<s>[INST] 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
$2^k=2^{k-1}+2^{k-2}+2^{k-2}$
$2^{k}+1=2^{k-1}+2^{k-1}+2^0$
 [/INST] We have $a\ge b\ge c\ge0$. But there is also a requirement that the representation is unique. So for example: 6 does not count because it is $4+1+1$ and also $2+2+2$.
Indeed, we have 
(1): $2^a+2^a+2^b=2^{a+1}+2^{b-1}+2^{b-1}$ for $b\ge1$ and
(2): $2^a+2^b+2^b=2^{a-1}+2^{a-1}+2^{b+1}$ for $a>b+1$. [For $a=b+1$ the two expressions are identical]
We have $2^{11}=2048$, so $a\le10$ and if $a=10$, then we need $b<10$. Also if $a=10,b=9$, then $c<9$.
So for $a=10$ we can have $b=9$ and then $0\le c\le 8$, giving 9 possibilities. If $b=8$ then (2) requires $0\le c<8$ giving 8 possibilities. And so on, down to $b=1$ for which we must have $c=0$. So in total $a=10$ gives $1+2+\dots+9=45$.
For $a=9$ and $b=9$ only $c=0$ is allowed because of (1). For $b=8$ we can have $0\le c\le 8$ (9 poss). For $b=7$ we can have $0\le c\le 6$ (because (2) rules out $c=7$), giving 7 poss. For $b=6$ we have 6 poss, and so on down to 1 poss for $b=1$. So in total $a=9$ gives $(1+2+\dots+7)+9+1=38$.
Similarly for $a=8$ we get $1+2+\dots+6+9=30$, for $a=7$ we get $1+2+\dots+5+8=23$. For $a=6$ we get $1+2+\dots+4+7=17$. For $a=5$ we get $1+2+3+6=12$. For $a=4$ we get $1+2+5=8$, for $a=3$ we get $1+4=5$. For $a=2$ we get 3 (check: $9=4+4+1,8=4+2+2,7=4+2+1$). For $a=1$ we get 2 and for $a=0$ we get 1. 
So grand total $45+38+30+23+17+12+8+5+3+2+1=184$.
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