Assume a process which repeatedly marks all of positive integers which belong to a Fibonacci sequence which start with the lowest unmarked x and x+c. For example, if c is one, the first Fibonacci sequence marks 1, 2, 3, 5, 8, 13, 21, etc. The process would start the second Fibonacci sequence at 6 and 7, marking thereafter 13, 20, 33, ..., the third one would mark 9, 10, 19, 29, etc. The unmarked numbers 4, 18, 22, 28, 32, 47, 54, 72, etc. are then what I call the anti-Fibonacci numbers for the given c = 1. For c = 2 the anti-Fibonacci numbers would be 5, 8, 9, 24, 34, 38, 45, 50, etc.One might assume that anti-Fibonacci number generation is a sufficiently complicated process to be essentially pseudo-random. My back-of-the-envelope calculation went as follows: Let p(i) be the probability that i is anti-Fibonacci. A single existing Fibonacci sequence grows sparser with a rate of phi/i where phi is the base of the golden ratio (sqrt(5) + 1) / 2 with which the Fibonacci values grow, but with a probability of p(i)^2 the i and i+c starts a new Fibonacci sequence and p(i) gets a nudge of 2/i upwards. Putting these in equilibrium and solving for p gives sqrt(sqrt(5) + 1) / 2, or roughly 0.899. I knew I had made some lofty approximations in the derivation above, especially in assuming that the density of multiple Fibonacci sequences would decrease as rapidly as that of one sequence, so I wrote a little program that computes some 100 million first anti-Fibonacci numbers. I was somewhat surprized to see that p is only approximately 0.856 for c = 1. For c = 2 the program gave 0.872, which as again surprizing because I didn't expect c to affect p. But when my computer spat out what looks like interference patterns my jaw really dropped.Having seen the non-randomness of the average of p over all i and observed how it depends on c, a change in coordinates lead me to wonder if conversely the average of p over c's depends on i. The answer is yes. Below I have a sample of p(i) estimated by c up to ten thousand.The moral of the story? Whenever playing with number theory, staple your jaw so it won't keep falling off constantly.