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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
1, 3, 7, 35, 104, 143, 371, 385, 2233, 6149, 16555, 17081
OFFSET
1,2
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.
EXAMPLE
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.
PROG
(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
CROSSREFS
Sequence in context: A179115 A299300 A047907 * A336012 A212417 A145874
KEYWORD
nonn,hard,more
AUTHOR
Jianing Song, Oct 14 2019
STATUS
approved