The question was asked by xport in the comments to Mike's solution as to whether this solution is unique.
There are no doubt many ways to case bash an answer to this, and here is one, proving that
Mike's solution is indeed unique.
Let $x_i$ be the number of times the digit $ i $ appears on the paper, then we must have
$$ \sum_{i=0}^9 x_i = 20 \quad \text{and} \quad \sum_{i=0}^9 ix_i = 45+20=65,$$
since there are $20$ digits on the completed paper and the first column sums to $45.$
All the digits appear at least once therefore $x_i \ge 1 $ for all $ i $ and $x_0=1.$
Thus from the above equations we obtain
$$\sum_{i=1}^9 x_i = 19 \quad (1) \quad \text{and} \quad \sum_{i=2}^9 (i-1)x_i = 46. \quad (2)$$
Since $\sum_{i=1}^8 i = 36$ it follows from $(2)$ that $(i-1)x_i \le 46 - (36-(i-1)),$ and so
$$ x_i \le \left \lfloor \frac{i+9}{i-1} \right \rfloor \quad \text{for} \quad i \ge 2.$$
Hence
$$ x_6 \le 3, \, x_7 \le 2, \, x_8 \le 2 \quad \text{and} \quad x_9 \le 2. \quad (3)$$
Now suppose $ x_i > 1 $ for some $ i > 2 $ then, since $i$ occurs in at least three places, we must have $x_j=i$ for some $j \ne i.$
And further suppose $j \ge 2,$ then from $(2)$ we obtain
$$(i-1)x_i + (j-1)i + (36-(i-1)-(j-1)) \le 46,$$
which gives
$$ x_i \le \frac{2i+8}{i-1} - j.$$
Hence when $j \ge 2$ we must have
$$x_6 \le 2 \quad \text{and} \quad x_7=x_8=x_9=1. \quad (4)$$
We now do some case bashing.
From $(3)$ we know $x_9 \le 2,$ so suppose $x_9=2$ then $(4)$
implies $x_1=9,$ which again by $(4)$ implies $x_7=x_8=1,$ since if $x_1=9,$ then $x_1$ cannot equal $7$ or $8.$
Putting $x_1=9, x_7=1, x_8=1$ and $x_9=2$ in $(1)$ and $(2)$ we obtain
$$x_2+x_3+ \cdots + x_6=6$$
and
$$x_2 + 2x_3 + \cdots + 5x_6 = 17.$$
The former clearly only has solution $\lbrace 1,1,1,1,2 \rbrace, $ which satisfies the latter equation only when $x_3=2$ and the other $x_i$ are $1.$ But $3$ does not appear twice in this "solution." Hence we have a contradiction and so
$$x_9=1.$$
Similary, suppose $x_8=2,$ then $(4)$
implies $x_1=8,$ which again by $(4)$ implies $x_7=x_9=1,$ since if $x_1=8,$ then $x_1$ cannot equal $7$ or $9.$
Putting $x_1=8, x_7=1, x_9=1$ and $x_8=2$ in $(1)$ and $(2)$ we obtain
$$x_2+x_3+ \cdots + x_6=7$$
and
$$x_2 + 2x_3 + \cdots + 5x_6 = 18.$$
The former equation only has solutions $\lbrace 1,1,1,1,3 \rbrace $ (which does not satisfy the latter equation) and $\lbrace 1,1,1,2,2 \rbrace,$ which only satisifes the latter
equation when $x_2=x_3=2$ and the other $x_i$ are $1.$ But again $3$ does not appear twice in this "solution." Hence we have a contradiction and so
$$x_8=1.$$
Similary, suppose $x_7=2,$ then $(4)$
implies $x_1=7,$ which again by $(4)$ implies $x_8=x_9=1,$ since if $x_1=7,$ then $x_1$ cannot equal $8$ or $9.$
Putting $x_1=7, x_8=1, x_9=1$ and $x_7=2$ in $(1)$ and $(2)$ we obtain
$$x_2+x_3+ \cdots + x_6=8$$
and
$$x_2 + 2x_3 + \cdots + 5x_6 = 19.$$
The former equation only has solutions $\lbrace 1,1,1,1,4 \rbrace $ (which does not satisfy the latter equation) and $\lbrace 1,1,2,2,2 \rbrace$ (which also does not satisfy the latter equation) and $\lbrace 1,1,1,2,3 \rbrace, $ which only satisifes the latter
equation when $x_2=3, x_3=2,x_4=1,x_5=1$ and $x_6=1,$ and this is the solution given in Mike's answer.
