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