... but I couldn't imagine the number would be doubled or tripled at any stage

It's fairly easily demonstrated. The string returned from scalar %hash is the no-of-buckets-used/no-of-buckets-allocated.

$buckets = %h; 
for ('aaaa'..'zzzz'){ 
    $h{$_}=undef; 
    ## Remove the conditional on the next line to see hat no bucket gr
+owth 
    ## occurs when an addition hashes to an existing value.
    print $buckets = %h if %h ne $buckets; 
};;

1/8
2/8
3/8
4/8
5/8        ##62%
6/16
7/16
8/16
9/16
10/16      ## 62%
15/32
16/32
17/32
18/32
19/32
20/32
21/32      ## 65%
25/64
26/64
27/64
28/64
29/64
30/64
31/64
32/64
33/64
34/64
35/64
36/64
37/64
38/64
39/64
40/64
41/64    ## 64%
51/128
52/128
53/128
54/128
55/128
56/128
57/128
58/128
59/128
60/128
61/128
62/128
63/128
64/128
65/128
66/128
67/128
68/128
69/128
70/128
71/128
72/128
73/128
74/128
75/128
76/128
77/128
78/128
79/128
80/128
81/128
82/128
83/128
84/128
85/128   ## 66%
106/256
[download]

Note that scalar %hash on a tied hash may do nothing or something completely different.