A320045 Smallest k such that (Z/kZ)* is isomorphic to (Z/nZ)*.

1, 1, 3, 3, 5, 3, 7, 8, 7, 5, 11, 8, 13, 7, 15, 15, 17, 7, 19, 15, 21, 11, 23, 24, 25, 13, 19, 21, 29, 15, 31, 32, 33, 17, 35, 21, 37, 19, 35, 40, 41, 21, 43, 33, 35, 23, 47, 40, 43, 25, 51, 35, 53, 19, 55, 56, 57, 29, 59, 40, 61, 31, 63, 51, 65, 33, 67, 51, 69, 35

Offset: 1

Author

Jianing Song, Oct 04 2018

Keywords

• nonn

Comments

a(n) = n iff n is a term in A296233.

Most terms are odd. Among the first 10000 terms there are 8837 odd ones. The even terms are divisible by 4 because (Z/kZ)* is isomorphic to (Z/(2k)Z)* for odd k.

A015126(n) <= a(n) <= n.

Examples

			The solutions to (Z/kZ)* = C_6 are k = 7, 9, 14 and 18, so a(7) = a(9) = a(14) = a(18) = 7.
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) = 55.
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) = 35.
		

Links

• Jianing Song, Table of n, a(n) for n = 1..10000
• Wikipedia, Multiplicative group of integers modulo n

Crossrefs

Cf. A015126, A296233, A320046.

Programs

PARI

a(n) = my(i=eulerphi(n)); while(znstar(i)[2]!=znstar(n)[2], i++); i