Very Good!
I started out with the following 28 primes:
I chose these because the order(2,p) for these primes was a factor of 720. The order(2,p) is the least m such that 2^m == 1 (mod p). Note that m will never be greater than p-1 becasue of Fermat's Little TheoremPrime Order-(2,p)
3 2
5 4
7 3
11 10
13 12
17 8
19 18
31 5
37 36
41 20
61 60
73 9
97 48
109 36
151 15
181 180
241 24
257 16
331 30
433 72
577 144
631 45
673 48
1321 60
23311 45
38737 72
54001 180
61681 40
which gives us 2^(p-1) == 1 (mod p) for all primes greater than 2.Let p be a prime which does not divide the integer a, then a^p-1 == 1 (mod p).
But I think that I ended up with far fewer "holes" than you did:
I have attached the details, all n (mod720) for which a prime p could exist such that p | k*2^n+1 aka "holes". The numbers in red are those n values (mod 720) that are common to two k values. There are 14 common values so we have 216 distinct values of n (mod 720) instead of the 230 that you would expect by totaling up the column above.K Holes
4847 18
5359 36
10223 36
19249 9
21181 24
22699 8
24737 18
27653 15
28433 12
33661 18
55459 28
67607 8
I don't know if this affects your analysis or not. If anything, it may enable you to refine your analysis.
EDIT: I just reread your post and saw that you talked about the factors of k, so just for the record:
K Factors/Prime
4847 37*131
5359 23*233
10223 Prime
19249 Prime
21181 59*359
22699 Prime
24737 29*853
27653 Prime
28433 Prime
33661 41*821
55459 31*1789
67607 Prime