I was also thinking about the D[t] - d. We know that at the correct t, D[t] - d = a(n-1).
Now we do know the equation for D[t] for both even and odd e's, but let's limit it to odd e's for now.
The equation for d in (e, 1) for odd e is:
2k*2 - 1 + (e + 1)/2
We know
2k2 - 1 + (e + 1)/2 - d = a(n-1) (for some k)
This can be rewritten to:
2k2 = a(n - 1) + d + 1 - (e + 1)/2.
Let's multiply both sides by 2:
4k2 = 2a(n - 1) + 2d + 2 - (e + 1)
4k2 = 2a(n - 1) + 2d + 2 - e + 1
4k2 = 2a(n - 1) + 2d + 1 - e
Looks familiar? 2*d + 1 - e? Yup f is here too.
4k2 = 2a*(n - 1) + f
And we can change it back:
4k2 - f = 2a*(n - 1)
Meaning D[t] - d = a(n-1) could also be expressed as (4t*2 - f)/2 = a(n-1).
Not sure if this is useful, obvious or neat.