bijection between number of partitions of 2n satisfying certain conditions with number of partitions of n Suppose $\lambda=(\lambda_1,\lambda_2,\ldots,\lambda_k)$ is a partition of $2n$ where $n\in\mathbb N$ satisfying the following conditions:


*

*(1) $\lambda_k=1$.

*(2) $\lambda_i−\lambda_{i+1}\leq 1$ for every $i \leq k−1$.

*(3) In the partition $\lambda$, the number of odd parts in odd places & the number of odd parts in even places are equal.
Here a part $\lambda_i$ is said to be in even place if $i$ is even, whereas $\lambda_i$ is said to be in odd place if $i$ is odd. $\lambda_i$ 's are called parts of $\lambda$ and $\lambda_i$ is called an odd part if it is odd & is called even part if it is even.
Now the question is to give a bijection between number of partitions of $2n$ satisfying the above conditions and number of partitions of $n$.
 A: I've finally found the proper interpretation of "conormal nodes", which is not the one I cited in the comment (even the corrected with "every node $A$" replaced by "every removable node $A$").
I gave a description of an algorithm in terms of Young diagrams that computes the correspondence in an answer to the corresponding MathOverflow question. What I would like to add here is a more direct description without using diagrams, which may be easier for pencil-and-paper (or computer) computations. For that purpose I will transpose (conjugate) the partitions, as this leads to an easier description.
Call a partition strict if all its nonzero parts are distinct, and black-white balanced if its Young diagram covers equally many black and white squares of a checkerboard pattern, which amounts to the equation
$$
  \sum_i(-1)^i(1-(-1)^{\lambda_i})=0,
$$
or to the condition (3) of the question (but the black-white formulation makes the transposition-invariance of the condition more evident).
Central in the description will be the notion of a cyclic parenthesis matching of a word according to attributes attached to its symbols partitioning them into "opening symbols", "closing symbols" and possibly "neutral symbols" (we apply this where symbols are integers, and the attributes are assigned according to parity). This is the usual pairing of matching parentheses, but where one imagines the word repeated periodically, so that an opening symbol at the end may match a closing symbol at the beginning. What will mostly be of interest are the opening (or closing) symbols that remain unmatched in this pairing; if there are $d\geq0$ more opening than closing symbols, there will always be that many unmatched opening symbols (and no unmatched closing symbols). For instance in the word $\def\red{\color{red}}BABB\red AABAABB\red AABAA$ the red letters are the unmatched ones if $A$ is considered opening symbol and $B$ a closing symbol (the two final $A$s respectively match the third and first $B$ due to cyclic repetition).
Proposition. The algorithm below establishes for every $n\in\Bbb N$ a bijection between the black-white balanced strict partitions of $2n$ and the partitions of $n$.
Algorithm. The algorithm operates on finite weakly decreasing sequences of natural numbers $(a_1,a_2,\ldots,a_l)$ that always end with $a_l=0$; there can however be more than one occurrence of $0$, and (unlike in partitions) the number of such occurrences is relevant. The algorithm also maintains a even/odd state, which is always (except maybe at the very beginning) the parity of the majority of the numbers $a_1,\ldots,a_l$ and therefore strictly speaking redundant, but its evolution is explicitly  indicated anyway. Given a black-white balanced strict partition $\lambda=(\lambda_1,\lambda_2,\ldots,\lambda_k)$ of $2n$ (with $\lambda_k>0$), one starts by setting $l=k+1$ and $a_i=\lambda_i-(l-i)$ for $0<i<l$, and of course $a_l=0$. The initial parity is that of $k$. If $k$ is even then condition (3) ensures that $(a_1,a_2,\ldots,a_k)$ has equally many even and odd terms, and $a_l=0$ gives the majority to the even terms; if $k$ is odd then $l$ is even and $(a_1,a_2,\ldots,a_l)$ has equally many even and odd terms (but the final $0$ will be considered neutral in the first step, giving the effective majority to the odd terms).


*

*In an even state, all even numbers are considered as opening symbols and all odd numbers as closing symbols; then


*

*if the final zero is matched, $1$ is added to all unmatched (even) symbols, and the new state is odd;

*if the final zero is unmatched, $1$ is subtracted from all matched symbols, an additional $0$ is added, and the new state remains even.


*In an odd state, all odd numbers are considered as opening symbols, the even numbers except the final $0$ are considered closing symbols, but the final $0$ is considered neutral; then $1$ is added to all unmatched (odd) symbols, and the new state is even.


The algorithm terminates when at some point all numbers have become even; at this point the sequence would not change anymore. The partition returned is obtained by dividing all parts by $2$.
Here is an example for the black-white balanced strict partition $(11,10,6,5,2)$ of $2n$ for $n=17$ the states are (with matching symbols indicated by parentheses, neutral ones by a dash, and unmatched ones by a vertical bar):
6 6 3 3 1 0 odd
) ) | ( ( -
6 6 4 3 1 0 even
| ( ( ) ) | 
6 5 3 2 0 0 0 even
( ) ) | | | (
6 5 3 3 1 1 0 odd
) | | | | ( -
6 6 4 4 2 1 0 even
| | | | ( ) |
6 6 4 4 1 0 0 0 even
| | | ( ) | | |
6 6 4 3 0 0 0 0 0 even
| | ( ) | | | | |
6 6 3 2 0 0 0 0 0 0 even
| ( ) | | | | | | |
6 5 2 2 0 0 0 0 0 0 0 even
( ) | | | | | | | | |
5 4 2 2 0 0 0 0 0 0 0 0 even
) | | | | | | | | | | (
5 5 3 3 1 1 1 1 1 1 1 0 odd
| | | | | | | | | | | -
6 6 4 4 2 2 2 2 2 2 2 0 terminal

The partition of $n$ obtained here is $(3,3,2,2,1,1,1,1,1,1,1)$.
