Numbers 'poetics' 1 and 2019

Question

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

$2^k=2^{k-1}+2^{k-2}+2^{k-2}$

$2^{k}+1=2^{k-1}+2^{k-1}+2^0$

Answer 1

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 $45+38+30+23+17+12+8+5+3+2+1=184$.

Answer 2

Hint : Since $a\ge b\ge c\ge0 $ , it follows that :

$$2019\ge2^a+2^b+2^c \ge 3\cdot2^c \implies 9\ge c$$

And trivial to notice that $a\le10$

We only have $10$ cases for $c$ which can be checked by hand .

Answer 3

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 $\displaystyle \binom{11}{3}+\binom{10}{1}+\binom{9}{1}=184$.