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# fibonnaci numbers to the 45 digit

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fibonnaci numbers to the 43rd digit..what is the exact number for the golden rule to break it down perfectly

1 6 6 7 3 5 1 2 8 2 7 1 6 2 2 6 6 3 4 8 6 2 9 3 8 3 7 2 0 5 2 5 0 8 3 5 5 7 4 2 8 2 7

Guest Jan 13, 2015

#2
+27056
+10

Are you asking about the Binet formula?  This gives the n'th standard Fibonacci number as

$$\frac{(1+\sqrt5)^n-(1-\sqrt5)^n}{2^n\sqrt5}$$

However, this doesn't reproduce the number you've written (yours lies between the 203rd and 204th standard Fibonacci numbers).

.

Alan  Jan 13, 2015
#1
+90023
+5

I do not understand your question.....

CPhill  Jan 13, 2015
#2
+27056
+10

Are you asking about the Binet formula?  This gives the n'th standard Fibonacci number as

$$\frac{(1+\sqrt5)^n-(1-\sqrt5)^n}{2^n\sqrt5}$$

However, this doesn't reproduce the number you've written (yours lies between the 203rd and 204th standard Fibonacci numbers).

.

Alan  Jan 13, 2015