Blue numbers

Blue numbers have one of these two forms:
2CN+C, C>0, N>=C (e.g. 2N+1, 4N+2, 6N+3)
(2C+1)N, C>0, N>=C+1 (e.g. 3N, 5N, 7N)
where C and N are integers.

The non-blue numbers are all that don't fit these forms. A sieve approach (similar to that used to identify prime numbers) will work easily in the range 1000 to 2000.

I get one: 1024.

All positive integers are blue if and only if they are not powers of 2.
Proof:

If a positive integer is blue, then it is not a power of 2. The first form (previous post) can be written as C(2N+1). The second form is (2C+1)N. Both forms have odd factors. Therefore they can't be powers of 2.

I still need to show that if a number is not a power of 2, then it is blue.

From form 1 we know that all odds greater than 1 are blue. From form 2 we know that almost all multiples of all odds greater than 1 are blue. The first multiples (1 through C times the odd number 2C+1) aren't necessarily blue. If these can be shown to be blue, then we're done because that would prove that all numbers with an odd factor (non powers of 2) are blue.

It turns out that we only have to bother with the even multiples of the multiples in question (the odd ones are blue because all odds are blue). Therefore, in form 2 let N=2M. These numbers are of the form (2C+1)2M = 4CM+2M.

There are two cases to consider:
1) M is odd
2) M is even

If M is odd, the number is of the form 4X+2. From the previous post, we know that 4N+2 is blue if N is greater than or equal to 2. It is also true for N=1 (6=2+3). Therefore, if M is odd, 4CM+2M is blue.

If M is even, the number is of the form 8X+4. From the previous post, we know that 8N+4 is blue if N is greater than or equal to 4. It is also true for N=1 (12=3+4+5), N=2 (20=2+3+4+5+6), and N=3 (28=1+2+3+4+5+6+7). Therefore, if M is even, 4CM+2M is blue.

So, if my statement that blue numbers have one of the two given forms is true, then all positive integers are blue if and only if they are not powers of 2.

I haven't proven the statement about the two forms, but it is easy to convince yourself that it is true by considering the sum of the first N numbers where N is even and odd. Then consider what happens to that sum if you add one to each number, then add two to each number, etc.

The sum of N consecutive numbers is:

S= (a+(a+N-1))N /2
= aN + (N^2-N) /2
= aN + N(N-1)/2

If N is un-even than S is a multiple of N
If N is even, than S is a multiple of N plus half of N

S can be:
every un-even number
every multiple of an un-even number

The remaining numbers are: 1 2 4 8 16 32 64 128 256 512 1024 2048 etc.
Powers of two. Between 1000 and 2000 there is only one. 1024

Whim

But this doesn't mean that all multiples of N are blue. For instance, the sum of five consecutive positive integers is a multiple of 5. However, not all multiples of 5 are the sum of five consecutive positive integers (5 and 10 aren't). You don't have to worry about 5 because it is odd. You do have demonstrate that 10 is blue. It is, of course, but not because it's a multiple of 5.