OFFSET
1,2
COMMENTS
These are the numbers k with the least proportion of primitive roots mod k to residues mod k coprime to k, of all m < k.
Let U(k) be the group of units of the ring Z/kZ, and Gen(k) the set of distinct single element generators of U(k). Then a(n) is equivalently those k for which floor(|U(k)|/|Gen(k)|) is a record value for those k with cyclic U(k).
Some data:
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n a(n) log(a(n)) phi(a(n)) max prime(r)#|phi(a(n)) r
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1 1 0 1 0 0
2 3 1.10... 2 2 1
3 7 1.96... 2*3 6 2
4 211 5.35... 2*3*5*7 210 4
5 43891 10.69... 2*3*5*7*11*19 2310 5
6 300690391 19.52... 2*3*5*7*11*13*17*19*31 9699690 8
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For n > 1, is a(n) necessarily prime? Furthermore, is a(n) necessarily a prime such that phi(a(n)) is squarefree? Lastly, is every phi(a(n)) divisible by a maximal primorial prime(r)# of length r <= omega(phi(a(n))), such that if a maximal prime(s)# divides phi(a(m)) and n < m, then r < s?
Based on the behavior of log(a(n)), we may expect a(7) to be found in the vicinity of floor(exp(40)) = 235385266837019985.
EXAMPLE
There are 2 residues mod 3 coprime to 3, and only 1 is a primitive root. 3 is the least k for which the floor of the ratio is 2, and so a(2) = 3.
There are 210 residues mod 211 coprime to 211, and 48 are primitive roots. Floor(210/48) = 4, and 211 is the least k for which the floor of the ratio is 4, and so a(4) = 211.
PROG
(PARI)
S=[1]; for(n=1, 100000, if(#znstar(n).cyc>1, next); f=eulerphi; if(floor(f(n)/f(f(n)))>floor(f(S[length(S)])/f(f(S[length(S)]))), S=concat(S, n))); print(S)
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Miles Englezou, Oct 26 2024
STATUS
approved