A certain operator on integer sequences Let $a\in\mathbb N^\mathbb Z$ be a bounded sequence of positive integers. Define another bounded sequence of positive integers $b=b(a)\in\mathbb N^\mathbb Z$ by
$$
b_n=\sum_{l=0}^\infty \max\{0,a_{n+l}-l\}. 
$$
For a constant sequence $a\equiv k\in\mathbb N$, then $b(a)\equiv \frac {k(k+1)}{2}$. Is the converse also true? That is, for every $k\in\mathbb N$, the equation $b(a)\equiv \frac {k(k+1)}{2}$ has a unique solution. More generally, is it true that for every $k\in\mathbb N$, the equation $b(a)=k$ has a unique solution up to a shift transformation?

Acknowledgement. This question originated from a puzzle that I heard from a friend who heard it from Ruth Lawrence
  https://en.wikipedia.org/wiki/Ruth_Lawrence. Apparently, the problem
  is known as "Bulgarian Solitaire," as pointed out by Brian Hopkins who
  wrote an essay on it
  https://www.researchgate.net/publication/259735333_30_Years_of_Bulgarian_Solitaire.
The original problem is told as follows: There are 66 cards arrange in
  a few stacks. On each turn, one card is from each stack is taken and
  the taken cards form a new stack. Prove that the process necessarily
  reaches the fixed point where there are stacks of one, two, three,...
  to eleven cards.

[I did not know how to tag this question. Since $b$ is an operator on $\ell_{\infty}$ I put it under "banach-spaces"]    
 A: Yes, this is true if specifically $b_n=k(k+1)/2$; however, in general, the sequences $a_n$ with $b_n=c$ are exactly the ones that take only two consecutive values $M,M-1$, repeated in an $M$ periodic pattern.
When we sum the $a_{n+j}-j$ over $j\ge 0$ to compute $b_n$, we can impose the restriction $a_{n+j}-j\ge 0$ or $\ge 1$. So
$$
b_{n+1}=\sum_{j\ge 1; a_{n+j}-j+1\ge 1} (a_{n+j}-j+1) = \sum_{j\ge 1; a_{n+j}-j\ge 0} (a_{n+j}-j+1) ,
$$
and this shows that $b_{n+1}-b_n=N_n - a_n$, where $N_n$ is the number of $j\ge 1$ with $a_{n+j}-j\ge 0$.
Now look at an $n$ with $a_n=M:=\max a_k$; let's say $n=0$ for convenience. Since we must have $N_0=M$, it follows that $a_M=M$. Now we can see that $a_n=M-1$ or $=M$ for all $n$: This is certainly true for $n=1$ since $N_1\ge M-1$. Moreover, $a_{M+1}\ge M-1$ by the same argument, and it then follows that $a_{M+1}=a_1$, or $N_1$ wouldn't come out right. This means that for $1\le j\le M-1$, we already know that $a_{2+j}\ge j$ (and $j=M$ may or may not satisfy this inequality), so we now see in the same way that $N_2=a_2=a_{M+2}\ge M-1$ also etc.
Conversely, any such $M$ periodic sequence $a_n$ that takes only the two values $M,M-1$ will lead to a constant $b_n$. This is obvious in fact since only $j=M$ can possibly fail to contribute to $N_n$, and which of these cases we are in depends on whether $a_{n+M}=a_n=M-1$ or $M$, as required.
Finally, if you take a pattern with the value $M-1$ taken $p$ times in the block $0\le j\le M$, then $b_n=M(M+1)/2-p$, and since $0\le p\le M-1$, we see that any constant value for $b_n$ can be obtained in this way. If specifically $b_n=k(k+1)/2$, then only $M=k$, $p=0$ works here, so in this case, $a_n$ must indeed be constant, as you suspected.
In fact, in general, both $M$ and $p$ are uniquely determined by the constant value of $c$, but of course how you distribute the $p$ values of $M-1$ within a period block is up to you.
A: In fact, the claim follows from the Bulgarian Solitaire one!
To see this, firstly we can visualize the process as follows. Assume that at the point $n$ of the real line there stands a stack of $a_n$ cells. From point $k\leq n$, you can see only those of these cells which are at least $n-k$ above the line. Then $b_k$ is the number of cells that are visible from point $k$.
Now consider what happens if you pass from point $k$ to $k+1$. Now you do not see the pile of $a_k$ cells, but you start seeing some more cells from the next piles --- $a_k$ in total, if you assume $b_{k+1}=b_k$. To each cell from the $k$th pile (ordered from the top to the bottom), put into correspondence the cells you start to see (ordered from the left to the right). Say that the newly seen cells are the images of those from the $k$th pile.
Notice that the cells in the $n$th pile (with $n$ sufficiently large) are the images of some cells from the $(n-1)$st, $(n-2)$nd, $\dots$, $(n-a_n)$th piles. Now, for every $n$, we mark as $n$-interesting the cells in the piles to the left of $n$ whose images are to the right of $n-1$ (one may notice that they are the lowest ones in each pile). There are $a_n$ such cells whose images are in the $n$th pile, $\max\{a_{n+1}-1,0\}$ such cells whose images are in the $(n+1)$st pile, and so on; so there are $b_n$ $n$-interesting cells.
How are $n$-interesting cells related to $(n+1)$-interesting ones? We forget about the $a_n$ cells in the $n$th pile, but take their pre-images instead, which are (to repeat) in the $(n-1)$st, $\dots$, $(n-a_n)$th piles. Thus, the process of forming $n$-interesting cells from the $(n+1)$-interesting ones  is exactly the Bulgarian solitaire: we take the rightmost pile (it is easily seen to be a largest one) and distribute its cells into $a_n$ previous ones.
Since we know what happens to the $k(k+1)/2$ cards in the Bulgarian solitaire in a bounded time (the process cycles on the piles with $k,k-1,\dots,1$), we also know that the largest pile of $n$-interesting cells (which contains $a_{n-1}$ cells) always contains $k$ cells (consider our procss starting far to the right). Similar description works for the other quantities.
