Numbers 'poetics' 1 and 2019 
One number is called 'poetic' when it can be represented of only one form how $2^a+2^b+2^c$ such that $a\ge b \ge c\ge 0.$ How many  poetic numbers are between $1$ and $2019?$

Attempt: I separate in cases, I find that every $2^k$ with $k\ge2 , 2^k + 1$ with $k\ge1$ and every case with $a>b>c$
My final answer was $184$. Check that please .
$a\ge b\ge c\ge 0$
Numbers of form $2^k, k\ge 2$ and $2^k+1,k\ge 1$ are also unique re presentable
$2^k=2^{k-1}+2^{k-2}+2^{k-2}$
$2^{k}+1=2^{k-1}+2^{k-1}+2^0$

As $2^{11}>2047>2020>2^{10}+2^9$, there are no poetic numbers from $2020$ to $2047$.
All poetic numbers from $1$ to $2047$ actually lie between $1$ and $2019$.
Integers in this range can be written as binary numbers of no more than $11$ digits.
A poetic number should have at most three $1$'s in the digits when it is expressed as a binary number.
A binary number with exactly three $1$'s in the digits when it is expressed as a binary number is clearly a poetic number.
A binary number with exactly two $1$'s in the digits when it is expressed as a binary number is a poetic number if and only if the last digit is $1$. [Note: $2^a+2^b=2^{a-1}+2^{a-1}+2^b=2^a+2^{b-1}+2^{b-1}$.]
A binary number with exactly one $1$ in the digits when it is expressed as a binary number is a poetic number if and only if the last  two digits are $0$'s. [Note: $2^a=2^{a-1}+2^{a-2}+2^{a-2}$.]
So the number of poetic numbers from $1$ to $2019$ is $\displaystyle \binom{11}{3}+\binom{10}{1}+\binom{9}{1}=184$.
