A320046 Largest k such that (Z/kZ)* is isomorphic to (Z/nZ)*.

2, 2, 6, 6, 10, 6, 18, 12, 18, 10, 22, 12, 26, 18, 30, 30, 34, 18, 54, 30, 42, 22, 46, 24, 50, 26, 54, 42, 58, 30, 62, 32, 66, 34, 90, 42, 74, 54, 90, 60, 82, 42, 98, 66, 90, 46, 94, 60, 98, 50, 102, 90, 106, 54, 150, 84, 114, 58, 118, 60, 122, 62, 126, 102, 130, 66, 134, 102, 138, 90

Offset: 1

Author

Jianing Song, Oct 04 2018

Keywords

• nonn

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.

Examples

			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.
		

Links

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

Crossrefs

Cf. A028476, A057635, A320045.

Programs

PARI

a(n) = if(abs(n)==1||abs(n)==2, 2, my(i=0, search_max = A057635(eulerphi(n))); for(j=eulerphi(n)+1, search_max, if(znstar(j)[2]==znstar(n)[2], i=j)); i) \\ search_max is the largest k such that phi(k) = phi(n). See A057635 for its program

Formula

n <= a(n) <= A028476(n).