A317993 Number of k such that (Z/kZ)* is isomorphic to (Z/nZ)*, where (Z/nZ)* is the multiplicative group of integers modulo n.
2, 2, 3, 3, 2, 3, 4, 2, 4, 2, 2, 2, 2, 4, 4, 4, 2, 4, 4, 4, 4, 2, 2, 1, 2, 2, 4, 4, 2, 4, 2, 1, 3, 2, 7, 4, 2, 4, 7, 3, 2, 4, 4, 3, 7, 2, 2, 3, 4, 2, 4, 7, 2, 4, 5, 3, 4, 2, 2, 3, 2, 2, 2, 4, 2, 3, 2, 4, 3, 7, 2, 3, 2, 2, 5, 4, 7, 7, 2, 1, 2, 2, 2, 3, 2, 4, 3
Offset: 1
Keywords
Examples
The solutions to (Z/kZ)* = C_6 are k = 7, 9, 14 and 18, so a(7) = a(9) = a(14) = a(18) = 4. 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) = 5. 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) = 7.
Links
- Jianing Song, Table of n, a(n) for n = 1..20000
- Wikipedia, Multiplicative group of integers modulo n
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