# Amitrano's formula (new)

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Binet's formula
---------------

f(n) = f(n-2) + f(n-1)
f(0) = 0
f(1) = 1

x = [1 + sqrt(5)] / 2
y = [1 - sqrt(5)] / 2

f(n) = (x^n - y^n) / sqrt(5)

Amitrano's formula
------------------

f(n) = a*f(n-2) + b*f(n-1)
For all a, b, f(0), f(1) are a member of R

x = [b + sqrt(b^2 + 4*a)] / 2
y = [b - sqrt(b^2 + 4*a)] / 2

f(n) = {a*f(0)*[x^(n-1) - y^(n-1)] + f(1)*[x^n - y^n]} / (x - y)

### Re: Amitrano's formula (new)

Hi drini,
I have found a document in a math's italian site.
www.matematicamente.it (Section: Number theory).

Best regards
Anna

P.S.
I'm italian

### Re: Amitrano's formula (new)

so? all I'm claiming is that such result is not new, is just a condensed form of the characteristic method for solving linear homogeneous recurrences
f
G -----> H G
p \ /_ ----- ~ f(G)
\ / f ker f
G/ker f

### Re: Amitrano's formula (new)

what's so new about that formula? isn't that just an application of the characteristic method I used to derive binet's formula, but using variables for boundary conditions instead of 0 and 1 ?

f
G -----> H G
p \ /_ ----- ~ f(G)
\ / f ker f
G/ker f