A Vector Arithmetic Function, Special Integer Partitions and Residue Sequences Consider the following arithmetic function, $f(a,b) = (a+1, \lceil a- \frac{a}{b} \rceil , \lceil a- \frac{2a}{b} \rceil, \dots, \lceil \frac{a}{b} \rceil, 1)$, where all terms except the first and last are computed by the application of a ceiling function. 
For any pair of positive integers $(a,b)$, the vector $f(a,b)$ consists of the terms of an integer partition of length $a+1$. Moreover, $f(a,b)$ and $f(b,a)$ represent dual partitions and have reflected Ferrers Diagrams. This can be proved by recalling the identity
\begin{eqnarray}
\sum_{j = 0}^{b-1} \left \lceil \tfrac{a(b-j)}{b} \right \rceil = \sum_{j = 0}^{a-1} \left \lceil \tfrac{b(a-j)}{a} \right \rceil.
\end{eqnarray}
(Going in the other direction, one can use the methods outlined here to give a neat combinatorial proof of this ceiling summation identity.)
In general, these integer partitions contain no $0$s, contain at least one $1$ and at most $a$ $1$s.
Question: Are these partitions a special (read named) subset of the standard partitions?
Problem 1: Determine all pairs $(a,b)$ such that $f(a,b)$ are the terms of an integer partition of $k$ for a fixed $k > 1$. Determine a simple way to find $k$ from $a$ and $b$ (and not by simply adding the terms of the defining relation).
Problem 2: Find other symmetries like $f(a,b) \leftrightarrow f(b,a)$.
(Motivation) For example, $f(2,3) = (4,2,1)$, $f(3,2) = (3,2,1,1)$, $f(1,5) = (6,1)$ and $f(5,1) = (2,1,1,1,1,1)$, which are dual integer partitions of $7$. 
Problem 3: Give an interpretation to the sequence {$f(a,b)$} in terms of $a$ and $b$ for fixed $a$ or $b$.
(Motivation) The sequence {$f(1,b)$} consists of terms of the form $(2,1, \dots, 1)$ with $b + 1$ $1$s, which are integer partitions of $b+2$; {$f(2,b)$} consists of terms of the form $(3, 1), (3,1,1), (3,2,1)$ or $(3, 2, \dots, 2, 1, \dots, 1)$ which are each integer partitions of those integers greater than 3 not divisible by $3$; {$f(3,b)$} consists of special integer partitions of integers congruent to {$1,2,5$} mod $6$. Similarly, {$f(4,b)$} consists of integer partitions of integers congruent to {$1, 2, 3$} mod $5$ or {$1,2,6,8$} mod $10$; {$f(5,b)$} consists of integer partitions of integers congruent to {$1, 2, 4$} mod $6$, etc.
Here is some Mathematica code that might be useful:
f[a_, b_] := 
  Flatten[{a + 1, Table[Ceiling[a - j a/b], {j, 1, b - 1}], 1}];
a = 2;
Table[Total@f[a, b], {b, 1, 100}]
Thanks!
 A: Here's an answer to the second part of Problem 1: $$k = \frac{(a+1)(b+1) - \gcd(a,b) + 3}{2}.$$
According to your definition,
$$k = 1 + (a+1) + \sum_{j=1}^{b-1} \left\lceil \frac{ja}{b} \right\rceil.$$ 
Now, $\left\lceil \frac{ja}{b} \right\rceil = \left\lfloor \frac{ja + b - 1}{b} \right\rfloor.$  (See, for example, Exercise 3.12, p.96, in Concrete Mathematics, 2nd edition.)
It is known that $$\sum_{j=0}^{b-1} \left\lfloor \frac{aj + x}{b} \right\rfloor = d \left\lfloor \frac{x}{d} \right\rfloor + \frac{(a-1)(b-1)}{2} + \frac{d-1}{2},$$
for any real number $x$, where $d = \gcd(a,b)$.  (See, for example, Eq. 3.32, p. 94, of Concrete Mathematics, 2nd edition.)  We have $x = b-1$.
Then $$d \left\lfloor \frac{b-1}{d} \right\rfloor = d\frac{b-1}{d} - d \left\{\frac{b-1}{d}\right\} = b-1 - d \left\{\frac{b}{d} - \frac{1}{d}\right\} = b - 1  - d \frac{d-1}{d} = b-d,$$
where $\{x\}$ denotes the fractional part of $x$.  (The second-to-last step holds because $d | b$.)
Putting everything together yields $$k = a + 2 + b - d + \frac{(a-1)(b-1)}{2} + \frac{d-1}{2},$$
and then combining terms gets the rest.

The elements in $f(a,b)$ are closely related to Spec$(\frac{a}{b}) = \{\lfloor \frac{a}{b}\rfloor, \lfloor\frac{2a}{b}\rfloor,  \lfloor\frac{3a}{b}\rfloor, \ldots \}$, although you want ceilings, and you want to stop when the multiplier is $b-1$.  You may find the discussion of Spec$(\alpha)$ in Concrete Mathematics helpful.  (In fact, when I looked at your problem I thought, "This reminds me of the discussion of spectra in Concrete Math," and then using results in there it wasn't too difficult to find the expression for $k$.)
