I have the following two formulas, which for $n > 1$, $n \in \Bbb N$ and $x \in \Bbb N$ lead to linear equations of the type $o(n) = mx + b$ if only n is set and to certain odd numbers if both are set ($o(n,x)$):
- $$o_c(n,x) = 2^{2n}x - \sum_{i=1}^{n-1} 2^{2i-1} - 1$$
- $$o_d(n,x) = 2^{2n+1}(x - 1) + \sum_{i=1}^{n-1} 2^{2i} + 1$$
My problem is the following:
Can every natural number $8m−3, m \in \Bbb N$ be written as either $o_c(n,x)$ or $o_d(n,x)$ for suitable $n,x$?
For instance, $5$ can be written as $o_d(2,1)$, $13$ can be written as $o_c(2,1)$, $21$ can be written as $o_c(3,1)$.
Unfortunately, I haven't been able to prove it myself and would really appreciate some help. Can someone please help me out on this?
[IMPORTANT EDIT] Here something that would really help me to solve this based on these examples:
- Set $n = 2$ for $o_c$: $o_c(2,x) = 16x - 3$. The numbers remaining to be proven is $16x - 11$.
- Set $n = 2$ for $o_d$: $o_d(2,x) = 32x - 27$. The numbers remaining to be proven is $32x - 11$.
- Set $n = 3$ for $o_c$: $o_c(3,x) = 64x - 11$. The numbers remaining to be proven is $64x - 43$.
- Set $n = 3$ for $o_d$: $o_d(3,x) = 128x - 107$. The numbers remaining to be proven is $128x - 43$.
- Set $n = 3$ for $o_c$: $o_c(4,x) = 256x - 43$. The numbers remaining to be proven is $256x - 171$.
- Set $n = 3$ for $o_d$: $o_d(4,x) = 512x - 427$. The numbers remaining to be proven is $512x - 171$.
- ...
One can see that only one half of the previously to be proven numbers remains.
Prove that the remaining numbers to be proven can be written as either:
- $$r_Oc(n,x) = 2^{2n}x - \sum_{i=0}^{n} 2^i$$ if $o_c$ is used to show that the next half of the previously remaining numbers belongs to the numbers proven now, or
- $$r_Oc(n,x) = 2^{2n-1}x - \sum_{i=0}^{n-1} 2^i$$ if $o_d$ is used to show that the next half of the previously remaining numbers belongs to the numbers proven now.
If this were proven I would be able to solve it.