Hence we have a unique solution when $x_7=2,$ otherwise we must have
$x_7=x_8=x_9=1.$
To complete the proof we need only show that when $x_7=1$ we must have $x_6=1.$ That would give us at least six $1's$ on the paper, implying that one of the numbers
$\lbrace 6,7,8,9 \rbrace$ must be repeated, which is a contradiction since we have
$x_6=x_7=x_8=x_9=1.$
So suppose $x_7=1$ and $x_6=3$ (the maximum possible from $(3)$) then $(4)$ implies
$x_1=6.$ And since $x_8=x_9=1,$ from $(1)$ and $(2)$ we obtain
$$x_2+ x_3 + x_4 + x_5=7$$
and
$$x_2 + 2x_3 + 3x_4 + 4x_5 = 10.$$
These have no solutions since the latter equation implies $x_2=x_3=x_4=x_5=1,$ which does not satisfy the former equation.
So suppose $x_7=1$ and $x_6=2,$ then since $x_8=x_9=1,$ from $(1)$ and $(2)$ we obtain
$$x_1+x_2+ x_3 + x_4 + x_5=14 \quad (5) $$
and
$$x_2 + 2x_3 + 3x_4 + 4x_5 = 15. \quad (6)$$
Now $(6)$ implies $x_5 \le 2,$ but if $x_5=2$ then we must have $x_2 \ge 3$ (since
$x_5=x_6=2$ there are at least $3$ occurrences of $2$) and so the LHS of $(6)$ must be at least $16,$ a contradiction. Hence $x_5=1.$
But now we have $x_0=1,x_5=1,x_6=2,x_7=1,x_8=1$ and $x_9=1,$ and since $x_6=2$ some number must occur 6 times but we do not have enough undetermined $x_i$ for this to occur unless
$x_1=6.$
Substituting $x_1=6$ and $x_5=1$ in $(5)$ and $(6)$ we obtain
$$x_2+ x_3 + x_4=7 \quad (7) $$
and
$$x_2 + 2x_3 + 3x_4 = 11. \quad (8)$$
and subtracting $(7)$ from $(8)$ we obtain
$$x_3+2x_4=4. \quad (9)$$
However, since we cannot have any more $1's$ (as $x_1=6$) we must have $x_3 \ge 2$ and
$x_4 \ge 2,$ which contradicts $(9).$
Thus we have shown that when $x_7=1$ we must have $x_6=1,$ which as noted above completes the proof that the unique solution is given by
$x_0=1,x_1=7,x_2=3,x_3=2,x_4=1,x_5=1,x_6=1,x_7=2,x_8=1$ and $x_9=1.$
Best Answer
For question three, we can simply compute. Besides n=1, the next solution is 199981, as generated by this PARI/GP code:
Define $r(n)$ to be the number of ones in the digits of $n$:
r(n) = nn=n;cc=0;while(nn>0,if((nn%10)==1,cc=cc+1);nn=floor(nn/10));return(cc)Then run a loop and output $i$ if the sum of $r(i)$ from 1 up to $i$ is equal to $i$:
yy=0;for(i=1,199981,yy=yy+r(i);if(yy==i,print(i)))
The output is this:
This shows $f(1)=1$ and $f(199981)=199981$ and there are no other solutions less than 199981.
Similarly for question one, we define $g(n)$ to be the sum of the digits of $n$:
g(n) = nn=n; cc=0; while(nn>0,cc=cc+(nn % 10);nn=floor(nn/10));return(cc)then calculate:
sum(i=1,1000000,g(i))which yields the sum 27000001.