Your sequence is powers of two, where the nth term is 2^n. I postulate here that the relationship described above is related to the exponetial curve. So I focused my exploration there thinking maybe this would apply to other exponential functions.

So first, I looked at the simplest one (this will be tricky to write without super/subcripts):

kx^0, kx^1, kx^2, … kx^n

where k is a constant coefficient and x raised to the nth power for each term. Applying the 1/3 & 2/3 condition, we get the following:

kx^n = (2/3)kx^(n-1) + (1/3)kx^(n+1)

Settles down to…

3kx = 2k + kx^2

k(x-1)(x-2) = 0

x = 1 or 2

Which leaves us with either all terms are equal (k) or x=2 (our original sequence). Both of these solutions work.

Next, I looked at the inverse to see if I could get any alternate variations

k, kx^-1, kx^-2, kx^-3, … kx^-n

Does this work? Let’s work it thorugh. First, the initial condition:

x^-n = (2/3)kx^1-n + (1/3)kx^(1+n)

Settles down to …

3kx = 2kx^2 +1k

k(1-x)(1-2x) = 0

x = -1 or -2.

x = -1 was already proved in the last problem, but just in case you didn’t get it:

sequence: k, k, k, k, …. k

k/3 + 2k/3 = k

Now let’s look at -2

k, k/2, k/-3, k/4, k/-5…

That doesn’t work because if x(n) is positive, x(n-1) and x(n+1) are both going to be negative. ]]>

4 = 2/3 * 2 + 1/3 * 8

8 = 2/3 * 4 + 1/3 * 16

I think we should talk about linear recurrences! One of those things we just need to say something about! This example isn’t in just that form, but as Steven Noble probably spotted, it can be (should be?) thought of in that way.

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Or have my brains completely fallen out? I suspect I’m missing something… ]]>