By AK, 12. August 2002, the first musings, concerning A007595.

  First, we apply Raney's lemma to case where we have
  a sequence of n +1's and n+2 -1's, and prove that exactly two
  of its cyclical shifts are such that all the proper partial
  sums are >= -2.

  (For information about Raney's lemma, see Graham, Knuth, Patashnik:
   "Concrete Mathematics", pages 345 & 346 (from the Chapter 7, "Generating
   Function")

  Raney's lemma says that

  "If <x1,x2,...,xm> is any sequence of integers whose sum is +1,
   then exactly one of the cyclic shifts
   <x1,x2,...,xm>, <x2,...,xm,x1>, ..., <xm,x1,...,x_m-1>
   has all of its partial sums positive."

  For the starters we change this to an equivalent fact:

  "If <x1,x2,...,xm> is any sequence of integers whose sum is -1,
   then exactly one of the cyclic shifts
   <x1,x2,...,xm>, <x2,...,xm,x1>, ..., <xm,x1,...,x_m-1>
   has all of its proper partial sums non-negative."

  which is based on the same geometric thinking as the original
  formulation, except that the aspect of the tangential line is now
  -(1/m) instead of 1/m. We will refer to this condition
  as (*1).

  Then, if we have n +1's and n+2 -1's, their total sum
  is -2. But if we divide all summands by 2, then the total
  sum is -1 again, and using the same geometric thinking as
  behind Raney's lemma, we can be sure that there is
  _at least one_ cyclic shift of the summands such that all of its
  proper partial sums are non-negative, thus there is at
  least one cyclic shift of n +1's and n+2 -1's such that
  all the proper partial sums are > -2. We will refer
  to this condition as (*2).

  We mark r(s,k) the sequence s cyclically shifted
  left by k steps.
  We let 'a' stand for the amount needed to shift
  s, so that it satisfies the condition (*2),
  so r(s,a) is the corresponding sequence.


  To prove that there exists only one other amount 'b',
  distinct from 'a' modulo 2n+2 (the length of the
  whole sequence), such that also r(s,b) satisfies
  the condition (*2), we proceed as follows:

  We start constructing from r(s,a) a rooted
  planar binary tree, by mapping +1's to branching nodes
  and 0's to leaves, in left-to-right, depth-first fashion.
  E.g. if r(s,a) is the sequence

    +1, +1, -1, +1, -1, -1, -1, +1, -1, +1, -1, -1

  we start constructing a binary tree (here we indicate
  branching nodes with 1's, corresponding to +1's in the
  above sequence, and leaves with 0's, corresponding to
  -1's):

  i.e. after consuming the three first terms, +1, +1 and -1,
  the tree looks like this, and there are two nodes open
  from which to continue it:

  0
   \ /
    1
     \ /
      1

  and when we continue we see that after "consuming" the first
  seven terms of the sequence, +1, +1, -1, +1, -1, -1, -1,
  the binary tree is ready, and there's no further
  "open nodes" from which to extend it.

    0   0
     \ /
  0   1
   \ /
    1   0
     \ /
      1


  Because the final sum of any cyclical shift is -2, including
  also ones which fulfill the condition (*2), and the absolute values
  of all summands are 1, there must be somewhere a point 'u'
  with the partial sum -1. Let's call the proper prefix of the
  the terms from the first one to the point 'u' a sequence 'p1',
  which we know to contain some k amount of +1's, and k+1 -1's.
  If the whole sequence satisfies the condition (*2), then
  this prefix must satisfy the condition (*1).

  Then, the terms after the point u, to the end of the
  cyclical shift satisfy also the condition (*1).

  This u is precisely the point where we need to cyclically
  shift s, to obtain another rotation satisfying the
  condition (*2): r(s,u) = r(s,b)

  Now, imagine that there were yet other point u' to which
  to rotate the sequence s, so that it would satisfy the
  condition (*2).
  
  Now, the prefix sub(r(s,...)) must form a valid binary tree
  by above construction algorithm. And the same applies to
  the suffix following it.

  Surely it begins by 1, so that 1 starts a subtree of
  either of the binary trees corresponding to the
  prefix x and suffix y. That means that if we consume ...
  at least when we reach the delimiting point 'u'
  between x and y, there will be no further open nodes left.
  ... there are no further "open nodes" left precisely
  at the delimiting point of x and y.
  So, next comes either x or y or one of their (other) subtrees,
  and when we have consumed that and built a second binary tree,
  we still have a few terms of the sequence left,
  but these cannot form a valid binary tree, because
  then we would have already a partial sum -3,
  so there are exactly one way to partition
...
  the rotation r(s,a) into 

  and that is a contradiction, because after constructing
  two plane binary trees, we already have a partial sum -2.

  Now, if construct the prefix x of the sequence
  in depth-first left-to-right fashion, and the suffix
  y of the sequence in depth-first but from RIGHT-TO-LEFT fashion,
  and then combine these binary trees with one extra node
  as their root, we have a bijection from (n,n+2) binary
  necklaces to the plane binary trees upto the reflection.





construct rooted plane
  binary trees