



2, 3, 4, 7, 8, 11, 12, 17, 30, 39, 52, 59
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OFFSET

1,1


COMMENTS

Companion sequence of A328420.
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) > a(n) seems impossible.


LINKS

Table of n, a(n) for n=1..12.
Wikipedia, Multiplicative group of integers modulo n


EXAMPLE

Let (Z/mZ)* be the multiplicative group of integers modulo m. We have (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 A317993(104) = 8 is a term.


PROG

(PARI) b(n) = if(abs(n)==1abs(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(b(k), ", "); t=b(k))) \\ Warning: program runs for about 30 min


CROSSREFS

Cf. A317993, A328420.
Sequence in context: A342028 A284937 A271441 * A003508 A225229 A239389
Adjacent sequences: A328418 A328419 A328420 * A328422 A328423 A328424


KEYWORD

nonn,hard,more


AUTHOR

Jianing Song, Oct 14 2019


STATUS

approved



