A sum of ordinals and a bijection Let $a=a_{i\in n}$ is an ordinal-indexed family of ordinals.
I call a structured sum of $a$ an order isomorphism from the lexicographically ordered set $\{ \left( i ; x \right) \; | \; i \in n, x\in a_i \}$ into $\sum a$.


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*I suspect that for every $a$ there exist exactly one structured sum. Right? I am also attempting to write an explicit formula for structured sum having specified $a$.

*Can you point me any reference where the thing which I call structured sums are studied? I wouldn't be happy to know that I am the person which first discovered structured sums :-)
 A: Let $$X=\bigcup_{i\in n}\Big(\{i\}\times a_i\Big)\;,$$ and let $\prec$ denote the lexicographic order on $X$. Then you’re simply looking at the order isomorphism $$\varphi:X\to\sum_{i\in n}a_i\;,$$ where the sum on the right is ordinal addition. Let $\alpha$ be this ordinal sum, and for $\langle i,\xi\rangle\in X$ let $$X(i,\xi)=\{\langle j,\zeta\rangle\in X:\langle j,\zeta\rangle\prec\langle i,\xi\rangle\}\;,$$ the set of all $\prec$-predecessors of $\langle i,\xi\rangle$ in $X$. Then clearly
$$\varphi\left(\langle i,\xi\rangle\right)=\inf\left(\alpha\setminus\varphi\left[X(i,\xi)\right]\right)\;.$$
I doubt that in general you can say much more than this and a few straightforward observations like $$\varphi\left(\langle i,\xi\rangle\right)=\varphi\left(\langle i,0\rangle\right)+\xi=\alpha_i+\xi\;,$$ where we define $\alpha_i=\sum_{j<i}a_j$.
A: Note that $\{ (i,x) : i \in x, x \in a_i \}$ is well-ordered by the lexicographic ordering, and therefore it is order isomorphic to a unique ordinal, which may be calculated to be $\sum_{i \in n} a_i$.  Note, also, that any order isomorphism between two well-ordered sets must be unique.  (This order isomorphism is included in Brian's answer.)
Because of this, there would not appear to be much use in studying these objects, as all of their basic properties should be expressible in terms of $( a_i )_{i \in n}$ (and $\sum_{i \in n} a_i$).
A: You also asked for references. I have no doubt that you can find this in many books. (Quite often they are left to exercises.) Using Google Books Search I was able to find a few references in textbooks.

When I searched Google Books for sum ordinals disjoint I found 
in the book Stoll R. Set theory and logic (Dover 1963, 1979) on p.313

The concept of the ordinal sum of two well-ordered sets extends
  directly to an arbitrary (well-ordered) family of well-ordered sets.
  First, a word about the notation for such families. In view of Lemma 7.6
  we may take the indexing set to be an ordinal number. We shall do
  this and use notation like 
  $$\{a_\xi; \xi\in\eta\}$$
  for such a family. If, then, $\{a_\xi; \xi\in\eta\}$ is a disjoint family of well-ordered
  sets, indexed by (the ordinal number) $\lambda$, we define its ordinal sum as 
  $\bigcup_\xi a_\xi$ ordered as follows: If $x$ and $y$ are members of the union and
  in the same set $a_\xi$ then the order in $a_\xi$ prevails; if $x\in a_\xi$ and $y\in a_\eta$
  where $\xi<\eta$, we take $x < y$.


When I searched Google Books for 
sum ordinals transfinite sequence I found that in the book
Peter Komjath, Vilmos Totik: Problems and Theorems in Classical Set Theory these notions are used and they are defined more generally for order types
on p.32:

If $\theta_i$ are order types for each $i\in I$ where $\langle I,<\rangle$ is an ordered set, then we define
  $\sum_{i\in I} \theta_i$ to be the order type of the ordered union of $\langle A_i, \prec_i\rangle$, $i\in I$, with respect to $\langle I,<\rangle$, where $\langle A_i, \prec_i\rangle$ are pairwise disjoint ordered sets with ordered type $\theta_i$.

Ordered union is defined on p.24:

Let $\langle A_i,<_i\rangle$, $i\in I$ be ordered sets with pairwise disjoint ground sets $A_i$
  and let the index set $I$ be also ordered by the relation $<$. The ordered union
  of $\langle A_i,<_i\rangle$, $i\in I$ with respect to the ordered set $\langle I,<\rangle$ is the ordered set
  $\langle B,\prec\rangle$ in which $B = \bigcup_{i\in I}A_i$, and for $a \in A_i$ and $b \in A_j$ the relation $a < b$
  holds if and only if $i < j$ or $i = j$ and $a <_i b$.

