Combinatorics inside of $GL(n,q)$ I'm studying conjugacy classes of subgroups of $GL(n,q)$ of the form $(\mathbb{Z}/p\mathbb{Z})^r$ where $q=p^d$ and $r$ is some non-negative integer.  I've been able to show that for $n=p=2$ and for $r=2$, there are $a_d$ such conjugacy classes, where $a_d$ is the sequence defined by the recurrence relation:
$$a_m=a_{m-1}+2a_{m-2}, \quad a_1=0, a_2=1$$
For $n=p=2$, and considering $r=3$, I am finding the following sequence (using Magma)
$$0,0,1,1,5,23,93,381,\ldots$$
and I would like to find a similar recurrence satisfied by these numbers.  The recurrence may not be linear as it is for the case $r=2$.  Can anybody help me find the relationship between the numbers I've listed here?  Thanks in advance.
 A: In principal, this is well-known, although some little tricks are helpful to make it tractable and conceptual, admittedly.
The $p$-Sylow subgroups of $G=GL(n,\mathbb F_q)$ with $q=p^d$ are all conjugate to the group $N$ of upper-triangular unipotent matrices, by counting the elements in $G$ and counting $N$. This is well-known, and not hard.
The normalizer of $N$ is $B$, the upper-triangular (not-necessarily-unipotent) matrices in $G$. Thus, the number of $p$-Sylow subgroups is the cardinality of $G/B$, that is, $\#G /\# B$... 
EDIT: and then how many subgroups of order smaller than the maximum? This I don't know off-hand, but/and is a question about $p$-groups of this particularly simple sort. The composition series is especially simple, but I'd need to work it out myself to approximate an answer to the general question... I would start from the point that every $p$-subgroup is a subgroup of a $p$-Sylow subgroup. Awkwardly, the normalizer of a $p$-Sylow subgroup is not at all the same as the normalizer of a $p^3$ (etc) subgroup... but, still, the way that smaller subgroups appear in the stratification of the derived/descending-series subgroups of the Sylow $p$-group of upper-triangular matrices should lend a teensy bit of order to the situation.
Clarify?
A: In Paul Garret's notation, subgroups of $N$ are conjugate in $G$ if and only if they are conjugate in $B$, so your question is equivalent to asking for the number of classes of subgroups of order $8$ in $B$. Now $B$ is the direct product of its centre, which is cyclic of order $2^n-1$, by a cyclic group of order $2^n-1$ that acts fixed point freely by conjugation on $N$.
The number of subgroups of order $8$ in $N$ is $n_d := (2^d-1)(2^d-2)(2^d-4)/168$. if $d$ is not a multiple of 3, then the $2^d-1$ cycle acts fixed-point-freely on these subgrous, so $a_d = n_d/(2^d-1)$. This agrees with your calculated numbers.
It is slightly more complicated when $3|d$, because then $7|(2^d-1)$ and some of the $n_d$ subgroups of order 8 are fixed by a subgroup of order 7. I think there are $(8^{d/3} - 1)/7$ fixed subgroups, which form a single orbit under the action of the $2^d-1$ cycles, so the answer in this case is 
$a_d = (n_d-(8^{d/3} - 1)/7)/(2^d-1)+1$.
A: using Derek's formula, and with Mathematica, I find
a(n)= 6 a(n-1)-8 a(n-2)+a(n-3)-6 a(n-4)+8 a(n-5) with initial values evidently
a(1)=a(2)=1 , a(3)=a(4)=1 and a(5)=5.
