technical question for oeis-entries

[Gottfried Helms]

Seqfans -

 just came across a sequence which I submitted earlier, and I would like to
 format the comments and examples a bit better, at least with linebreaks,
 but don't want to experiment with and possibly spoil actual contents of the OEIS.
 Example:

ID Number: A096773
URL:       http://www.research.att.com/projects/OEIS?Anum=A096773
Sequence:  0,3,1,13,5,53,21,213,85,853,341,3413,1365,13653,5461,54613,
           21845,218453,87381,873813,349525,3495253,1398101,13981013,
           5592405,55924053,22369621,223696213,89478485,894784853,
           357913941,3579139413,1431655765
Name:      a(n+2) = 4*a(n) + 1; a(1)=0, a(2) = 3.
Comments:  Remainders for classes m of integers n (mod 2^(m+1)) which divide the
              integers in 2 groups after one Collatz (3x+1)-transformation.
           With one 3x+1-transformation T(x;p) := x'= (3x+1)/2^p all numbers of
              the form x = i*2^N +a(N) -> x' = i*6 + 1 for odd N>2, -> x' = i*6 + 5
              for even N. x = 4*i + 3 -> x' = 6*i + 5, x = 8*i + 1 -> x' = 6*i + 1, x =16*i +13
              -> x' = 6*i + 5, x = 32*i + 1 -> x' = 6*i + 1, x =64*i +53 -> x' = 6*i + 5, x
              =128*i + 1 -> x' = 6*i + 1, with i a free parameter >=0
Formula:   a(2m) = (5*2^(2m-1) - 1)/3, a(2m-1) = (2^(2m-2)-1)/3.
Example:   a(1) = (2^0-1)/3 = 0, a(2) = (5*2^1 - 1) / 3 = 3, a(3) = (2^2-1)/3 = 1, a(4) =
              (5*2^3 - 1) / 3 =13, a(5) = (2^4-1)/3 = 5, a(6) = (5*2^5 - 1) / 3 =53, a(7) =
              (2^6-1)/3 =21.

I would like to format this at least like
==================================================================================
Comments:  Remainders for classes m of integers n (mod 2^(m+1)) which divide the
              integers in 2 groups after one Collatz (3x+1)-transformation.

           With one 3x+1-transformation

                 T(x;p) := x'= (3x+1)/2^p

          all numbers of the form
               x = i*2^N +a(N)
                           ->   x' = i*6 + 1 for odd N>2,
                           ->   x' = i*6 + 5 for even N.
           Thus
               x = 4*i + 3 -> x' = 6*i + 5,        x =  8*i + 1 -> x' = 6*i + 1,
               x =16*i +13 -> x' = 6*i + 5,        x = 32*i + 5 -> x' = 6*i + 1,
               x =64*i +53 -> x' = 6*i + 5,        x =128*i +21 -> x' = 6*i + 1,
           ....
            all with "i" as a free parameter >=0  covering all natural numbers

Example:   a(1) = (2^0-1)/3 = 0,                 a(2) = (5*2^1 - 1) / 3 = 3,
           a(3) = (2^2-1)/3 = 1,                 a(4) = (5*2^3 - 1) / 3 =13,
           a(5) = (2^4-1)/3 = 5,                 a(6) = (5*2^5 - 1) / 3 =53,
           a(7) = (2^6-1)/3 =21.
           ....
==================================================================================

but I am not sure I recall right how to fix linebreaks and to adjust indentation
(is linebreak ctrl-return or shift-return or... and hard spaces at the left?)
(windows OS, standard german keyboard) - I think I must have overseen something,
when I submitted this comment/example.

Just a hint?

Gottfried Helms