<s>[INST] if $G_{4n+1}=2G_{2n+1}-G_{n},G_{4n+3}=3G_{2n+1}-2G_{n}$,Find $G_{n}=n?$ 
let sequence $\{G_{n}\}$ such
  $$G_{1}=1,G_{3}=3,G_{2n}=G_{n}$$
  $$G_{4n+1}=2G_{2n+1}-G_{n},G_{4n+3}=3G_{2n+1}-2G_{n}$$

If such $G_{n}=n$, then we said $n$ is 'good'.
How many 'good' numbers $n$, such that $n<2^{100}?$
My try:
since
$$\begin{eqnarray}G_{1}&=&1,\\
G_{2}&=&1,\\
G_{3}&=&3,\\
G_{4}&=&G_{2}=1\\
G_{5}&=&2G_{3}-G_{1}=5,\\
G_{6}&=&G_{3}=3,\\
G_{7}&=&3G_{3}-2G_{1}=9-2=7\\
G_{8}&=&G_{4}=1,\\
G_{9}&=&2G_{5}-G_{2}=10-1=9,\\
G_{10}&=&G_{5}=5,\\
G_{11}&=&3G_{5}-2G_{2}=15-2=13,
G_{12}&=&G_{6}=3,\\
G_{13}&=&2G_{7}-G_{3}=14-3=11,\\
G_{14}&=&G_{7}=7,\\
G_{15}&=&3G_{7}-2G_{3}=21-6=15,\end{eqnarray}$$
so when $n=1,3,5,7,9,,15,\cdots$ is 'good"
But How find numbers? when $n<2^{100}?$
Thank you
 [/INST] This is sequence A030101 in the OEIS.  That is, $G(n)$ is the number obtained by reversing the digits of $n$ when written base $2$, e.g. $G(25)=G(11001_2)=10011_2=19$.
This is easy to check:  If $n=d_0+2d_1+\cdots+2^rd_r$, then
$$
\begin{align}
G(n)&=d_r+\cdots+2^rd_0\\
G(2n)&=d_r+\cdots+2^rd_0+2^{r+1}\cdot0\\
G(2n+1)&=d_r+\cdots+2^rd_0+2^{r+1}\cdot1\\
G(4n+1)&=d_r+\cdots+2^rd_0+2^{r+1}\cdot0+2^{r+2}\cdot1\\
G(4n+3)&=d_r+\cdots+2^rd_0+2^{r+1}\cdot1+2^{r+2}\cdot1
\end{align}
$$
from which the identities $G(2n)=G(n)$, $G(4n+1)=2G(2n+1)-G(n)$, and $G(4n+3)=3G(2n+1)-2G(n)$ are easily verified.
The upshot is that the OP's "good" numbers are those that are palindromes when written in binary.
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