0 for all i, and m>=0 is the prime factorization of n, then:
...a(n) = Product_{1<=i<=k} (p_i^(r_i)+1) if m=0,
...a(n) = 2*Product_{1<=i<=k} (p_i^(r_i)+1) if m=1,
...a(n) = 6*Product_{1<=i<=k} (p_i^(r_i)+1) if m=2,
...a(n) = (4+2^(m-1)+2^m)*Product_{1<=i<=k} (p_i^(r_i)+1) if m>=3.
a(n) = Sum_{a in row(n) of A228179} gcd(a+1,n).
Sum_{k=1..n} a(k) ~ c * n^2, where c = zeta(2)/(2*zeta(3)) = 0.684216... (A335005). - Amiram Eldar, Dec 05 2022
Comments