A380578 Number of nonisomorphic groups appearing as the group of units of the ring Z/kZ for every k such that phi(k) = n.

1, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 2, 0, 0, 0, 3, 0, 1, 0, 2, 0, 1, 0, 2, 0, 0, 0, 1, 0, 1, 0, 4, 0, 0, 0, 3, 0, 0, 0, 3, 0, 1, 0, 1, 0, 1, 0, 3, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 2, 0, 0, 0, 5, 0, 1, 0, 0, 0, 1, 0, 5, 0, 0, 0, 0, 0, 1, 0, 3, 0, 1, 0, 1, 0, 0, 0, 3, 0, 0, 0

Offset: 1

Author

Miles Englezou, Mar 26 2025

Keywords

• nonn

Comments

Every group of units is abelian.

Examples

			a(4) = 2 because of the 4 distinct k such that phi(k) = 4 there are 2 nonisomorphic group of units Z/kZ*: C_4, and C_2 x C_2.
a(40) = 3 because of the 9 distinct k such that phi(k) = 40 there are 3 nonisomorphic group of units Z/kZ*: C_40, C_20 x C_2, and C_10 x C_2 x C_2.
a(41) = 0 because there are no k such that phi(k) = 41.
		

Crossrefs

Cf. A000010, A000688, A005277, A007617, A014197, A049283, A057635.

Programs

PARI

groupcount(n) = b=[]; if(n==1, b=concat(b,2), forstep(k=floor(exp(Euler)*n*log(log(n^2))+2.5*n/log(log(n^2))), n, -1, if(eulerphi(k)==n, b=concat(b,k)); if(k==n, b=concat(b,0)))); Z=[]; if(istotient(n)==0, return(0), for(m=2, b[1], if(eulerphi(m)<>n, next, W=[]; U=[]; D=divisors(eulerphi(m)); lambda=lcm(znstar(m)[2]); for(k=1, m-1, if(gcd(k,m)==1, U=concat(U, k))); for(j=1, length(D), if(D[j]>lambda, break); S=[]; for(r=1, eulerphi(m), if(znorder(Mod(U[r], m))==D[j], S=concat(S, U[r]))); W=concat(W, length(S)))); Z=concat(Z,[W]); Z=Set(Z)); return(length(Z)))

Formula

a(n) <= A014197(n).

a(n) = 0 for every n belonging to A007617.