A328420 Numbers k such that A317993(k) sets a new record; numbers k such that (Z/mZ)* = (Z/kZ)* has more solutions for m than all k' < k, where (Z/mZ)* is the multiplicative group of integers modulo m.

1, 3, 7, 35, 104, 143, 371, 385, 2233, 6149, 16555, 17081

Offset: 1

Author

Jianing Song, Oct 14 2019

Keywords

• nonn
• hard
• more

Comments

It seems that this sequence is infinite (i.e., A317993 is unbounded), but for each n, to really construct a number k such that A317993(k) > A317993(a(n)) seems impossible.

Examples

			For k = 104: (Z/mZ)* = (Z/104Z)* has 8 solutions, namely m = 104, 105, 112, 140, 144, 156, 180, 210; for all k' < 104, (Z/mZ)* = (Z/k'Z)* has fewer than 8 solutions. So 104 is a term.
		

Links

• Wikipedia, Multiplicative group of integers modulo n

Crossrefs

Cf. A317993, A328421.

Programs

PARI

b(n) = if(abs(n)==1||abs(n)==2, 2, my(i=0, k=eulerphi(n), N=floor(exp(Euler)*k*log(log(k^2))+2.5*k/log(log(k^2)))); for(j=k+1, N, if(znstar(j)[2]==znstar(n)[2], i++)); i)
my(t=0); for(k=1, 20000, if(b(k)>t, print1(k, ", "); t=b(k))) \\ Warning: program runs for about 30 min