√(n² + 2n - 1) = [n;1,n-1,1,2n]

√d = n + (√d - n) mit d = n² + 2n - 1

a0 = n

    1          1      (√d + n)   (√d + n)   2n - 1   1 + (√d - n)       (√d - (n - 1))
-------- = -------- × -------- = -------- = ------ + ------------ = 1 + --------------
(√d - n)   (√d - n)   (√d + n)    2n - 1    2n - 1      2n - 1              2n - 1

a1 = 1

   (2n - 1)         (2n - 1)      (√d + (n - 1))   (2n - 1) × (√d + (n - 1))   (√d + (n - 1))
-------------- = -------------- × -------------- = ------------------------- = --------------
(√d - (n - 1))   (√d - (n - 1))   (√d + (n - 1))          2 × (2n - 1)                2

  n + (√d - n) + n - 1   2n - 2   1 + (√d - n)           (√d - (n - 1))
= -------------------- = ------ + ------------ = n - 1 + --------------
            2               2           2                       2

a2 = n - 1

       2                2         (√d + (n - 1))   2 × (√d + (n - 1))   (√d + (n - 1))
-------------- = -------------- × -------------- = ------------------ = --------------
(√d - (n - 1))   (√d - (n - 1))   (√d + (n - 1))      2 × (2n - 1)          2n - 1

  n + (√d - n) + n - 1   2n - 1   (√d - n)       (√d - n)
= -------------------- = ------ + -------- = 1 + --------
         2n - 1          2n - 1    2n - 1         2n - 1

a3 = 1

(2n - 1)   (2n - 1)   (√d + n)   (2n - 1) × (√d + n)   (√d + n)   2n + (√d - n)        (√d - n)
-------- = -------- × -------- = ------------------- = -------- = ------------- = 2n + --------
(√d - n)   (√d - n)   (√d + n)         (2n - 1)            1            1                  1

a4 = 2n

Fehlerhinweise, Kommentare und Anregungen sind mir herzlich willkommen.

Last Update: 2012-08-15