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A376349
Number of isomorphism classes k of groups G of order p*2^n when G contains a unique Sylow p subgroup and the maximal 2^m dividing p-1 is such that 2^m >= 2^n.
1
1, 2, 5, 15, 54, 247, 1684, 21820, 1118964
OFFSET
0,2
COMMENTS
A Sylow p subgroup is a subgroup of order p^r that necessarily exists when r is a maximal power of p. It is not necessarily unique, but when it is unique it is normal in G.
The condition that G of order p*2^n contains a unique Sylow p subgroup places an upper bound on the number of isomorphism classes of G; it is equivalent to stating that the minimal 2^r such that 2^r == 1 (mod p) be such that 2^r > 2^n. The condition that the maximal 2^m dividing p-1, i.e. for p == 1 (mod 2^m), is such that 2^m >= 2^n ensures a lower bound which is equal to the upper bound. See the Miles Englezou link for a proof.
If we relax the two conditions and just consider an arbitrary odd prime p and the number of isomorphism classes for |G| = p*2^n, it is likely that the set of such numbers is unique to p. Since every odd prime has a minimal 2^r such that 2^r == 1 (mod p) (a consequence of Fermat's little theorem), when 2^r = 2^n for |G| = p*2^n, the number of isomorphism classes will differ from a(n) due to the existence of groups where the Sylow p subgroup is not unique.
LINKS
FORMULA
a(n) = A000001(p*2^(n)) for every p satisfying the two conditions mentioned in Comments.
EXAMPLE
a(2) = 5 since D_(p*2^2), C_(p*2^2), C_(p*2^1) x C_2, and two semidirect products C_p : C_4 are all the groups of order p*2^2 for p satisfying the two conditions.
Table showing minimal 2^r and maximal 2^m (as defined in the Comments) for some primes:
---------------------------------------------------------------------------
p | Minimal 2^r == 1 (mod p) | Maximal 2^m, p == 1 (mod 2^m) |
---------------------------------------------------------------------------
2 | 2^0 = 1 | 2^0 = 1 |
3 | 2^2 = 4 | 2^1 = 2 |
5 | 2^4 = 16 | 2^2 = 4 |
7 | 2^3 = 8 | 2^1 = 2 |
11| 2^10 = 1024 | 2^1 = 2 |
13| 2^12 = 4096 | 2^2 = 4 |
17| 2^8 = 256 | 2^4 = 16 |
19| 2^18 = 262144 | 2^1 = 2 |
23| 2^11 = 2048 | 2^1 = 2 |
29| 2^28 = 268435456 | 2^2 = 4 |
31| 2^5 = 32 | 2^1 = 2 |
37| 2^36 = 68719476736 | 2^2 = 4 |
---------------------------------------------------------------------------
Table of primes satisfying 2^r > 2^n, and 2^m >= 2^n:
-------------------------------------------------------------------------------
2^n | primes | a(n) |
-------------------------------------------------------------------------------
2^0 = 1 | all primes = A000040 | 1 |
2^1 = 2 | all primes > 2 = A065091 | 2 |
2^2 = 4 | 5, 13, 17, 29, 37, 41, 53, ... = A002144 | 5 |
2^3 = 8 | 17, 41, 73, 89, 97, 113, 137, ... = A007519 | 15 |
2^4 = 16 | 17, 97, 113, 193, 241, 257, 337 ... = A094407 | 54 |
2^5 = 32 | 97, 193, 257, 353, 449, 577, 641, ... = A133870 | 247 |
2^6 = 64 | 193, 257, 449, 577, 641, 769, 1153, ... = A142925 | 1684 |
2^7 = 128| 257, 641, 769, 1153, 1409, 2689, 3329, ... = A208177 | 21820 |
2^8 = 256| 257, 769, 3329, 7937, 9473, 14081, 14593 ... = A105131 | 1118964 |
-------------------------------------------------------------------------------
PROG
(GAP)
S:=[];
for i in [0..8] do
n:=7681*2^i; # 7681 is an appropriate prime for reproducing up to a(8)
S:=Concatenation(S, [NrSmallGroups(n)]);
od;
Print(S);
KEYWORD
nonn,more,hard
AUTHOR
Miles Englezou, Sep 19 2024
STATUS
approved