OFFSET
1,1
COMMENTS
All terms are even because (Z/kZ)* is isomorphic to (Z/(2k)Z)* for odd k. Most terms are congruent to 2 modulo 4. Among the first 10000 terms there are 8980 ones congruent to 2 modulo 4.
LINKS
Jianing Song, Table of n, a(n) for n = 1..10000
Wikipedia, Multiplicative group of integers modulo n
FORMULA
n <= a(n) <= A028476(n).
EXAMPLE
The solutions to (Z/kZ)* = C_6 are k = 7, 9, 14 and 18, so a(7) = a(9) = a(14) = a(18) = 18.
The solutions to (Z/kZ)* = C_2 X C_20 are k = 55, 75, 100, 110 and 150, so a(55) = a(75) = a(100) = a(110) = a(150) = 150.
The solutions to (Z/kZ)* = C_2 X C_12 are k = 35, 39, 45, 52, 70, 78 and 90, so a(35) = a(39) = a(45) = a(52) = a(70) = a(78) = a(90) = 90.
PROG
CROSSREFS
KEYWORD
nonn
AUTHOR
Jianing Song, Oct 04 2018
STATUS
approved