Define "rarer"? There's an infinite number of integers that can be written in consecutive powers of 2. So in some sense, there's exactly as many of those as there are of integers!
For fun, see if you can write a bijection between the integers and the "consecutive ones". Hint: realize that a "consecutive power of 2" number can be split into two numbers: a number of consecutive ones greater than zero followed by a number of consecutive zeroes (you can have zero of these), then think about diagonalization.
That said, I suppose you could define some sort of "density" of numbers with consecutive ones in a fixed-size range. Say that you want to know how many of them are between 0 and 2^x (i.e. how many such numbers are x bits long). Well, you get sum(1..x-1) different consecutive numbers (for 2^8, there are 7 ways to have a pair of ones, 6 ways to have 3 consecutive ones, as you can see by imagining sliding the pair or triplet: 11100000, 01110000, 00111000, ... 00000111) and there are 2^x possibilities total.
Add x-1 + x-2 + ... + 2 + 1 to itself backwards and you get:
(x-1 + 1) + (x-2 + 2) + ... where there are x-1 terms in the series, which allows us to rewrite it as x * (x-1). This is double the original sum (because we added it to itself), so divide it by two and we've shown that sum(1..x-1) == (x^2 - x)/2. Now, divide that by 2^x and simplify to get D(x) = (x^2 - x)/2^(x+1) for the fraction of x-bit numbers that are consecutive powers of two.
It doesn't take much analysis to see that it is decreasing when x grows after increasing initially:
Density of consecutive ones in 1-bit numbers = 0
Density of consecutive ones in 2-bit numbers = 0.25
Density of consecutive ones in 3-bit numbers = 0.375
Density of consecutive ones in 4-bit numbers = 0.375
Density of consecutive ones in 5-bit numbers = 0.3125
Density of consecutive ones in 6-bit numbers = 0.234375
Density of consecutive ones in 7-bit numbers = 0.1640625
Density of consecutive ones in 8-bit numbers = 0.109375
In short, "much rarer" depends on how big a number we're dealing with and the measure won't work for infinitely large numbers, because there are a countably infinite number of consecutive-power numbers, so they can be put into 1-to-1 correspondence with the integers, even though our D(x) decreases.
TL;DR: Math is crazy and relies on precise definitions. Be wary of intuition.
For fun, see if you can write a bijection between the integers and the "consecutive ones". Hint: realize that a "consecutive power of 2" number can be split into two numbers: a number of consecutive ones greater than zero followed by a number of consecutive zeroes (you can have zero of these), then think about diagonalization.
That said, I suppose you could define some sort of "density" of numbers with consecutive ones in a fixed-size range. Say that you want to know how many of them are between 0 and 2^x (i.e. how many such numbers are x bits long). Well, you get sum(1..x-1) different consecutive numbers (for 2^8, there are 7 ways to have a pair of ones, 6 ways to have 3 consecutive ones, as you can see by imagining sliding the pair or triplet: 11100000, 01110000, 00111000, ... 00000111) and there are 2^x possibilities total.
Add x-1 + x-2 + ... + 2 + 1 to itself backwards and you get: (x-1 + 1) + (x-2 + 2) + ... where there are x-1 terms in the series, which allows us to rewrite it as x * (x-1). This is double the original sum (because we added it to itself), so divide it by two and we've shown that sum(1..x-1) == (x^2 - x)/2. Now, divide that by 2^x and simplify to get D(x) = (x^2 - x)/2^(x+1) for the fraction of x-bit numbers that are consecutive powers of two.
It doesn't take much analysis to see that it is decreasing when x grows after increasing initially:
Density of consecutive ones in 1-bit numbers = 0 Density of consecutive ones in 2-bit numbers = 0.25 Density of consecutive ones in 3-bit numbers = 0.375 Density of consecutive ones in 4-bit numbers = 0.375 Density of consecutive ones in 5-bit numbers = 0.3125 Density of consecutive ones in 6-bit numbers = 0.234375 Density of consecutive ones in 7-bit numbers = 0.1640625 Density of consecutive ones in 8-bit numbers = 0.109375
In short, "much rarer" depends on how big a number we're dealing with and the measure won't work for infinitely large numbers, because there are a countably infinite number of consecutive-power numbers, so they can be put into 1-to-1 correspondence with the integers, even though our D(x) decreases.
TL;DR: Math is crazy and relies on precise definitions. Be wary of intuition.