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A270120 Number of k with k^n=1 (mod n) and k^k=k (mod n); related to some groups of order n 1


%S 0,1,1,2,1,2,1,4,3,2,1,4,1,2,1,6,1,4,1,6,3,2,1,8,5,2,3,4,1,4,1,6,1,2,

%T 1,8,1,2,3,12,1,8,1,4,3,2,1,12,7,6,1,6,1,4,5,8,3,2,1,12,1,2,9,10,1,4,

%U 1,6,1,4,1,16,1,2,5,4,1,8,1,20,9,2,1,16,1,2,1,8,1,8,1,4,3,2,1,12,1,8,3,14

%N Number of k with k^n=1 (mod n) and k^k=k (mod n); related to some groups of order n

%C Given integers n and k, consider the operation

%C o_k: Z_n x Z_n -> Z_n, (a, b) -> a + k^a * b (mod n).

%C (Z_n, o_k) is a group if k^n == 1 (mod n) and k^k == k (mod n).

%C The first condition is necessary to get the definition well-defined.

%C The second condition is necessary for the associative property.

%C a(n) gives the number of different k out of {1, 2, ..., n-1} that comply the conditions.

%C E.g., for n = 4, k = -1 (or, what is the same, k = 3) results in the Klein four-group. (a o_3 b := a + (-1)^a * b (mod 4).)

%C Note that different k can result in groups that are isomorphic to each other.

%C The neutral element is always 0.

%C The inverse element to a is always -a*k^(-a) (mod n).

%H Antti Karttunen, <a href="/A270120/b270120.txt">Table of n, a(n) for n = 1..65537</a>

%H Alfred Heiligenbrunner, <a href="/A270120/a270120_1.txt">A270120 Further examples and how the small groups were named</a>

%H OEIS Wiki, <a href="http://oeis.org/wiki/Number_of_groups_of_order_n">OEIS Wiki Groups of order n</a>

%e a(4) = 2, because in Z_4, k == 1 and k == 3 are the only number out of {0, 1, 2, 3} with conditions k^k==k mod n and k^n==1 mod n.

%e a(8) = 4, because k can be out of {1, 3, 5, 7}.

%e a(18) = 4, because k can be out of {1, 7, 13, 17}.

%e If n is even, k == -1 (or, equivalently, k == n-1) is always to be counted. This group is isomorphic to the Dihedral group D_(n/2), with generating elements -1 and 2.

%e The following table shows the first results with n, k and the name of the group (due to A. D. Thomas and G. V. Wood: 'Group Tables', found by comparing the element-orders).

%e Note that for n=8, k=1 and k=5 result in Z8. None of the k results in Z2 x Z4 or in Z2 x Z2 x Z2.

%e Note that for n=9 all k are isomorphic to Z9, none to Z3 x Z3.

%e n=2, k=1: Z2

%e n=3, k=1: Z3

%e n=4, k=1: Z4

%e n=4, k=3: Z2 x Z2

%e n=5, k=1: Z5

%e n=6, k=1: Z6

%e n=6, k=5: D3

%e n=7, k=1: Z7

%e n=8, k=1: Z8

%e n=8, k=3: Q4

%e n=8, k=5: Z8

%e n=8, k=7: D4

%e n=9, k=1: Z9

%e n=9, k=4: Z9

%e n=9, k=7: Z9

%e n=10, k=1: Z10

%e n=10, k=9: D5

%e ...

%t Table[Length[ Select[Range[1, n-1], ((GCD[n, # - 1] > 1) && (PowerMod[#, n, n] == 1) && (PowerMod[#, # - 1, n] == 1)) &]], {n, 1, 100}]

%o (PARI) a(n) = sum(k=1, n-1, (Mod(k,n)^n == 1) && (Mod(k,n)^k == k)); \\ _Michel Marcus_, Mar 12 2016

%Y Cf. A000001 (number of groups of order n).

%K nonn,easy

%O 1,4

%A _Alfred Heiligenbrunner_, Mar 11 2016

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Last modified April 20 20:26 EDT 2021. Contains 343137 sequences. (Running on oeis4.)